Prove that if in a group $(ab)^2= a^2 b^2$ then the group is commutative.  I am having a hard time doing this. Here is what I have so far: Proof:  $a^2 b^2= a^1 a^1 b^1 b^1$ =$aa^{-1}bb$ =ebb Hence,$aa^{-1}=e$  I am stuck, I do not know if this is the right process in proving this
122407
I want to show that $\forall x$,$y\in G$, $(xy)^2=x^2y^2 \iff G $ is an abelian group.  Proof of the backward direction seems easy enough:  Assume $G$ is an abelian group. Then consider $(xy)^2=(xy)(xy)$. By associativity, $(xy)(xy)=xyxy$. Since $G$ is abelian, $xyxy=xxyy=x^2y^2$.   Proof of the forward direction I attempt by contrapositive:  That is, assume $G$ is not an abelian group. Then, consider $(xy)^2=xyxy$ as above. Yet, $xyxy\neq xxyy=x^2y^2$ since we no longer have commutativity.  Is this proof by contrapositive completely correct?
47706
From Dr. Pinter's "A Book of Abstract Algebra":     Given $a$ and $b$ are in $G$ and $ab=ba$, we say that $a$ and $b$ commute.         Prove $a^2$ commutes with $b^2$      I tried:  $$ ab=ba $$ $$ a^{2}b^{2}=b^{2}a^{2} \text{ // to prove that $a^2$ commutes with $b^2$} $$ $$ aabb=bbaa $$  Then, knowing that $$a=b^{-1}ab,$$ I substituted that '$a$' into '$aabb=bbaa$'.  But that simplified to '$aabb=bbaa$', i.e. I got nowhere with it.  How can I prove that $$aabb=bbaa$$ per this problem?
44724
Let $G$ be a group, and suppose that $a= {a^{-1}}$ for every $a\in G$. Prove that $G$ is Abelian.    I know that I need to prove that ${a^{-1}}$ = a and use right multiplication. I also know that I need to declare a, b $\in$ G and go from there, but I'm confused when I get to   (${a^{-1}}$${b^{-1}}$) = ab.
59790
Let $G$ be a group and $a,b \in G$. Prove that if $a^{2}=e$ and $ab^{4}a=b^{7}$, then $b^{33}=e$, where $e$ is the identity of a group $G$.
61648
Given matrices $A$ and $B$, where that $AB = A + B$, prove $AB = BA$. I keep coming up with AB = AB. It seems like basic algebra, but for the life of me, I'm getting nowhere :/. Someone help please?
83231
Let $A$ be a normal subgroup of a group $G$. Let $B$ be a subgroup of   $A$ with the property that whenever $B_1 \leq A$ is conjugate to $B$   in $G$ then it is conjugate to $B$ in $A$. Prove that $G=AN_G(B)$   Help me some hints.
96709
Let $G$ be a group and let $f:G\to G$ be an isomorphism. Let $H = \{ a \in G \; | \; f(a)=a^{-1}\}$ Prove that $H$ is a subgroup of $G$ if and only if $G$ is abelian.  My attempt: I already know that $H$ is a subset of $G$ since $a$ is in $G$. And that $H$ is $a$ group if and only if for every $a$ and $b$ in $H$, $ab^-1$ is in $H$. That is, if $f(a)=a^{-1}$ and $f(b)=b^{-1}$ then $f(ab^-1)=(ab^{-1})^{-1}=(b^{-1})a$.  But I don't know how to derive this nor the converse based on the abelian property of $G$ that for every $a$ and $b$ in $G$, $ab=ba$.  Please help some. Thanks!
41628
I've been working on this problem listed in Herstein's Topics in Algebra (Chapter 2.3, problem 4):     If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, show that $G$ is abelian.   I managed to prove it, but I'm not very happy with my result (I think there's a neater way to prove this). Anyway, I'm just looking to see if there's a different approach to this.   My approach:  Let $j=i+1, k=i+2$ for some $i\in \mathbb{Z}$.  Then we have that $(ab)^i = a^ib^i$, $(ab)^j = a^jb^j$ and $(ab)^k = a^kb^k$.  If $(ab)^k = a^kb^k$, then $a^jb^jab =a^jab^jb$.   We cancel on the left and right and we have $b^ja = ab^j$, that is $b^iba = ab^j$.  Multiply both sides by $a^i$ on the left and we get $a^ib^iba = a^jb^j$, so $(ab)^iba = (ab)^j$. But that is $(ab)^iba = (ab)^iab$.   Cancelling on the left yields $ab=ba$, which holds for all $a,b \in G$, and therefore, $G$ is abelian.  Thanks!
114260
Let $G$ be a group. If $(ab)^n=a^nb^n$ $\forall a,b \in G$ and $(|G|, n(n-1))=1$ then prove that $G$ is abelian.  What I have proven is that:  If $G$ is a group such that $(ab)^i = a^ib^i$ for three consecutive integers $i$ for all $a, b\in G$, then $G$ is abelian.  A proof of this can be found in the answers to  old question.
107609
Let $(S, \cdot)$ be a semigroup. There are natural numbers $p,q \geq 2$ such that $a^pb^q=ba$ for all $a,b \in S$. Prove that $S$ is commutative.   I wrote   $$\begin{align} a^{p+1}b^{q+1} &amp;=b^{(q+1)p}a^{(p+1)q} \\  &amp;=b^{p}\cdot(b^q)^p \cdot (a^p)^q\cdot a^q \\  &amp;=b^p\cdot a^p \cdot b^q \cdot a^q \\  &amp;= b^p\cdot b \cdot a \cdot a^q \\  &amp;=b^{p+1}a^{q+1}. \end{align}$$  From the given identity I also got $a^{p+1}b^{q+1}=abab$. Using $a^{p+1}b^{q+1}=b^{p+1}a^{q+1}$ I then got $abab=baba$.  Making $a=b$ in the statement gives $a^{p+q}=a^2$. I don't know what to do from there.
76779
Let $R$ be a ring such that for all $a,b$ in $R$, $(a^2-a)b=b(a^2-a)$. Then $R$ is commutative.   I am having difficulties making the connection. Any help would be appreciated.
34805
Let $G$ be a group and $a,b\in G$.  Show $(ab)^n=a^nb^n$ for all $n\in\mathbb{Z}$ if and only if $ab=ba$.  I don't known where to start. It seems trivially.
41419
$G$ is a group and $f:G \rightarrow G$ is a function defined as $f(a)=a^{-1}$ where $a^{-1}$ is the inverse of $a$ under the group operation. Prove that $f$ is an isomorphism if and only if $G$ is abelian.    I understand that I have to prove $f(ab)=(ab)^{-1}=b^{-1}a^{-1}$. How might I do that?  Reference: Fraleigh p. 49 Question 4.40 in A First Course in Abstract Algebra
24873
Let G be a group, where $(ab)^3=(a^3)(b^3)$ and $(ab)^5=(a^5)(b^5)$. Prove that $G$ is abelian group.  Thank you in advance. Any help is appreciated.
94331
Problem: Hi, I'm trying to solve the following proof:  If $a &gt; b \geq 0$, Prove that $a^2 &gt; b^2$.  But I'm a bit new with proofs and I don't manage to write a proper one. I would be very grateful if someone could explain me which are the steps required to demonstrate the proof. Thank you.
62639
Let $G$ be a group in which $a^2=e$ for all elements of $a$ of $G$.   Show that $G$ is Abelian.  I need help on this problem.  Appreciated!
109914
Let $G$ be a group. If for all $a,b,c\in G$, $ab=ca\Rightarrow b=c$, then phow can I prove that  $G$ is abelian?