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I don't understand this step (this is from some proof of a lemma from a book):

Let $g$ and $h$ be elements of a group $G$, where all elements are self-inverse.

so $gg = e$ and $hh = e$.

We can then write: $(gh)(gh) = e$.

By what operation do we reach this last step?

ThePhi
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    You might want to check the specifics of the situation described in the book. You will likely find that you are told that every element of the group has order $2$, as this is a standard beginner exercise. It is not true that the product of involutions (elements of order $2$) is always an involution, nor is it true that all involutions in a general group commute ($S_3$ the non-abelian group of order $6$ is a simple counterexample). – Mark Bennet Jun 30 '22 at 07:39
  • @MarkBennet My mistake sorry. Indeed, all elements are self-inverse in fact (I've edited). – ThePhi Jun 30 '22 at 07:41
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    Then the element $gh$ is self-inverse, which means that $(gh)(gh)=e$ – Mark Bennet Jun 30 '22 at 07:42
  • Ah ok! So it's a direct property, and no algebraic proof is needed. Thank you. – ThePhi Jun 30 '22 at 07:48
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    For such groups with this property see also the posts here (it is a popular homework question), e.g., here. – Dietrich Burde Jun 30 '22 at 08:23
  • Which book are you referring to? – Shaun Jun 30 '22 at 11:38
  • @Shaun It's not a published book, it's the book accompanying a course from my university. – ThePhi Jun 30 '22 at 15:24

2 Answers2

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Let $g$ and $h$ be elements of a group $G$, where all elements are self-inverse.

...

We can then write: $(gh)(gh) = e$.

By what operation do we reach this last step?

All elements are self-inverse, and it's a group, so $gh\in G$, so $gh$ is self-inverse, in other words $ghgh=e$.

Suzu Hirose
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This addresses the original version of the question, before the extra condition "where all elements are self-inverse" set in. So, suppose there are $g,h\in G$ such that $g^2=h^2=e$; then $(hg)(gh)=e$, and hence $(gh)(gh)=e$ only if $hg=gh$, namely $h\in C_G(g)$ (or, equivalently, $g\in C_G(h)$).