There exist pairs of finite groups $G$ and $H$ such that $G$ and $H$ are not isomorphic, yet they have the same number of elements of each order. For example, if $p$ is an odd prime, then the group $$H_{p} = \left\{\begin{pmatrix} 1 & a & b \\ 0 & 1 & c \\ 0 & 0 & 1\end{pmatrix} : a,b,c\in\mathbb{Z}_{p}\right\}$$
and the group $\mathbb{Z}_{p}^{3}$ both have exponent equal to $p$ and order $p^{3}$. Also, for any such pair $G$ and $H$, at least one of $G$ and $H$ must be non-commutative. My question is this:
Do there exist three groups $A$, $B$ and $C$, of the same finite order, such that no two of them are isomorphic and such that all three of $A$, $B$ and $C$ have the same number of elements of each order?
Ideally, I'd like a nice, concrete description of any examples that might exist (preferably, the smallest such), or a reference to a proof that there is no such example.
Best Answer
Let $G$ and $H$ be the two groups of your previous example. Then $G \times G$, $G \times H$ and $H \times H$ give an example.