# Let $G$ be a group in which $aba = b \; \forall a,b \in G$. Prove that $G$ is Abelian.

I was wondering if someone could tell me if my method is correct. I feel like the substation $$b=e$$ is a stretch and that there is a better way to do it.

We know that $$G$$ is Abelian $$\iff (ab)^{-1} = a^{-1}b^{-1}$$. So I know that we will have to manipulate $$aba = b$$ into $$(ab)^{-1} = a^{-1}b^{-1}$$.

But this is where I get stuck. $$aba = b \implies ab = ba^{-1} \implies (ab)^{-1} = (ba^{-1})^{-1} \implies (ab)^{-1}=ab^{-1}$$ This is close, but not what I want. One thing that I did notice is that if $$b=e$$, then $$aba=b \implies a^2 = e \implies a=a^{-1}$$. Since $$a^{-1}$$ is a unique element we know that $$a=a^{-1}$$ regardless of what $$b$$ is. So if we make that substitution into out previous equation we get $$(ab)^{-1} = a^{-1}b^{-1}$$ Therefore $$G$$ is Abelian.

• Notice that $ab=ba^{-1}$ and $aa=e$ for $b=a$ and just substitute the value of $a^{-1}$
– Gio
Jan 31, 2021 at 21:50
• The substitution $b=e$ is correct, since $aba=b$ holds for every $a$ and $b$ in $G$. Jan 31, 2021 at 21:50
• Why is setting $b = e$ a "stretch"? It is a completely standard result that if $g^2 = e$ for all elements $g$ in a group that the group is abelian. Therefore it's really quite natural to set $b = e$ to get $a^2 = e$ for all $a$ in the group.
– KCd
Jan 31, 2021 at 21:50

You're correct.

If $$a^2=e$$ for all $$a\in G$$, then for any $$g,h\in G$$,

\begin{align} (gh)^2&=ghgh\\ &=e\\ &=ee\\ &=g^2h^2\\ &=gghh, \end{align}

from which it follows that $$gh=hg$$. Hence $$G$$ is abelian.

The hypothesis implies an exponent of two for the group. But then it is abelian, since $$ab=(ab)^{-1}=b^{-1}a^{-1}=ba$$, for any two elements of the group $$a$$ and $$b$$.

Choose

$$a = b; \tag 1$$

then

$$aba = b \tag 2$$

becomes

$$a^3 = a, \tag 3$$

which implies

$$a^2 = e, \tag 4$$

$$e$$ being the identity element of $$G$$; thus,

$$a = a^{-1}, \tag 5$$

whence (2) becomes

$$aba^{-1} = b, \; \forall a, b \in G, \tag 6$$

or

$$ab = ba, \tag 7$$

and we see that $$G$$ is abelian. $$OE\Delta$$.