Question on abelian groups

Question

I am solving an exercise in Artin's textbook that asks us to assume, within some group $G$, that $xyz = 1$ and asks if this implies that $yxz = 1$. I've found a counterexample, but want to be sure that I'm using the language correctly.

To motivate finding the counterexample, I supposed for a contradiction that $yxz = 1$. I was able to prove that $xz = zx$. It was easy to, within the non-abelian group $S_3$ under composition, find an example where $xyz = 1$ but $yxz \neq 1$.

Here is my question: is finding that $yxz = 1$ implies that $xz = zx$ the same as finding that $yxz = 1$ implies that the group is abelian? $x$ and $z$ are arbitrary, surely, but I've added an additional assumption that $xyz = 1$. It isn't clear, for example, that I also have that $xy = yx$.

My understanding at the moment is that I have not found that the group is abelian, but rather that these particular $x$ and $z$ live in the center of the group, $Z(G)$. Is this correct?

Answer 1

No, it is not the same as proving the group is abelian.

What you’ve done is show that if $xyz$ and $yxz$ are both trivial, then $x$ commutes with $z$. Nothing more, and nothing else. (Edit: Though you may be able to derive other properties about $x$, $y$, and $z$, as user750041 says in comments; but just about these elements and how they interact with each other). It does not assert that $y$ and $x$ commute, nor that $y$ and $z$ commute. You have not even shown that $x$ and $z$ are central, only that they commute with each other, not with everything in $G$.

Note also that “$x$ and $z$ are arbitrary” is inaccurate: they must satisfy both $xyz=1$ and $yxz=1$. That makes them not arbitrary!

Answer 2

You are given the following condition:

$$\exists x,y,z \in G\mid xyz=1 \tag 1$$

Then, by $(1)$:

\begin{alignat}{1} yxz=1 &\Rightarrow yx=z^{-1}\stackrel{(1)}{=}xy \tag 2\\ \end{alignat}

Likewise, again by $(1)$:

\begin{alignat}{1} yxz=1 &\Rightarrow y=(xz)^{-1}=z^{-1}x^{-1} \\ &\stackrel{(1)}{\Rightarrow}1=xyz=xz^{-1}x^{-1}z=(xz^{-1})(z^{-1}x)^{-1} \\ &\Rightarrow xz^{-1}=z^{-1}x \\ &\Rightarrow zx=xz \\ \tag 3 \end{alignat}

Therefore:

$$(xyz=1 \Rightarrow yxz=1) \Rightarrow (xy=yx \wedge xz=zx)\tag 4$$

and

$$(xy=yx \vee xz=zx)\Rightarrow (xyz=1 \Rightarrow yxz=1) \tag 5$$

This has nothing to do with abelianiness, since no $\forall$ quantifier enters the discussion.