question stringlengths 3 474 | answer stringlengths 0 3.93k |
|---|---|
Let $\|\cdot\|$ be any norm on $\mathbb{R}^{m}$ and let $B=\left\{x \in \mathbb{R}^{m}:\|x\| \leq 1\right\}$. Prove that $B$ is compact. | \begin{proof}
Let us call $\|\cdot\|_E$ the Euclidean norm in $\mathbb{R}^m$. We start by claiming that there exist constants $C_1, C_2>0$ such that
$$
C_1\|x\|_E \leq\|x\| \leq C_2\|x\|_E, \forall x \in \mathbb{R}^m .
$$
Assuming (1) to be true, let us finish the problem. First let us show that $B$ is bounded... |
Show that if $S$ is connected, it is not true in general that its interior is connected. | \begin{proof}
Consider $X=\mathbb{R}^2$ and
$$
A=([-2,0] \times[-2,0]) \cup([0,2] \times[0,2])
$$
which is connected, while $\operatorname{int}(A)$ is not connected.
To see this consider the continuous function $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ is defined by $f(x, y)=x+y$. Let $U=f^{-1}(0,+\infty)$ whi... |
Suppose that $E$ is an uncountable subset of $\mathbb{R}$. Prove that there exists a point $p \in \mathbb{R}$ at which $E$ condenses. | \begin{proof}
I think this is the proof by contrapositive that you were getting at.
Suppose that $E$ has no limit points at all. Pick an arbitrary point $x \in E$. Then $x$ cannot be a limit point, so there must be some $\delta>0$ such that the ball of radius $\delta$ around $x$ contains no other points of $E$ :
... |
Prove that $\sqrt{n+1}-\sqrt{n} \rightarrow 0$ as $n \rightarrow \infty$. | \begin{proof}
$$
\sqrt{n+1}-\sqrt{n}=\frac{(\sqrt{n+1}-\sqrt{n})(\sqrt{n+1}+\sqrt{n})}{\sqrt{n+1}+\sqrt{n}}=\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2 \sqrt{n}}
$$
\end{proof} |
Prove that $\sum 1/k(\log(k))^p$ diverges when $p \leq 1$. | \begin{proof}
Using the integral test, for a set $a$, we see
$$
\lim _{b \rightarrow \infty} \int_a^b \frac{1}{x \log (x)^c} d x=\lim _{b \rightarrow \infty}\left(\frac{\log (b)^{1-c}}{1-c}-\frac{\log (a)^{1-c}}{1-c}\right)
$$
which goes to infinity if $c \leq 1$ and converges if $c>1$. Thus,
$$
\sum_{n=2}^... |
Show that a group of order 5 must be abelian. | \begin{proof}
Suppose $G$ is a group of order 5 which is not abelian. Then there exist two non-identity elements $a, b \in G$ such that $a * b \neq$ $b * a$. Further we see that $G$ must equal $\{e, a, b, a * b, b * a\}$. To see why $a * b$ must be distinct from all the others, not that if $a *$ $b=e$, then $a$ an... |
If $G$ is a finite group, prove that there is an integer $m > 0$ such that $a^m = e$ for all $a \in G$. | \begin{proof}
Let $n_1, n_2, \ldots, n_k$ be the orders of all $k$ elements of $G=$ $\left\{a_1, a_2, \ldots, a_k\right\}$. Let $m=\operatorname{lcm}\left(n_1, n_2, \ldots, n_k\right)$. Then, for any $i=$ $1, \ldots, k$, there exists an integer $c$ such that $m=n_i c$. Thus
$$
a_i^m=a_i^{n_i c}=\left(a_i^{n_i}\r... |
Let $G$ be a group in which $(a b)^{3}=a^{3} b^{3}$ and $(a b)^{5}=a^{5} b^{5}$ for all $a, b \in G$. Show that $G$ is abelian. | \begin{proof}
We have
$$
\begin{aligned}
& (a b)^3=a^3 b^3, \text { for all } a, b \in G \\
\Longrightarrow & (a b)(a b)(a b)=a\left(a^2 b^2\right) b \\
\Longrightarrow & a(b a)(b a) b=a\left(a^2 b^2\right) b \\
\Longrightarrow & (b a)^2=a^2 b^2, \text { by cancellation law. }
\end{aligned}
$$
Again,
$$
\b... |
If $G$ is a group and $a, x \in G$, prove that $C\left(x^{-1} a x\right)=x^{-1} C(a) x$ | \begin{proof}
Note that
$$
C(a):=\{x \in G \mid x a=a x\} .
$$
Let us assume $p \in C\left(x^{-1} a x\right)$. Then,
$$
\begin{aligned}
& p\left(x^{-1} a x\right)=\left(x^{-1} a x\right) p \\
\Longrightarrow & \left(p x^{-1} a\right) x=x^{-1}(a x p) \\
\Longrightarrow & x\left(p x^{-1} a\right)=(a x p) x^... |
If $a > 1$ is an integer, show that $n \mid \varphi(a^n - 1)$, where $\phi$ is the Euler $\varphi$-function. | \begin{proof}
Proof: We have $a>1$. First we propose to prove that
$$
\operatorname{Gcd}\left(a, a^n-1\right)=1 .
$$
If possible, let us assume that
$\operatorname{Gcd}\left(a, a^n-1\right)=d$, where $d>1$.
Then
$d$ divides $a$ as well as $a^n-1$.
Now,
$d$ divides $a \Longrightarrow d$ divides $a^n$.
Thi... |
Suppose that $|G| = pm$, where $p \nmid m$ and $p$ is a prime. If $H$ is a normal subgroup of order $p$ in $G$, prove that $H$ is characteristic. | \begin{proof}
Let $G$ be a group of order $p m$, such that $p \nmid m$. Now, Given that $H$ is a normal subgroup of order $p$. Now we want to prove that $H$ is a characterestic subgroup, that is $\phi(H)=H$ for any automorphism $\phi$ of $G$. Now consider $\phi(H)$. Clearly $|\phi(H)|=p$. Suppose $\phi(H) \neq H$,... |
If $G$ is a nonabelian group of order 6, prove that $G \simeq S_3$. | \begin{proof}
Suppose $G$ is a non-abelian group of order 6 . We need to prove that $G \cong S_3$. Since $G$ is non-abelian, we conclude that there is no element of order 6. Now all the nonidentity element has order either 2 or 3 . All elements cannot be order 3 .This is because except the identity elements there ... |
Prove that a group of order $p^2$, $p$ a prime, has a normal subgroup of order $p$. | \begin{proof}
We use the result from problem 40 which is as follows: Suppose $G$ is a group, $H$ is a subgroup and $|G|=n$ and $n \nmid\left(i_G(H)\right) !$. Then there exists a normal subgroup $K \neq \{ e \}$ and $K \subseteq H$.
So, we have now a group $G$ of order $p^2$. Suppose that the group is cyclic, t... |
If $G$ is an abelian group and if $G$ has an element of order $m$ and one of order $n$, where $m$ and $n$ are relatively prime, prove that $G$ has an element of order $mn$. | \begin{proof}
Let $G$ be an abelian group, and let $a$ and $b$ be elements in $G$ of order $m$ and $n$, respectively, where $m$ and $n$ are relatively prime. We will show that the product $ab$ has order $mn$ in $G$, which will prove that $G$ has an element of order $mn$.
To show that $ab$ has order $mn$, let $k$ be... |
Prove that any two nonabelian groups of order 21 are isomorphic. | \begin{proof}
By Cauchy's theorem we have that if $G$ is a group of order 21 then it has an element $a$ of order 3 and an element $b$ of order 7. By exercise 2.5.41 we have that the subgroup generated by $b$ is normal, so there is some $i=0,1,2,3,4,5,6$ such that $a b a^{-1}=b^i$. We know $i \neq$ 0 since that imp... |
If $G_1$ and $G_2$ are cyclic groups of orders $m$ and $n$, respectively, prove that $G_1 \times G_2$ is cyclic if and only if $m$ and $n$ are relatively prime. | \begin{proof}
The order of $G \times H$ is $n$. $m$. Thus, $G \times H$ is cyclic iff it has an element with order n. $m$. Suppose $\operatorname{gcd}(n . m)=1$. This implies that $g^m$ has order $n$, and analogously $h^n$ has order $m$. That is, $g \times h$ has order $n$. $m$, and therefore $G \times H$ is cycli... |
If $P$ is a $p$-Sylow subgroup of $G$ and $P \triangleleft G$, prove that $P$ is the only $p$-Sylow subgroup of $G$. | \begin{proof}
let $G$ be a group and $P$ a sylow-p subgroup. Given $P$ is normal. By sylow second theorem the sylow-p subgroups are conjugate. Let $K$ be any other sylow-p subgroup. Then there exists $g \in G$ such that $K=g P g^{-1}$. But since $P$ is normal $K=g P g^{-1}=P$. Hence the sylow-p subgroup is unique.... |
Show that any subgroup of order $p^{n-1}$ in a group $G$ of order $p^n$ is normal in $G$. | \begin{proof}
Proof: First we prove the following lemma.
\textbf{Lemma:} If $G$ is a finite $p$-group with $|G|>1$, then $Z(G)$, the center of $G$, has more than one element; that is, if $|G|=p^k$ with $k\geq 1$, then $|Z(G)|>1$.
\textit{Proof of the lemma:} Consider the class equation
$$
|G|=|Z(G)|+\sum_{a \n... |
Show that there is an infinite number of solutions to $x^2 = -1$ in the quaternions. | \begin{proof}
Let $x=a i+b j+c k$ then
$$
x^2=(a i+b j+c k)(a i+b j+c k)=-a^2-b^2-c^2=-1
$$
This gives $a^2+b^2+c^2=1$ which has infinitely many solutions for $-1<a, b, c<1$.
\end{proof} |
Let $R$ be a ring in which $x^3 = x$ for every $x \in R$. Prove that $R$ is commutative. | \begin{proof}
To begin with
$$
2 x=(2 x)^3=8 x^3=8 x .
$$
Therefore $6 x=0 \quad \forall x$.
Also
$$
(x+y)=(x+y)^3=x^3+x^2 y+x y x+y x^2+x y^2+y x y+y^2 x+y^3
$$
and
$$
(x-y)=(x-y)^3=x^3-x^2 y-x y x-y x^2+x y^2+y x y+y^2 x-y^3
$$
Subtracting we get
$$
2\left(x^2 y+x y x+y x^2\right)=0
$$
Multiply the ... |
Let $p$ be an odd prime and let $1 + \frac{1}{2} + ... + \frac{1}{p - 1} = \frac{a}{b}$, where $a, b$ are integers. Show that $p \mid a$. | \begin{proof}
First we prove for prime $p=3$ and then for all prime $p>3$.
Let us take $p=3$. Then the sum
$$
\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{(p-1)}
$$
becomes
$$
1+\frac{1}{3-1}=1+\frac{1}{2}=\frac{3}{2} .
$$
Therefore in this case $\quad \frac{a}{b}=\frac{3}{2} \quad$ implies $3 \mid a$, i.e. $p... |
Let $R$ be the ring of $2 \times 2$ matrices over the real numbers; suppose that $I$ is an ideal of $R$. Show that $I = (0)$ or $I = R$. | \begin{proof}
Suppose that $I$ is a nontrivial ideal of $R$, and let
$$
A=\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)
$$
where not all of $a, b, c d$ are zero. Suppose, without loss of generality -- our steps would be completely analogous, modulo some different placement of 1 s in our matrices... |
Let $F = \mathbb{Z}_p$ be the field of integers $\mod p$, where $p$ is a prime, and let $q(x) \in F[x]$ be irreducible of degree $n$. Show that $F[x]/(q(x))$ is a field having at exactly $p^n$ elements. | \begin{proof}
In the previous problem we have shown that any for any $p(x) \in F[x]$, we have that
$$
p(x)+(q(x))=a_{n-1} x^{n-1}+\cdots+a_1 x+a_0+(q(x))
$$
for some $a_{n-1}, \ldots, a_0 \in F$, and that there are $p^n$ choices for these numbers, so that $F[x] /(q(x)) \leq p^n$. In order to show that equality... |
If $p$ is a prime, show that $q(x) = 1 + x + x^2 + \cdots x^{p - 1}$ is irreducible in $Q[x]$. | \begin{proof}
Lemma: Let $F$ be a field and $f(x) \in F[x]$. If $c \in F$ and $f(x+c)$ is irreducible in $F[x]$, then $f(x)$ is irreducible in $F[x]$.
Proof of the Lemma: Suppose that $f(x)$ is reducible, i.e., there exist non-constant $g(x), h(x) \in F[x]$ so that
$$
f(x)=g(x) h(x) .
$$
In particular, then w... |
Show that there is an infinite number of integers a such that $f(x) = x^7 + 15x^2 - 30x + a$ is irreducible in $Q[x]$. | \begin{proof}
Via Eisenstein's criterion and observation that 5 divides 15 and $-30$, it is sufficient to find infinitely many $a$ such that 5 divides $a$, but $5^2=25$ doesn't divide $a$. For example $5 \cdot 2^k$ for $k=0,1, \ldots$ is one such infinite sequence.
\end{proof} |
Let $V$ be a vector space over an infinite field $F$. Show that $V$ cannot be the set-theoretic union of a finite number of proper subspaces of $V$. | \begin{proof}
Assume that $V$ can be written as the set-theoretic union of $n$ proper subspaces $U_1, U_2, \ldots, U_n$. Without loss of generality, we may assume that no $U_i$ is contained in the union of other subspaces.
Let $u \in U_i$ but $u \notin \bigcup_{j \neq i} U_j$ and $v \notin U_i$. Then, we have $... |
Prove that $\cos 1^{\circ}$ is algebraic over $\mathbb{Q}$. | \begin{proof}
Since $\left(\cos \left(1^{\circ}\right)+i \sin \left(1^{\circ}\right)\right)^{360}=1$, the number $\cos \left(1^{\circ}\right)+i \sin \left(1^{\circ}\right)$ is algebraic. And the real part and the imaginary part of an algebraic number are always algebraic numbers.
\end{proof} |
Prove that $x^3 - 3x - 1$ is irreducible over $\mathbb{Q}$. | \begin{proof}
Let $p(x)=x^3-3 x-1$. Then
$$
p(x+1)=(x+1)^3-3(x+1)-1=x^3+3 x^2-3
$$
We have $3|3,3| 0$ but $3 \nmid 1$ and $3^2 \nmid 3$. Thus the polynomial is irreducible over $\mathbb{Q}$ by 3 -Eisenstein criterion.
\end{proof} |
Let $H$ be the subgroup generated by two elements $a, b$ of a group $G$. Prove that if $a b=b a$, then $H$ is an abelian group. | \begin{proof}
Since $a$ and $b$ commute, for any $g, h\in H$ we can write $g=a^ib^j$ and $h = a^kb^l$. Then $gh = a^ib^ja^kb^l = a^kb^la^ib^j = hg$. Thus $H$ is abelian.
\end{proof} |
Prove that if a group contains exactly one element of order 2 , then that element is in the center of the group. | \begin{proof}
Let $x$ be the element of order two. Consider the element $z=y^{-1} x y$, we have: $z^2=\left(y^{-1} x y\right)^2=\left(y^{-1} x y\right)\left(y^{-1} x y\right)=e$. So: $z=x$, and $y^{-1} x y=x$. So: $x y=y x$. So: $x$ is in the center of $G$.
\end{proof} |
Prove that a group of even order contains an element of order $2 .$ | \begin{proof}
Pair up if possible each element of $G$ with its inverse, and observe that
$$
g^2 \neq e \Longleftrightarrow g \neq g^{-1} \Longleftrightarrow \text { there exists the pair }\left(g, g^{-1}\right)
$$
Now, there is one element that has no pairing: the unit $e$ (since indeed $e=e^{-1} \Longleftrigh... |
Let $V$ be a vector space which is spanned by a countably infinite set. Prove that every linearly independent subset of $V$ is finite or countably infinite. | \begin{proof}
Let $A$ be the countable generating set, and let $U$ be an uncountable linearly independent set. It can be extended to a basis $B$ of the whole space. Now consider the subset $C$ of elements of $B$ that appear in the $B$-decompositions of elements of $A$.
Since only finitely many elements are involv... |
Let $Z$ be the center of a group $G$. Prove that if $G / Z$ is a cyclic group, then $G$ is abelian and hence $G=Z$. | \begin{proof}
We have that $G / Z(G)$ is cyclic, and so there is an element $x \in G$ such that $G / Z(G)=\langle x Z(G)\rangle$, where $x Z(G)$ is the coset with representative $x$. Now let $g \in G$
We know that $g Z(G)=(x Z(G))^m$ for some $m$, and by definition $(x Z(G))^m=x^m Z(G)$.
Now, in general, if $H \... |
Prove that no group of order $p^2 q$, where $p$ and $q$ are prime, is simple. | \begin{proof}
We may as well assume $p<q$. The number of Sylow $q$-subgroups is $1 \bmod q$ and divides $p^2$. So it is $1, p$, or $p^2$. We win if it's 1 and it can't be $p$, so suppose it's $p^2$. But now $q \mid p^2-1$, so $q \mid p+1$ or $q \mid p-1$.
Thus $p=2$ and $q=3$. But we know no group of order 36 is ... |
Prove that two elements $a, b$ of a group generate the same subgroup as $b a b^2, b a b^3$. | \begin{proof}
Let $H = \langle bab^2, bab^3\rangle$. It is clear that $H\subset \langle a, b\rangle$. Note that $(bab^2)^{-1}(bab^3)=b$, therefore $b\in H$. This then implies that $b^{-1}(bab^2)b^{-2}=a\in H$. Thus $\langle a, b\rangle\subset H$.
\end{proof} |
Prove that in the ring $\mathbb{Z}[x],(2) \cap(x)=(2 x)$. | \begin{proof}
Let $f(x) \in(2 x)$. Then there exists some polynomial $g(x) \in \mathbb{Z}$ such that
$$
f(x)=2 x g(x)
$$
But this means that $f(x) \in(2)$ (because $x g(x)$ is a polynomial), and $f(x) \in$ $(x)$ (because $2 g(x)$ is a polynomial). Thus, $f(x) \in(2) \cap(x)$, and
$$
(2 x) \subseteq(2) \cap(x... |
Let $I, J$ be ideals in a ring $R$. Prove that the residue of any element of $I \cap J$ in $R / I J$ is nilpotent. | \begin{proof}
If $x$ is in $I \cap J, x \in I$ and $x \in J . R / I J=\{r+a b: a \in I, b \in J, r \in R\}$. Then $x \in I \cap J \Rightarrow x \in I$ and $x \in J$, and so $x^2 \in I J$. Thus
$$
[x]^2=\left[x^2\right]=[0] \text { in } R / I J
$$
\end{proof} |
Let $R$ be a ring, with $M$ an ideal of $R$. Suppose that every element of $R$ which is not in $M$ is a unit of $R$. Prove that $M$ is a maximal ideal and that moreover it is the only maximal ideal of $R$. | \begin{proof}
Suppose there is an ideal $M\subset I\subset R$. If $I\neq M$, then $I$ contains a unit, thus $I=R$. Therefore $M$ is a maximal ideal.
Suppose we have an arbitrary maximal ideal $M^\prime$ of $R$. The ideal $M^\prime$ cannot contain a unit, otherwise $M^\prime =R$. Therefore $M^\prime \subset M$. But... |
Prove that $x^3 + 6x + 12$ is irreducible in $\mathbb{Q}$. | \begin{proof}
Apply Eisenstein's criterion with $p=3$.
\end{proof} |
Prove that $x^2+1$ is irreducible in $\mathbb{F}_7$ | \begin{proof}
If $p(x)=x^2+1$ were reducible, its factors must be linear. But no $p(a)$ for $a\in\mathbb{F}_7$ evaluates to 0, therefore $x^2+1$ is irreducible.
\end{proof} |
Let $p$ be a prime integer. Prove that the polynomial $x^n-p$ is irreducible in $\mathbb{Q}[x]$. | \begin{proof}
Straightforward application of Eisenstein's criterion with $p$.
\end{proof} |
Prove that if a prime integer $p$ has the form $2^r+1$, then it actually has the form $2^{2^k}+1$. | \begin{proof}
In particular, we have
$$
\frac{x^a+1}{x+1}=\frac{(-x)^a-1}{(-x)-1}=1-x+x^2-\cdots+(-x)^{a-1}
$$
by the geometric sum formula. In this case, specialize to $x=2^{2^m}$ and we have a nontrivial divisor.
\end{proof} |
Prove that the addition of residue classes $\mathbb{Z}/n\mathbb{Z}$ is associative. | \begin{proof}
We have
$$
\begin{aligned}
(\bar{a}+\bar{b})+\bar{c} &=\overline{a+b}+\bar{c} \\
&=\overline{(a+b)+c} \\
&=\overline{a+(b+c)} \\
&=\bar{a}+\overline{b+c} \\
&=\bar{a}+(\bar{b}+\bar{c})
\end{aligned}
$$
since integer addition is associative.
\end{proof} |
Prove that for all $n>1$ that $\mathbb{Z}/n\mathbb{Z}$ is not a group under multiplication of residue classes. | \begin{proof}
Note that since $n>1, \overline{1} \neq \overline{0}$. Now suppose $\mathbb{Z} /(n)$ contains a multiplicative identity element $\bar{e}$. Then in particular,
$$
\bar{e} \cdot \overline{1}=\overline{1}
$$
so that $\bar{e}=\overline{1}$. Note, however, that
$$
\overline{0} \cdot \bar{k}=\overlin... |
Let $x$ be an element of $G$. Prove that $x^2=1$ if and only if $|x|$ is either $1$ or $2$. | \begin{proof}
$(\Rightarrow)$ Suppose $x^2=1$. Then we have $0<|x| \leq 2$, i.e., $|x|$ is either 1 or 2 .
( $\Leftarrow$ ) If $|x|=1$, then we have $x=1$ so that $x^2=1$. If $|x|=2$ then $x^2=1$ by definition. So if $|x|$ is 1 or 2 , we have $x^2=1$.
\end{proof} |
Let $x$ and $y$ be elements of $G$. Prove that $xy=yx$ if and only if $y^{-1}xy=x$ if and only if $x^{-1}y^{-1}xy=1$. | \begin{proof}
If $x y=y x$, then $y^{-1} x y=y^{-1} y x=1 x=x$. Multiplying by $x^{-1}$ then gives $x^{-1} y^{-1} x y=1$.
On the other hand, if $x^{-1} y^{-1} x y=1$, then we may multiply on the left by $x$ to get $y^{-1} x y=x$. Then multiplying on the left by $y$ gives $x y=y x$ as desired.
\end{proof} |
If $x$ and $g$ are elements of the group $G$, prove that $|x|=\left|g^{-1} x g\right|$. | \begin{proof}
First we prove a technical lemma:
{\bf Lemma.} For all $a, b \in G$ and $n \in \mathbb{Z},\left(b^{-1} a b\right)^n=b^{-1} a^n b$.
The statement is clear for $n=0$. We prove the case $n>0$ by induction; the base case $n=1$ is clear. Now suppose $\left(b^{-1} a b\right)^n=b^{-1} a^n b$ for som... |
Prove that if $x^{2}=1$ for all $x \in G$ then $G$ is abelian. | \begin{proof}
Solution: Note that since $x^2=1$ for all $x \in G$, we have $x^{-1}=x$. Now let $a, b \in G$. We have
$$
a b=(a b)^{-1}=b^{-1} a^{-1}=b a .
$$
Thus $G$ is abelian.
\end{proof} |
If $x$ is an element of infinite order in $G$, prove that the elements $x^{n}, n \in \mathbb{Z}$ are all distinct. | \begin{proof}
Solution: Suppose to the contrary that $x^a=x^b$ for some $0 \leq a<b \leq n-1$. Then we have $x^{b-a}=1$, with $1 \leq b-a<n$. However, recall that $n$ is by definition the least integer $k$ such that $x^k=1$, so we have a contradiction. Thus all the $x^i$, $0 \leq i \leq n-1$, are distinct. In part... |
Prove that the multiplicative groups $\mathbb{R}-\{0\}$ and $\mathbb{C}-\{0\}$ are not isomorphic. | \begin{proof}
Isomorphic groups necessarily have the same number of elements of order $n$ for all finite $n$.
Now let $x \in \mathbb{R}^{\times}$. If $x=1$ then $|x|=1$, and if $x=-1$ then $|x|=2$. If (with bars denoting absolute value) $|x|<1$, then we have
$$
1>|x|>\left|x^2\right|>\cdots,
$$
and in parti... |
Let $G$ be any group. Prove that the map from $G$ to itself defined by $g \mapsto g^{-1}$ is a homomorphism if and only if $G$ is abelian. | \begin{proof}
$(\Rightarrow)$ Suppose $G$ is abelian. Then
$$
\varphi(a b)=(a b)^{-1}=b^{-1} a^{-1}=a^{-1} b^{-1}=\varphi(a) \varphi(b),
$$
so that $\varphi$ is a homomorphism.
$(\Leftarrow)$ Suppose $\varphi$ is a homomorphism, and let $a, b \in G$. Then
$$
a b=\left(b^{-1} a^{-1}\right)^{-1}=\varphi\left(... |
Prove that $G$ cannot have a subgroup $H$ with $|H|=n-1$, where $n=|G|>2$. | \begin{proof}
Solution: Under these conditions, there exists a nonidentity element $x \in H$ and an element $y \notin H$. Consider the product $x y$. If $x y \in H$, then since $x^{-1} \in H$ and $H$ is a subgroup, $y \in H$, a contradiction. If $x y \notin H$, then we have $x y=y$. Thus $x=1$, a contradiction. Th... |
Prove that if $H$ is a subgroup of $G$ then $H$ is generated by the set $H-\{1\}$. | \begin{proof}
If $H=\{1\}$ then $H-\{1\}$ is the empty set which indeed generates the trivial subgroup $H$. So suppose $|H|>1$ and pick a nonidentity element $h \in H$. Since $1=h h^{-1} \in\langle H-\{1\}\rangle$ (Proposition 9), we see that $H \leq\langle H-\{1\}\rangle$. By minimality of $\langle H-\{1\}\rangle... |
Show that the subgroup of all rotations in a dihedral group is a maximal subgroup. | \begin{proof}
Fix a positive integer $n>1$ and let $H \leq D_{2 n}$ consist of the rotations of $D_{2 n}$. That is, $H=\langle r\rangle$. Now, this subgroup is proper since it does not contain $s$. If $H$ is not maximal, then by the previous proof we know there is a maximal subset $K$ containing $H$. Then $K$ must... |
Let $A$ be an abelian group and let $B$ be a subgroup of $A$. Prove that $A / B$ is abelian. | \begin{proof}
Lemma: Let $G$ be a group. If $|G|=2$, then $G \cong Z_2$.
Proof: Since $G=\{e a\}$ has an identity element, say $e$, we know that $e e=e, e a=a$, and $a e=a$. If $a^2=a$, we have $a=e$, a contradiction. Thus $a^2=e$. We can easily see that $G \cong Z_2$.
If $A$ is abelian, every subgroup of $A$ ... |
Prove that the intersection of an arbitrary nonempty collection of normal subgroups of a group is a normal subgroup (do not assume the collection is countable). | \begin{proof}
Let $\left\{H_i \mid i \in I\right\}$ be an arbitrary collection of normal subgroups of $G$ and consider the intersection
$$
\bigcap_{i \in I} H_i
$$
Take an element $a$ in the intersection and an arbitrary element $g \in G$. Then $g a g^{-1} \in H_i$ because $H_i$ is normal for any $i \in H$
By the... |
Let $H \leq K \leq G$. Prove that $|G: H|=|G: K| \cdot|K: H|$ (do not assume $G$ is finite). | \begin{proof}
Proof. Let $G$ be a group and let $I$ be a nonempty set of indices, not necessarily countable. Consider the collection of subgroups $\left\{N_\alpha \mid \alpha \in I\right\}$, where $N_\alpha \unlhd G$ for each $\alpha \in I$. Let
$$
N=\bigcap_{\alpha \in I} N_\alpha .
$$
We know $N$ is a subgro... |
Prove that $\mathbb{Q}$ has no proper subgroups of finite index. | \begin{proof}
Solution: We begin with a lemma.
Lemma: If $D$ is a divisible abelian group, then no proper subgroup of $D$ has finite index.
Proof: We saw previously that no finite group is divisible and that every proper quotient $D / A$ of a divisible group is divisible; thus no proper quotient of a divisible g... |
Prove that if $G$ is an abelian simple group then $G \cong Z_{p}$ for some prime $p$ (do not assume $G$ is a finite group). | \begin{proof}
Solution: Let $G$ be an abelian simple group.
Suppose $G$ is infinite. If $x \in G$ is a nonidentity element of finite order, then $\langle x\rangle<G$ is a nontrivial normal subgroup, hence $G$ is not simple. If $x \in G$ is an element of infinite order, then $\left\langle x^2\right\rangle$ is a no... |
Prove that subgroups of a solvable group are solvable. | \begin{proof}
Let $G$ be a solvable group and let $H \leq G$. Since $G$ is solvable, we may find a chain of subgroups
$$
1=G_0 \unlhd G_1 \unlhd G_2 \unlhd \cdots \unlhd G_n=G
$$
so that each quotient $G_{i+1} / G_i$ is abelian. For each $i$, define
$$
H_i=G_i \cap H, \quad 0 \leq i \leq n .
$$
Then $H_i \... |
Prove that if $H$ is a nontrivial normal subgroup of the solvable group $G$ then there is a nontrivial subgroup $A$ of $H$ with $A \unlhd G$ and $A$ abelian. | \begin{proof}
Suppose $H$ is a nontrivial normal subgroup of the solvable group $G$.
First, notice that $H$, being a subgroup of a solvable group, is itself solvable. By exercise $8, H$ has a chain of subgroups
$$
1 \leq H_1 \leq \ldots \leq H
$$
such that each $H_i$ is a normal subgroup of $H$ itself and $H_... |
Let $G$ be a transitive permutation group on the finite set $A$ with $|A|>1$. Show that there is some $\sigma \in G$ such that $\sigma(a) \neq a$ for all $a \in A$. | \begin{proof}
Let $G$ be a transitive permutation group on the finite set $A,|A|>1$. We want to find an element $\sigma$ which doesn't stabilize anything, that is, we want a $\sigma$ such that
$$
\sigma \notin G_a
$$
for all $a \in A$.
Since the group is transitive, there is always a $g \in G$ such that $b=g ... |
Let $G$ be a finite group of composite order $n$ with the property that $G$ has a subgroup of order $k$ for each positive integer $k$ dividing $n$. Prove that $G$ is not simple. | \begin{proof}
Solution: Let $p$ be the smallest prime dividing $n$, and write $n=p m$. Now $G$ has a subgroup $H$ of order $m$, and $H$ has index $p$. Then $H$ is normal in $G$.
\end{proof} |
Prove that characteristic subgroups are normal. | \begin{proof}
Let $H$ be a characterestic subgroup of $G$. By definition $\alpha(H) \subset H$ for every $\alpha \in \operatorname{Aut}(G)$. So, $H$ is in particular invariant under the inner automorphism. Let $\phi_g$ denote the conjugation automorphism by $g$. Then $\phi_g(H) \subset H \Longrightarrow$ $g H g^{-... |
If $H$ is the unique subgroup of a given order in a group $G$ prove $H$ is characteristic in $G$. | \begin{proof}
Let $G$ be group and $H$ be the unique subgroup of order $n$. Now, let $\sigma \in \operatorname{Aut}(G)$. Now Clearly $|\sigma(G)|=n$, because $\sigma$ is a one-one onto map. But then as $H$ is the only subgroup of order $n$, and because of the fact that a automorphism maps subgroups to subgroups, w... |
Prove that if $P \in \operatorname{Syl}_{p}(G)$ and $H$ is a subgroup of $G$ containing $P$ then $P \in \operatorname{Syl}_{p}(H)$. | \begin{proof}
If $P \leq H \leq G$ is a Sylow $p$-subgroup of $G$, then $p$ does not divide $[G: P]$. Now $[G: P]=[G: H][H: P]$, so that $p$ does not divide $[H: P]$; hence $P$ is a Sylow $p$-subgroup of $H$.
\end{proof} |
Prove that a group of order 312 has a normal Sylow $p$-subgroup for some prime $p$ dividing its order. | \begin{proof}
Since $|G|=351=3^{2}.13$, $G$ has $3-$Sylow subgroup of order $9$, as well as $13-$Sylow subgroup of order $13$. Now, we count the number of such subgroups. Let $n_{13}$ be the number of $13-$Sylow subgroup and $n_{3}$ be the number of $3-$Sylow subgroup. Now $n_{13}=1+13k$ where $1+13k|9$. The choi... |
Let $|G|=p q r$, where $p, q$ and $r$ are primes with $p<q<r$. Prove that $G$ has a normal Sylow subgroup for either $p, q$ or $r$. | \begin{proof}
Let $|G|=p q r$. We also assume $p<q<r$. We prove that $G$ has a normal Sylow subgroup of $p$, $q$ or $r$. Now, Let $n_p, n_q, n_r$ be the number of Sylow-p subgroup, Sylow-q subgroup, Sylow-r subgroup resp. So, we have $n_r=1+r k$ such that $1+r k \mid p q$. So, in this case as $r$ is greatest $n_r$... |
Prove that a group of order 200 has a normal Sylow 5-subgroup. | \begin{proof}
Let $G$ be a group of order $200=5^2 \cdot 8$. Note that 5 is a prime not dividing 8 . Let $P \in$ $S y l_5(G)$. [We know $P$ exists since $S y l_5(G) \neq \emptyset$ by Sylow's Theorem]
The number of Sylow 5-subgroups of $G$ is of the form $1+k \cdot 5$, i.e., $n_5 \equiv 1(\bmod 5)$ and $n_5$ di... |
Prove that if $|G|=1365$ then $G$ is not simple. | \begin{proof}
Since $|G|=1365=3.5.7.13$, $G$ has $13-$Sylow subgroup of order $13$. Now, we count the number of such subgroups. Let $n_{13}$ be the number of $13-$Sylow subgroup. Now $n_{13}=1+13k$ where $1+13k|3.5.7$. The choices for $k$ is $0$. Hence, there is a unique $13-$Sylow subgroup and hence is normal. so... |
Prove that if $|G|=132$ then $G$ is not simple. | \begin{proof}
Since $|G|=132=2^{2}.3.11$, $G$ has $2-$Sylow subgroup of order $4$, as well as $11-$Sylow subgroup of order $11$, and $3-$Sylow subgroup of order $3$. Now, we count the number of such subgroups. Let $n_{11}$ be the number of $11-$Sylow subgroup and $n_{3}$ be the number of $3-$Sylow subgroup. Now ... |
Let $G$ be a group of order 105. Prove that if a Sylow 3-subgroup of $G$ is normal then $G$ is abelian. | \begin{proof}
Given that $G$ is a group of order $1575=3^2 .5^2 .7$. Now, Let $n_p$ be the number of Sylow-p subgroups. It is given that Sylow-3 subgroup is normal and hence is unique, so $n_3=1$. First we prove that both Sylow-5 subgroup and Sylow 7-subgroup are normal. Let $P$ be the Sylow3 subgroup. Now, Consid... |
Prove that a subgroup $H$ of $G$ is normal if and only if $[G, H] \leq H$. | \begin{proof}
$H \unlhd G$ is equivalent to $g^{-1} h g \in H, \forall g \in G, \forall h \in H$. We claim that holds if and only if $h^{-1} g^{-1} h g \in H, \forall g \in G, \forall h \in H$, i.e., $\left\{h^{-1} g^{-1} h g: h \in H, g \in G\right\} \subseteq H$. That holds by the following argument:
If $g^{-1}... |
Prove that if $R$ is an integral domain and $x^{2}=1$ for some $x \in R$ then $x=\pm 1$. | \begin{proof}
Solution: If $x^2=1$, then $x^2-1=0$. Evidently, then,
$$
(x-1)(x+1)=0 .
$$
Since $R$ is an integral domain, we must have $x-1=0$ or $x+1=0$; thus $x=1$ or $x=-1$.
\end{proof} |
A ring $R$ is called a Boolean ring if $a^{2}=a$ for all $a \in R$. Prove that every Boolean ring is commutative. | \begin{proof}
Solution: Note first that for all $a \in R$,
$$
-a=(-a)^2=(-1)^2 a^2=a^2=a .
$$
Now if $a, b \in R$, we have
$$
a+b=(a+b)^2=a^2+a b+b a+b^2=a+a b+b a+b .
$$
Thus $a b+b a=0$, and we have $a b=-b a$. But then $a b=b a$. Thus $R$ is commutative.
\end{proof} |
Let $G=\left\{g_{1}, \ldots, g_{n}\right\}$ be a finite group. Prove that the element $N=g_{1}+g_{2}+\ldots+g_{n}$ is in the center of the group ring $R G$. | \begin{proof}
Let $M=\sum_{i=1}^n r_i g_i$ be an element of $R[G]$. Note that for each $g_i \in G$, the action of $g_i$ on $G$ by conjugation permutes the subscripts. Then we have the following.
$$
\begin{aligned}
N M &=\left(\sum_{i=1}^n g_i\right)\left(\sum_{j=1}^n r_j g_j\right) \\
&=\sum_{j=1}^n \sum_{i=1}... |
An ideal $N$ is called nilpotent if $N^{n}$ is the zero ideal for some $n \geq 1$. Prove that the ideal $p \mathbb{Z} / p^{m} \mathbb{Z}$ is a nilpotent ideal in the ring $\mathbb{Z} / p^{m} \mathbb{Z}$. | \begin{proof}
First we prove a lemma.
Lemma: Let $R$ be a ring, and let $I_1, I_2, J \subseteq R$ be ideals such that $J \subseteq I_1, I_2$. Then $\left(I_1 / J\right)\left(I_2 / J\right)=I_1 I_2 / J$.
Proof: ( $\subseteq$ ) Let
$$
\alpha=\sum\left(x_i+J\right)\left(y_i+J\right) \in\left(I_1 / J\right)\left(I... |
Let $N$ be a positive integer. Let $M$ be an integer relatively prime to $N$ and let $d$ be an integer relatively prime to $\varphi(N)$, where $\varphi$ denotes Euler's $\varphi$-function. Prove that if $M_{1} \equiv M^{d} \pmod N$ then $M \equiv M_{1}^{d^{\prime}} \pmod N$ where $d^{\prime}$ is the inverse of $d \bmod... | \begin{proof}
Note that there is some $k \in \mathbb{Z}$ such that $M^{d d^{\prime}} \equiv M^{k \varphi(N)+1} \equiv\left(M^{\varphi(N)}\right)^k \cdot M \bmod N$. By Euler's Theorem we have $M^{\varphi(N)} \equiv 1 \bmod N$, so that $M_1^{d^{\prime}} \equiv M \bmod N$.
\end{proof} |
Prove that if an integer is the sum of two rational squares, then it is the sum of two integer squares. | \begin{proof}
Let $n=\frac{a^2}{b^2}+\frac{c^2}{d^2}$, or, equivalently, $n(b d)^2=a^2 d^2+c^2 b^2$. From this, we see that $n(b d)^2$ can be written as a sum of two squared integers. Therefore, if $q \equiv 3(\bmod 4)$ and $q^i$ appears in the prime power factorization of $n, i$ must be even. Let $j \in \mathbb{N... |
Prove that the quotient ring $\mathbb{Z}[i] /(1+i)$ is a field of order 2. | \begin{proof}
Let $a+b i \in \mathbb{Z}[i]$. If $a \equiv b \bmod 2$, then $a+b$ and $b-a$ are even and $(1+i)\left(\frac{a+b}{2}+\frac{b-a}{2} i\right)=a+b i \in\langle 1+i\rangle$. If $a \not \equiv b \bmod 2$ then $a-1+b i \in\langle 1+i\rangle$. Therefore every element of $\mathbb{Z}[i]$ is in either $\langle ... |
Prove that $(x, y)$ is not a principal ideal in $\mathbb{Q}[x, y]$. | \begin{proof}
Suppose, to the contrary, that $(x, y)=p$ for some polynomial $p \in \mathbb{Q}[x, y]$. From $x, y \in$ $(x, y)=(p)$ there are $s, t \in \mathbb{Q}[x, y]$ such that $x=s p$ and $y=t p$.
Then:
$$
\begin{aligned}
& 0=\operatorname{deg}_y(x)=\operatorname{deg}_y(s)+\operatorname{deg}_y(p) \text { so... |
Prove that if $f(x)$ and $g(x)$ are polynomials with rational coefficients whose product $f(x) g(x)$ has integer coefficients, then the product of any coefficient of $g(x)$ with any coefficient of $f(x)$ is an integer. | \begin{proof}
Let $f(x), g(x) \in \mathbb{Q}[x]$ be such that $f(x) g(x) \in \mathbb{Z}[x]$.
By Gauss' Lemma there exists $r, s \in \mathbb{Q}$ such that $r f(x), s g(x) \in \mathbb{Z}[x]$, and $(r f(x))(s g(x))=r s f(x) g(x)=f(x) g(x)$. From this last relation we can conclude that $s=r^{-1}$.
Therefore for an... |
Prove that $x^6+30x^5-15x^3 + 6x-120$ is irreducible in $\mathbb{Z}[x]$. | \begin{proof}
$$
x^6+30 x^5-15 x^3+6 x-120
$$
The coefficients of the low order.: $30,-15,0,6,-120$
They are divisible by the prime 3 , but $3^2=9$ doesn 't divide $-120$. So this polynomial is irreducible over $\mathbb{Z}$.
\end{proof} |
Prove that $\frac{(x+2)^p-2^p}{x}$, where $p$ is an odd prime, is irreducible in $\mathbb{Z}[x]$. | \begin{proof}
$\frac{(x+2)^p-2^p}{x} \quad \quad p$ is on add pprime $Z[x]$
$$
\frac{(x+2)^p-2^p}{x} \quad \text { as a polynomial we expand }(x+2)^p
$$
$2^p$ cancels with $-2^p$, every remaining term has $x$ as $a$ factor
$$
\begin{aligned}
& x^{p-1}+2\left(\begin{array}{l}
p \\
1
\end{array}\right) x^{p-2}... |
Prove that $x^2+y^2-1$ is irreducible in $\mathbb{Q}[x,y]$. | \begin{proof}
$$
p(x)=x^2+y^2-1 \in Q[y][x] \cong Q[y, x]
$$
We have that $y+1 \in Q[y]$ is prime and $Q[y]$ is an UFD, since $p(x)=x^2+y^2-1=x^2+$ $(y+1)(y-1)$ by the Eisenstein criterion $x^2+y^2-1$ is irreducibile in $Q[x, y]$.
\end{proof} |
If $r$ is rational $(r \neq 0)$ and $x$ is irrational, prove that $rx$ is irrational. | \begin{proof}
If $r x$ were rational, then $x=\frac{r x}{r}$ would also be rational.
\end{proof} |
Let $E$ be a nonempty subset of an ordered set; suppose $\alpha$ is a lower bound of $E$ and $\beta$ is an upper bound of $E$. Prove that $\alpha \leq \beta$. | \begin{proof}
Since $E$ is nonempty, there exists $x \in E$. Then by definition of lower and upper bounds we have $\alpha \leq x \leq \beta$, and hence by property $i i$ in the definition of an ordering, we have $\alpha<\beta$ unless $\alpha=x=\beta$.
\end{proof} |
Prove that no order can be defined in the complex field that turns it into an ordered field. | \begin{proof}
By Part (a) of Proposition $1.18$, either $i$ or $-i$ must be positive. Hence $-1=i^2=(-i)^2$ must be positive. But then $1=(-1)^2$, must also be positive, and this contradicts Part $(a)$ of Proposition 1.18, since 1 and $-1$ cannot both be positive.
\end{proof} |
If $z_1, \ldots, z_n$ are complex, prove that $|z_1 + z_2 + \ldots + z_n| \leq |z_1| + |z_2| + \cdots + |z_n|$. | \begin{proof}
We can apply the case $n=2$ and induction on $n$ to get
$$
\begin{aligned}
\left|z_1+z_2+\cdots z_n\right| &=\left|\left(z_1+z_2+\cdots+z_{n-1}\right)+z_n\right| \\
& \leq\left|z_1+z_2+\cdots+z_{n-1}\right|+\left|z_n\right| \\
& \leq\left|z_1\right|+\left|z_2\right|+\cdots+\left|z_{n-1}\right|+\... |
If $z$ is a complex number such that $|z|=1$, that is, such that $z \bar{z}=1$, compute $|1+z|^{2}+|1-z|^{2}$. | \begin{proof}
$|1+z|^2=(1+z)(1+\bar{z})=1+\bar{z}+z+z \bar{z}=2+z+\bar{z}$. Similarly $|1-z|^2=(1-z)(1-\bar{z})=1-z-\bar{z}+z \bar{z}=2-z-\bar{z}$. Hence
$$
|1+z|^2+|1-z|^2=4 \text {. }
$$
\end{proof} |
Prove that $|\mathbf{x}+\mathbf{y}|^{2}+|\mathbf{x}-\mathbf{y}|^{2}=2|\mathbf{x}|^{2}+2|\mathbf{y}|^{2}$ if $\mathbf{x} \in R^{k}$ and $\mathbf{y} \in R^{k}$. | \begin{proof}
The proof is a routine computation, using the relation
$$
|x \pm y|^2=(x \pm y) \cdot(x \pm y)=|x|^2 \pm 2 x \cdot y+|y|^2 .
$$
If $\mathrm{x}$ and $\mathrm{y}$ are the sides of a parallelogram, then $\mathrm{x}+\mathrm{y}$ and $\mathbf{x}-\mathrm{y}$ are its diagonals. Hence this result says tha... |
If $k = 1$ and $\mathbf{x} \in R^{k}$, prove that there does not exist $\mathbf{y} \in R^{k}$ such that $\mathbf{y} \neq 0$ but $\mathbf{x} \cdot \mathbf{y}=0$ | \begin{proof}
Not true when $k=1$, since the product of two nonzero real numbers is nonzero.
\end{proof} |
If $A$ and $B$ are disjoint closed sets in some metric space $X$, prove that they are separated. | \begin{proof}
We are given that $A \cap B=\varnothing$. Since $A$ and $B$ are closed, this means $A \cap \bar{B}=\varnothing=\bar{A} \cap B$, which says that $A$ and $B$ are separated.
\end{proof} |
Prove that every compact metric space $K$ has a countable base. | \begin{proof}
$K$ can be covered by a finite union of neighborhoods of radius $1 / n$, and this shows that this implies that $K$ is separable.
It is not entirely obvious that a metric space with a countable base is separable. To prove this, let $\left\{V_n\right\}_{n=1}^{\infty}$ be a countable base, and let $x... |
Suppose $E\subset\mathbb{R}^k$ is uncountable, and let $P$ be the set of condensation points of $E$. Prove that at most countably many points of $E$ are not in $P$. | \begin{proof}
If $x \in W^c$, and $O$ is any neighborhood of $x$, then $x \in V_n \subseteq O$ for some n. Since $x \notin W, V_n \cap E$ is uncountable. Hence $O$ contains uncountably many points of $E$, and so $x$ is a condensation point of $E$. Thus $x \in P$, i.e., $W^c \subseteq P$.
Conversely if $x \in W$, ... |
Prove that every open set in $\mathbb{R}$ is the union of an at most countable collection of disjoint segments. | \begin{proof}
Let $O$ be open. For each pair of points $x \in O, y \in O$, we define an equivalence relation $x \sim y$ by saying $x \sim y$ if and only if $[\min (x, y), \max (x, y)] \subset$ 0 . This is an equivalence relation, since $x \sim x([x, x] \subset O$ if $x \in O)$; if $x \sim y$, then $y \sim x$ (sinc... |
Prove that $\lim_{n \rightarrow \infty}\sqrt{n^2 + n} -n = 1/2$. | \begin{proof}
Multiplying and dividing by $\sqrt{n^2+n}+n$ yields
$$
\sqrt{n^2+n}-n=\frac{n}{\sqrt{n^2+n}+n}=\frac{1}{\sqrt{1+\frac{1}{n}}+1} .
$$
It follows that the limit is $\frac{1}{2}$.
\end{proof} |
For any two real sequences $\left\{a_{n}\right\},\left\{b_{n}\right\}$, prove that $\limsup _{n \rightarrow \infty}\left(a_{n}+b_{n}\right) \leq \limsup _{n \rightarrow \infty} a_{n}+\limsup _{n \rightarrow \infty} b_{n},$ provided the sum on the right is not of the form $\infty-\infty$. | \begin{proof}
Since the case when $\limsup _{n \rightarrow \infty} a_n=+\infty$ and $\limsup _{n \rightarrow \infty} b_n=-\infty$ has been excluded from consideration, we note that the inequality is obvious if $\limsup _{n \rightarrow \infty} a_n=+\infty$. Hence we shall assume that $\left\{a_n\right\}$ is bounded... |
Prove that the convergence of $\Sigma a_{n}$ implies the convergence of $\sum \frac{\sqrt{a_{n}}}{n}$ if $a_n\geq 0$. | \begin{proof}
Since $\left(\sqrt{a_n}-\frac{1}{n}\right)^2 \geq 0$, it follows that
$$
\frac{\sqrt{a_n}}{n} \leq \frac{1}{2}\left(a_n^2+\frac{1}{n^2}\right) .
$$
Now $\Sigma a_n^2$ converges by comparison with $\Sigma a_n$ (since $\Sigma a_n$ converges, we have $a_n<1$ for large $n$, and hence $\left.a_n^2<a_n... |
Prove that the Cauchy product of two absolutely convergent series converges absolutely. | \begin{proof}
Since both the hypothesis and conclusion refer to absolute convergence, we may assume both series consist of nonnegative terms. We let $S_n=\sum_{k=0}^n a_n, T_n=\sum_{k=0}^n b_n$, and $U_n=\sum_{k=0}^n \sum_{l=0}^k a_l b_{k-l}$. We need to show that $U_n$ remains bounded, given that $S_n$ and $T_n$ ... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.