# Thread: Problem #12: Introductory Group Theory

1. ## Problem #12: Introductory Group Theory

Here's this week's puzzle.

Let G be a group. If we have the following two conditions:
1) $(xy)^2 = (yx)^2 ~~\forall x, y \in G$

2) There exists no element of G of order 2.

Prove that G is Abelian.

(Source: Nick's Mathematical Puzzles)

-Dan

2. ## Re: Problem #12: Introductory Group Theory

Spoiler:
\begin{align*}\left(x^{-1}(yx)\right)^2 & = \left((yx)x^{-1}\right)^2 \quad \text{ by Property (1)} \\ & = \left(y(xx^{-1})\right)^2 \quad \text{ by Associativity of }G \\ & = \left(ye\right)^2 \\ & = y^2\end{align*}

Also, for any $x,y\in G$, we have $\left(x^{-1}(yx)\right)^2 = \left(x^{-1}(yx)\right)\left(x^{-1}(yx)\right) = x^{-1}y(xx^{-1})yx = x^{-1}y^2x$.

Hence, for all $x,y\in G$, we have $x^{-1}y^2x = y^2$.

This implies $y^2 \in Z(G)$ for all $y \in G$. Let $G_0 = \{x \in G \mid \text{ord}_G(x) \equiv 0 \pmod{2}\}$ be the elements of $G$ with even (or infinite) order and $G_1 = \{x \in G \mid \text{ord}_G(x) \equiv 1 \pmod{2}\}$ be the elements with odd order. Let $x \in G_1$. Then there exists $k \in \Bbb{Z}$ with $k\ge 0$ such that $x^{2k+1} = e$. Hence, $x = (x^{-k})^2 \in Z(G)$.

So, all that is left is to show $G_0 \subseteq Z(G)$. I haven't quite figured out how to show that yet.

3. ## Re: Problem #12: Introductory Group Theory

Spoiler:
Oh, if $G$ is finite, then since no element of $G$ has order 2, then $2 {\not |} |G|$ (by Cauchy's Theorem). Hence, $G_0 = \emptyset$, so $G = G_1 = Z(G)$, and the group is abelian.

If $G$ is infinite, then I am not sure if it can have elements with finite even order, nor am I sure how to complete the proof.

4. ## Re: Problem #12: Introductory Group Theory

Spoiler tags!!

Much more advanced than my solution, but mine doesn't have to involve whether the group is finite or not. You may be over-thinking the problem?

-Dan

Edit: On the other hand you seem to have proven something that I suspected but didn't see how to do. Such a group G is always isomorphic to a cyclic group, isn't it?

5. ## Re: Problem #12: Introductory Group Theory

Originally Posted by topsquark
Spoiler tags!!

Much more advanced than my solution, but mine doesn't have to involve whether the group is finite or not. You may be over-thinking the problem?

-Dan

Edit: On the other hand you seem to have proven something that I suspected but didn't see how to do. Such a group G is always isomorphic to a cyclic group, isn't it?
I believe $\Bbb{Z}/3\Bbb{Z} \times \Bbb{Z}/3\Bbb{Z}$ satisfies the two conditions, yet is not isomorphic to a cyclic group. I have not checked it, though. I will continue thinking about the problem to see if I can extend the solution to the case where $G$ might be infinite.

6. ## Re: Problem #12: Introductory Group Theory

Dan,
Here's a solution, perhaps not entirely elementary. By the way, any abelian group of odd order clearly satisfies your two conditions. So such a group need not be cyclic.

7. ## Re: Problem #12: Introductory Group Theory

Here is my solution (I didn't look at the other ones, so there may be duplication, here):

Spoiler:

Suppose $(xy)^2 = (yx)^2$. Then:

$xyxy = yxyx$

$xyx^{-1}y^{-1} = yx$

$xyx^{-1}y^{-1} = y^{-1}x^{-1}yx$, or in commutator notation:

$[x,y] = [y^{-1},x^{-1}]\dots(1)$. <--This is an equivalent re-formulation of the original condition.

Now, note that $[x,y]^{-1} = (xyx^{-1}y^{-1})^{-1} = yxy^{-1}x^{-1} = [y,x]\dots(2)$

Hence $[x,y] = [y^{-1},x^{-1}]$ by (1)

$= [x^{-1},y^{-1}]^{-1}$ by (2),

$= [(y^{-1})^{-1},(x^{-1})^{-1}]$ by (1),

$= [y,x] = [x,y]^{-1}$ by (2) again.

Since $G$ has no element of order 2, it must be that for every $x,y \in G$, we have $[x,y] = e$, that is: $xyx^{-1}y^{-1} = e$, so that $xy = yx$ and $G$ is abelian.

8. ## Re: Problem #12: Introductory Group Theory

Thanks to everyone for their solutions. Mine has elements similar to most of the work shown.

For reference:
1) [tex](xy)^2 = (yx)^2

2) There exists no element of G of order 2.
Let e be the identity in G.

Then
$x^2y = \left [ \left ( xy^{-1} \right ) y \right ] ^2y$

$= \left [ y \left ( xy^{-1} \right) \right ]^2y$ by 1)

$= \left [ yxy^{-1}yxy^{-1} \right ] y$

Thus
3) $x^2y = yx^2$

Now
$x^{-1}y^{-1}x = \left [ x \left ( x^{-1} \right )^2 \right ]y^{-1}x$

$= x \left [ \left ( x^{-1} \right ) ^2 y^{-1} \right ]x$

$= x \left [ y^{-1} \left ( x^{-1} \right )^2 \right ] x$ by 3)

Thus
4) $x^{-1}y^{-1}x = xy^{-1}x^{-1}$

Good so far. Next:
$\left ( xyx^{-1}y^{-1} \right ) ^2 = xy \left [ x^{-1} y^{-1}x \eflt ] yx^{-1}y^{-1}$

$=xy \left [ xy^{-1}x^{-1} \right ] yx^{-1}y^{-1}$ by 4)

$= xyx \left [ y^{-1}x^{-1}y \right ] x^{-1}y^{-1}$

$= xyx \left [ yx^{-1}y^{-1} \right ] x^{-1}y^{-1}$ by 4)

$= (xy)^2 \left ( x^{-1}y^{-1} \right )^2$

$= (xy)^2 \left ( y^{-1} x^{-1} \right ) ^2$ by 1)

$= xyxyy^{-1}x^{-1}y^{-1}x^{-1}$

Thus $\left ( xyx^{-1}y^{-1} \right ) ^2 = e$

Thus we have an element of G that has order 2. But we can't have that by 2). So $xyx^{-1}y^{-1} = e$

Finally(!) we have that xy = yx. Thus G is Abelian.

-Dan