$G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$

Question

Suppose that $G$ is a group that there exists no element $x \neq e$ such that $x^2=e$. Moreover, for every $a,b \in G$ we have $(ab)^2=(ba)^2$. Prove that $G$ is Abelian.

Well, I attempted to prove that $(aba^{-1}b^{-1})^2=e$ because then if I could prove it that would imply $aba^{-1}b^{-1}=e$ which is the same as $ab=ba$. In other words I wanted to show the order of $[a,b]=aba^{-1}b^{-1}$ is two for every $a$ and $b$ in $G$. I spent almost an hour trying to show that by anything that came to my mind. I tried to brute force the problem by writing any possible equation that I could come up with but I failed :/

Any ideas?

Boris Novikov · Answer 1 · 2013-11-14T21:00:28.303

13

Set $b=a^{-1}x$. We have $x^2=a^{-1}x^2a$, i.e. $ax^2=x^2a$ for all $x,a$. Since $x^2$ runs all group, then $G$ is Abelian.

Correction: This proof is valid only for a finite group. Thanks to DonAntonio.

Addendum: I am not sure that this assertion is true for infinite groups. A candidate \for a counter-example is $G=\langle a,b|a^2=b^2, (ab)^2=(ba)^2\rangle$.

edited Nov 14 '13 at 21:00

answered Nov 14 '13 at 16:44

Boris Novikov

17,470

Why does $x^2$ run the whole group? – rfauffar Nov 14 '13 at 16:45
Actually he's right: $x^2=y^2$ if and only if $(xy^{-1})^2=e$, if and only if $x=y$. – rfauffar Nov 14 '13 at 16:46
So $x\mapsto x^2$ is at least injective. If the group is finite then it is also surjective by cardinality. – rfauffar Nov 14 '13 at 16:47
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@PedroTamaroff, $2\mathbb{Z}\neq\mathbb{Z}$. – rfauffar Nov 14 '13 at 16:50
@PedroTamaroff what does your notation mean? – rfauffar Nov 14 '13 at 16:51
But $G^2=G$ with your notation is a trivial statement, since $g=ge\in G^2$. – rfauffar Nov 14 '13 at 16:53
Nevermind, I will remove comments and add an answer (maybe), this is getting too long. – Pedro Nov 14 '13 at 16:54
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@RobertAuffarth: $;x^2=y^2\iff x^2y^{-2}=1;$ . How do you pass from this to $;(xy^{-1})^2=1;$ ?? – DonAntonio Nov 14 '13 at 17:05
@DonAntonio: You're right. I guess Boris Novikov's answer needs to include why $x^2$ exhausts the group. – user66733 Nov 14 '13 at 17:12
And yet he's right, @some1.new4u...at least when the group is finite: since the order of the group is clearly odd, we can put $;|G|=2n-1;,;n\in\Bbb N;$ , and then $$x^2=y^2\implies x=x\cdot x^{2n-1}=x^{2n}=(x^2)^n=(y^2)^n=y^{2n}=y^{2n-1}\cdot y=y$$ – DonAntonio Nov 14 '13 at 17:20
@DonAntonio: Yes, but that only settles the case when $|G|$ is finite. What if $|G|$ is infinite? – user66733 Nov 14 '13 at 17:28

Giorgio Mossa · Answer 2 · 2013-11-16T16:31:05.550

7

Edit: Andreas Caranti have pointed out, this proof work iff the all the elements of the group have finite order. So it doesn't answer completely but is more a partial result.

Let $x,a \in G$ then by hypothesis we have that there's $b \in G$ such that $ab=x$ then $$a^{-1}x^2a=a^{-1}(ab)^2a=a^{-1}ababa=baba=(ba)^2=(ab)^2=x^2$$ So $x^2 \in Z(G)$ for all $x \in G$.

So we get the quotient $G/Z(G)$ in which all the elements have order two.

If there's an $x \in G \setminus Z(G)$ then $xZ(G) \ne Z(G)$ should have order $2$, and by properties of homomorphisms of groups the order of $x$ should be divided by $2$. That's absurd since by hypothesis $G$ can't have elements of order $2$ hence it can't have either elements which have order a multiple of $2$.

edited Nov 16 '13 at 16:31

answered Nov 14 '13 at 16:38

Giorgio Mossa

18,150

$abab\neq a^2b^2$, unfortunately. – vadim123 Nov 14 '13 at 16:38
@vadim123 My bad I've misread the question. I'm gonna edit soon. – Giorgio Mossa Nov 14 '13 at 16:39
That's another problem you have in mind I guess. – user66733 Nov 14 '13 at 16:40
@some1.new4u I was confused, now I've edited :) – Giorgio Mossa Nov 14 '13 at 16:57
@vadim123 thanks for pointing out my mistake :) – Giorgio Mossa Nov 14 '13 at 16:57
@GiorgioMossa: This is a really good answer, but I think there must be a more elementary solution to this problem. The problem has been put before the chapter that discusses homomorphisms or the center of group in my book. – user66733 Nov 14 '13 at 17:33
@some1.new4u it's possible, anyway till now I haven't come up with nothing simpler. Btw could you tell me which book are you referring to? – Giorgio Mossa Nov 14 '13 at 17:35
@GiorgioMossa: It's a non-English book. It hasn't been published yet, so I can't actually call it a book. It's rather a set of lecture notes and problems discussed in the class + homework and other materials covered during a typical Algebra I course. – user66733 Nov 14 '13 at 17:41
@some1.new4u if you tell me exactly what are the instrument you could use, I could try to come up with some simpler solution :) – Giorgio Mossa Nov 14 '13 at 17:43
@GiorgioMossa: Well, ideally we should use only the definition of an Abelian group and theorems that we know about groups, semi-groups, monoids and such basic stuff. :| I spent a long time trying to solve it by elementary ways, but I failed. – user66733 Nov 14 '13 at 17:57
I don't see why $x$ could not, in principle, be an element of infinite order, with $x^2 \in Z(G)$. But maybe you are assuming $G$ finite. – Andreas Caranti Nov 14 '13 at 18:57
@AndreasCaranti no, I'm not assuming that the group have finite order. Apparently I've assumed that the elements have finite order. I'm gonna edit the answer and add this detail. Thanks for pointing out. – Giorgio Mossa Nov 14 '13 at 21:05

score 2 · Answer 3 · answered Oct 10 '21 at 03:36

The following solution is the same as in the duplicate, just made shorter. As in the answer of Boris Novikov, for every $x\in G$ the element $x^2$ is in the centralizer of $Z(G)$ of $G$, by using $(ab)^2=(ba)^2$ for $b=a^{-1}x$: $$ x^2 =(ab)^2=(ba)^2=(a^{-1}xa)^2=a^{-1}x^2a\ . $$ Let now $s,t\in G$ be two elements. We show $st=ts$. Consider $$a = s^{-2}t^{-2}\; stst\ . $$ One easily shows $a^2=1$ by using $s^{\pm 2},t^{\pm 2}\in Z(G)$: $$ \begin{aligned} a^2 &= s^{-4}t^{-4}\cdot stst\cdot stst\\ &= s^{-4}t^{-4}\cdot stst\cdot tsts\\ &= s^{-4}t^{-4}\cdot sts\cdot t^2\cdot sts\\ &= s^{-4}t^{-2}\cdot st\cdot s^2\cdot ts\\ &= s^{-2}t^{-2}\cdot s\cdot t^2\cdot s\\ &= s^{-2}\cdot s^2\\ &=1\ . \end{aligned} $$ From the assumption, $a=1$, i.e. $$ 1=s^{-2}\; stst\; t^{-2}=s^{-1}\; ts\; t^{-1}\ , $$ so after multiplying from left with $s$, and from right with $t$ we get $st=ts$.

$\square$

medicu · Answer 4 · 2013-11-16T19:41:04.183

Elementary solution to this problem.

The problem is equivalent to that the $x^2 \in Z(G)$ for any $x\in G$, as stated in a previous response.

Under these conditions we show that $(xyx^{-1}y^{-1})^4=e.$ $$(xyx^{-1}y^{-1})^4=(xyx^{-1}y^{-1})^2(xyx^{-1}y^{-1})(xyx^{-1}y^{-1})=$$ $$=(xyx^{-1})(xyx^{-1}y^{-1})^2(y^{-1})(xyx^{-1}y^{-1})=$$ $$= xy(x^{-1}x)(yx^{-1}y^{-1})(xyx^{-1}y^{-1})(y^{-1})(xyx^{-1}y^{-1})=$$ $$=xy^2(x^{-1}y^{-1})(xyx^{-1})(y^{-1})^2(xyx^{-1}y^{-1})=$$ $$=y^2(xx^{-1})(y^{-1}xy(y^{-1})^2)(xyx^{-1}y^{-1})=(yxy^{-1})(yx^{-1}y^{-1})=e.$$ Considering that the group no has elements of order $ 2 $ result, step by step, that $(xyx^{-1}y^{-1})^2=e$, $(xyx^{-1}y^{-1})=e$ and $xy=yx$ and therefore the group is commutative.

It can be shown that in a group the following statements are equivalent: a) $(ab)^n=(ba)^n$, for any $a,b \in G$; b)$a^nb=ba^n$, for any $a,b \in G$. — medicu, Nov 17 '13 at 16:45

score 0 · Answer 5 · answered Apr 12 '22 at 12:49

$\forall x,a\in G,ax,a^{-1} \in G\Rightarrow ax^2a^{-1}=x^2\Rightarrow ax^2=x^2a$

$\forall x,y\in G$

$xyxy=yxyx\Rightarrow x^{-1}y^{-1}xy=yxy^{-1}x^{-1}$

$(xyx^{-1}y^{-1})^2=xy(x^{-1}y^{-1}xy)x^{-1}y^{-1}=xy^2xy^{-1}(x^{-1})^2y^{-1}$=$x^2y^2y^{-2}x^{-2}=e$

$order(xyx^{-1}y^{-1})\neq 2\Rightarrow xyx^{-1}y^{-1}=e$

$xy=yx$

score 0 · Answer 6 · answered Mar 20 '24 at 15:10

Another way:

$(aba^{-1}b^{-1})^4=(aba^{-1}b^{-1})(aba^{-1}b^{-1})^2(aba^{-1}b^{-1})\\$ $=(aba^{-1}b^{-1})(a^{-1}b^{-1}ab)^2(aba^{-1}b^{-1})\\$ $=aba^{-1}b^{-1}a^{-1}b^{-1}aba^{-1}b^{-1}ababa^{-1}b^{-1}\\$ $=ab(a^{-1}b^{-1})^2aba^{-1}b^{-1}(ab)^2a^{-1}b^{-1}\\$ $=ab(b^{-1}a^{-1})aba^{-1}b^{-1}(ba)^2a^{-1}b^{-1}\\$ $=abb^{-1}a^{-1}b^{-1}a^{-1}aba^{-1}b^{-1}baba^{-1}b^{-1}\\$ $=eeee=e\implies (aba^{-1}b^{-1})^2=e\implies (aba^{-1}b^{-1})=e\implies ab=ba$

$G$ is Abelian if it has no element of order $2$ and $(ab)^2=(ba)^2$

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