Group G with $\left(\forall g \in G: g^2=1_G\right) \implies \text{G is abelian}$

Asked Nov 10 '15 at 04:16

Active Nov 10 '15 at 04:16

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I have the following task:

Be $G$ a group with $\forall g \in G: g^2=1_G$

Prove that $G$ is abelian.

I proved it this way:

$\forall g \in G: g^2=1_G$ implies $\forall g \in G: g=g^{-1}$

Be $g,h \in G$.

It is $gh \in G$ and $(gh)^{-1} = h^{-1}g^{-1}$ since $(g(hh^{-1})g^{-1})=1_G=(h^{-1}(g^{-1}g)h)$

Hence:

$$gh = (gh)^{-1} = h^{-1}g^{-1} = hg \quad \square$$

This is the first proof with groups I've ever done in my life, hence I am a bit unsure if everything is correct. It would be nice when someone could tell me if my proof is complete.

edited Jun 12 '20 at 10:38

Community

asked Nov 10 '15 at 04:16

Harald Meisner

4

This looks good. You don't even need the middle two lines. The demonstration that $g$ = $g^{-1}$ and the last line should do it. – John Douma Nov 10 '15 at 04:19
1

http://math.stackexchange.com/questions/252362/prove-that-a-group-g-such-that-every-element-g-in-g-satisfies-the-equality-g – Asaf Karagila Nov 10 '15 at 04:25
1

Your proof is valid. You can also note that gh= g(gh.gh)h=(gg)hg(hh)=hg. – DanielWainfleet Nov 10 '15 at 04:48

Group G with $\left(\forall g \in G: g^2=1_G\right) \implies \text{G is abelian}$

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