If $g^p$ is identity for all $g$ in a group $G$ ($p$ prime), is $G$ then abelian?

Question

Let $p$ be a prime and $G$ a finite group and $e$ is the identity element, such that $g^p = e$ for all $g \in G$. Does this imply $G$ is abelian? (if it does it is then very easy to show $G$ is a direct product of $\Bbb Z/p\Bbb Z$'s) I can easily show answer is yes when $p=2$. I don't have any clue on how to proceed. Any help would be appreciated.

Answer 1

No, unless $p=2$. The classical example is the group of matrices of the form $$\pmatrix{1&a&b\\0&1&c\\0&0&1}$$ where the entries are from the integers modulo $p$.