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It's been a while since I did things with groups. I have a certain group which has the following property:

$$g_1 * g_2 * g_1 =g_2$$

for every $g_1,g_2$ in the group

I am trying to prove its abelian. But I got a bit tangled here.

Let's say I multiply both sides from right by $g_2 ^{-1}$

Then I get:

$$g_1 * g_2 * g_1 * g_2^{-1}=1$$

So I know that

$$(g_1*g_2)^{-1} = g_1*g_2^{-1}$$

But I am not sure where this leads me.

Another option I tried is to look at $g_2^{2}$, but it did not lead me anywhere.

I feel I am going in circles and that it is not something complicated, only that I miss some thing.

Shaun
  • 44,997
Marik S.
  • 761

2 Answers2

8

If you always have $g_1g_2g_1=g_2$, then, in particular, $g_1eg_1=e$, which means that $g_1^{\,2}=e$. So, $g_1^{\,-1}=g_1$, for each $g_1\in G$. So, if $g_1,g_2\in G$,$$g_2=g_1g_2g_1=g_1g_2g_1^{\,-1},$$and therefore $g_1g_2=g_2g_1$.

José Carlos Santos
  • 427,504
0

Let $a\in G$. Then $aea=e$ by the property, so that $a^2=e$. Now for any $x,y\in G$, we have

$$\begin{align} (xy)^2&=e\\ &=ee\\ &=x^2y^2, \end{align}$$

so if we multiply on the left by $x^{-1}$ and on the right by $y^{-1}$, we get

$$yx=xy.$$

Hence $G$ is abelian.

Shaun
  • 44,997