How can I justify that $a^{2}=e$ and $b^{3}=e$ for an Abelian group of order 6?

Question

I've been reading a proof that an Abelian group of order 6 is cyclic and I am not quite sure how we can justify that in the computations $a^{2}=e$ and $b^{3}=e$. What theorem or proposition is being used to justify that claim? I feel like this is fundamental about groups that I lost somewhere along the way.

Answer 1

You can use Cauchy's theorem, that if a prime $p$ divides the group order, so if $p\mid 6$, then the group has an element of order $p$. Apply this for $p=2$ and for $p=3$ and you will obtain the claim.