In one of the simplest ways of proving that two groups aren't isomorphic, we can compare the orders of each element of both groups. If we find an order $\ell$ such that one of the groups has more elements of this order than the other group, we're done.

I am interested in finite groups where this fails --- ones that aren't isomorphic, but the counts for each order are the same:

$$\DeclareMathOperator\ord{ord} G \not\cong H \quad \land \quad \forall\ell.\ |\{g \in G \mid \ord g = \ell\}| = |\{h \in H \mid \ord h = \ell\}|$$

With some SageMath code, I have computed that the smallest groups where this happens are of order 16. However, I am more interested in how one could methodically find such a counterexample, apart from just trying out groups you can think of until something sticks. How would you find such a pair of groups without a computer?

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    $\begingroup$ I'm guessing that you are secretly thinking only about finite groups? Otherwise, $\mathbb Z$ and $\mathbb Z^2$ are a counterexample: each have $1$ element of order $0$; no elements of any finite order other than $0$; and a countable infinity of elements of infinite order. But $\mathbb Z$ and $\mathbb Z^2$ are not isomorphic. $\endgroup$
    – Lee Mosher
    Dec 29, 2020 at 0:06
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    $\begingroup$ You can construct nonabelian groups of order $p^n$, $n\geq 3$, and exponent $p$ for every odd prime $p$, and then an abelian group of order $p^n$ and exponent $p$. In both cases, the numbers of elements of order $k$ are $0$ for $k\neq 0,p$; $1$ for $k=0$; and $p^n-1$ for $k=p$. $\endgroup$ Dec 29, 2020 at 0:07
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    $\begingroup$ @ArturoMagidin Thanks. It is the first time I encounter the exponent of a group, but I can see that such groups would indeed work. I could construct the abelian group with $Z_p^n$, but I don't see how I would do the non-abelian one. Care to elaborate? $\endgroup$
    – Maya
    Dec 29, 2020 at 0:16
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    $\begingroup$ @NieDzejkob: Any group that classifies groups of small order will construct the nonabelian group of order $p^3$ and exponent $p$; it is also known as the "Heisenberg group" or "Heisenberg group mod $p$". Also the extra-special $p$-groups have the property, and they exist for all odd $n\geq 3$. And once you have one such nonabelian $p$-group $A$, then $A\times\mathbb{Z}_p^k$ and a suitable power of $\mathbb{Z}_p$ will give you yet another example. $\endgroup$ Dec 29, 2020 at 0:17
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    $\begingroup$ (In my initial comment, replace order $0$ with order $1$; it's too late to edit it) $\endgroup$ Dec 29, 2020 at 0:18

1 Answer 1


Here is one family of such groups.

The nonabelian group of order $p^3$ and exponent $p$ can be realized as the multiplicative group of $3\times 3$ unipotent matrices with coefficients in $\mathbb{Z}/p\mathbb{Z}$, i.e., matrices of the form $$\left(\begin{array}{ccc} 1 & a & c\\ 0 &1 & b\\ 0 & 0 & 1 \end{array}\right),\qquad a,b,c\in \mathbb{Z}/p\mathbb{Z}.$$ In this group, every nontrivial element has order $p$, and so you get the same count as you do for the abelian (an dhence non-isomorphic) group $(\mathbb{Z}/p\mathbb{Z})^3$.

You can also construct further examples, known as extra-special $p$-groups (of exponent $p$), by taking central products of copies of the group above. Let us denote the group above by $\mathbf{E}_3$ ($E$ for "extra-special"; it's a technical term). Note that the center is cyclic of order $3$, given by the matrices with $a=b=0$. Now, we construct $\mathbf{E}_5$ by taking $\mathbf{E}_3\times\mathbf{E}_3/N$, where $N$ is group of elements of the form $(z,z^{-1})$ with $z\in Z(\mathbf{E}_3)$ (this "identifies" the centers of each copy of $\mathbf{E}_3$). This is again a nonabelian group of exponent $p$, with center of order $p$, of order $p^5$.

Assuming we have defined $\mathbf{E}_{2n+1}$, nonabelian group of order $p^{2n+1}$, exponent $p$, and center of order $p$, we construct $\mathbf{E}_{2n+3}$ by identifying generators $w$ and $z$ of $Z(\mathbf{E}_{2n+1})$ and $Z(\mathbb{E}_3)$, and constructing $\mathbf{E}_{2n+1}\times\mathbf{E}_3/N$, where $N$ consists of the elements of the form $(w^k,z^{-k})$. This will be a nonabelian group of order $p^{2n+3}$, exponent $p$, with center of order $p$.

Then $\mathbf{E}_{2k+1}$ and $(\mathbb{Z}/p\mathbb{Z})^{2k+1}$ will have the desired property, for all $k\geq 1$.

Now, say you have (finite) groups $G$ and $H$ as desired (nonisomorphic, but for each $k$ the number of elements of order $k$ in $G$ and $H$ are the same). Let $K$ be any finite group. I claim that $G\times K$ and $H\times K$ also satisfy the desired properties. That they are not isomorphic is not trivial, but nonetheless it is true (e.g., it follows from the Remak-Krull-Schmidt Theorem). Now I claim that for every $r$, the number of elements of order $r$ in $G\times K$ and in $H\times K$ is the same.

An element $(g,k)\in G\times K$ has order $r$ if and only if (i) the order of $g$ divides $r$; (ii) the order of $k$ divides $r$; and (iii) at least one of the orders of $g$ and $k$ is exactly equal to $r$. Thus, letting $G(d)$ be the set of elements of $G$ of order exactly $d$, and similarly for $H(d)$ and $K(d)$, we have that $$(G\times K)(r) = \left(\sum_{d|r}|G(d)||K(r)|\right) + \left(\sum_{d|r}|G(r)||K(d)|\right) - |G(d)||K(d)|.$$ Similarly, $$(H\times K)(r) = \left(\sum_{d|r}|H(d)||K(r)|\right) + \left(\sum_{d|r}|H(r)||K(d)|\right) - |H(d)||K(d)|.$$ Since $|G(m)|=|H(m)|$ for all $m$ by assumption, it follows that $|(G\times K)(r)|=|(H\times K)(r)|$.

An interesting question might be to determine all $n$ for which there exist two non-isomorphic groups $G$ and $H$ of order $n$ for which the count of elements of order $k$ is equal for all $k$. From the above, we see that if $n$ divisible by the cube of an odd prime, or by $16$, then such groups exist. We also know that if $n$ is cubefree and a nilpotent number, then every group of order $n$ is abelian, so no such examples exist. This reduces the problem to $n$ that is cubefree but not a nilpotent number, and to numbers of the form $2^km$ where $m$ is odd and cubefree, $k=3$. I expect counterexamples exist in that case as well.


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