Is it true that if two groups have same number of elements of order n, then they are isomorphic?

Question

I know that if two groups are isomorphic, they have same number of elements of order n. But will the converse holds? I think it won't as there are many group characteristicss like abelian or cyclic will also matter. But I can't come up with example.

Answer 1

Not in general. The group $\mathbf{C}_3\times\mathbf{C}_3\times\mathbf{C}_3$, where $\mathbf{C}_3$ is the cyclic group of order $3$, has $27$ elements, of which one has order $1$ and 26 have order $3$.

Now consider the group of all $3\times 3$ matrices of the form $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right)$$ with $a,b,c$ taken in $\mathbb{Z}/3\mathbb{Z}$ (this is a “Heisenberg group”). It is not hard to verify that $$\left(\begin{array}{ccc} 1 & a & c\\ 0 & 1 & b\\ 0 & 0 & 1 \end{array}\right)\left(\begin{array}{ccc} 1 & x & z\\ 0 & 1 & y\\ 0 & 0 & 1 \end{array}\right) = \left(\begin{array}{ccc} 1 & a+x & c+z+ay\\ 0 & 1 & b+y\\ 0 & 0 & 1 \end{array}\right)$$ and therefore that every element has order either $1$ or $3$. So it has exactly one element of order $1$, and 26 of order $3$.

But this group is not abelian, so it cannot be isomorphic to $\mathbf{C}_3\times\mathbf{C}_3\times\mathbf{C}_3$.

In fact, there are examples of order $p^n$ for any odd prime $p$ and $n\geq 3$, and examples of order $2^n$ for any $n\geq 4$. With these, you can construct examples for groups of any order that is divisible by the cube of an odd prime or by $16$.