Let $G$ be a group such that $a^2 = a$ for all $a \in G$, Is $G$ an abelian group?

Question

I tried to solve this by following:

Since $G$ is a group, an inverse exists for every element in $G$.
Multiply by inverse to $a$ on both sides of $a^2 = a$.
We will get $a = i$, where $i$ is the identity element.
This holds true for all $a*$, which implies $G$ contains only one distinct element i.e. $i.$
Hence $G$ is abelian.

Is my approach correct?

Correct, but it seems too easy for an exercise. Are you sure the question didn't say, let $G$ be a group such that $a^2=i$ (the identity element) for all $a\in G,$ is $G$ an Abelian group? — bof, Aug 22 '17 at 03:42
I solved for a^2 = i. This I could solve easily so I got confused that there might be some subtle point I am missing. Thank you — sourav, Aug 22 '17 at 03:46
If the goal of your question is mainly to ask about correctness of your proof (as opposed to asking for any proof of this fact), you should add ([tag:proof-verification]) tag to make this clear. — Martin Sleziak, Aug 22 '17 at 05:30
Related: Trick to proving a group has exactly one idempotent element - Fraleigh p. 48 4.31. — Martin Sleziak, Aug 22 '17 at 05:33
If the question was actually $a^2=e$ (the identity element) then it would be a duplicate of this one — rschwieb, Aug 22 '17 at 13:26

score 2 · Accepted Answer · answered Aug 22 '17 at 03:59

2

Yes, I think you are right.

Also, we can get the following.

$$abab=ab$$ gives $$aba=a,$$ which gives $$ab=ba=e.$$ Done!

answered Aug 22 '17 at 03:59

Michael Rozenberg

1

@1ENİGMA1 Yes of course. There are very many similar easy problems. – Michael Rozenberg Aug 22 '17 at 07:31

1 Answers1