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

I tried to solve this by following:

1. Since $G$ is a group, an inverse exists for every element in $G$.
2. Multiply by inverse to $a$ on both sides of $a^2 = a$.

3. We will get $a = i$, where $i$ is the identity element.

4. This holds true for all $a*$, which implies $G$ contains only one distinct element i.e. $i.$
5. Hence $G$ is abelian.

Is my approach correct?

• you are correct Aug 22, 2017 at 3:39
• 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, 2017 at 3:42
• 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 Aug 22, 2017 at 3:46
• 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 (proof-verification) tag to make this clear. Aug 22, 2017 at 5:30
• Aug 22, 2017 at 5:33

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