69
$\begingroup$

Suppose $G$ is a finite group and I know for every $k \leq |G|$ that exactly $n_k$ elements in $G$ have order $k$. Do I know what the group is? Is there a counterexample where two groups $G$ and $H$ have the same number of elements for each order, but $G$ is not isomorphic to $H$? I suspect that there is, but I haven't thought of one.

$\endgroup$
9
  • 4
    $\begingroup$ Two finite abelian groups with the same number of elements of each order are isomorphic, but in general the answer is no. $\endgroup$
    – KCd
    May 24, 2015 at 15:57
  • 1
    $\begingroup$ It links to an answer for finite groups in general groupprops.subwiki.org/wiki/…. It appears that the answer is no. $\endgroup$
    – user208649
    May 24, 2015 at 15:58
  • 2
    $\begingroup$ See also math.stackexchange.com/questions/693163/… - this has been discussed before. $\endgroup$ May 24, 2015 at 16:39
  • 6
    $\begingroup$ It is generally believed that there is no list of properties of finite groups (such as orders of elements, orders of terms of central series, character tables, etc) which is sufficient to characterize groups up to isomorphism. $\endgroup$
    – Derek Holt
    May 24, 2015 at 17:38
  • 3
    $\begingroup$ @DerekHolt, look at arxiv.org/pdf/math/9210219.pdf. $\endgroup$
    – KCd
    May 25, 2015 at 0:23

3 Answers 3

80
$\begingroup$

Take $G=\mathbb{Z}/4\times \mathbb{Z}/4$, and $H=Q_8\times \mathbb{Z}/2$ of order $16$, where $Q_8$ denotes the quaternion group. Both groups have exactly $1$ element of order $1$, $3$ elements of order $2$ and $12$ elements of order $4$.

Edit: I understood the question as follows: Is there a counterexample where two groups $G$ and $H$ have the same number of elements for each order, but $G$ is not isomorphic to $H$ ? Is it really required, that all elements different from $1$ in $G$ have the same order ?

$\endgroup$
2
  • 1
    $\begingroup$ You understood correctly. I wanted to know if the orders of the group elements fully characterise the group. I think you may have provided the smallest counterexample so far. $\endgroup$
    – Stanley
    May 24, 2015 at 18:07
  • 2
    $\begingroup$ The smallest counterexamples are of order 16, hence this is smallest possible (interestingly enough, there are 2 trios of 3 different groups that share same order statistics among the 14 groups of order 16, as well as another pair that share order statistics). $\endgroup$ Apr 30, 2017 at 21:24
35
$\begingroup$

Here is an example with two groups of order $27$. Consider the group $G$ which is elementary abelian (all elements $x \in G$ satisfy $x^{3} = 1$), of order $27$. And then $H$ the non-abelian group of order $27$ and exponent $3$ (once more, $x^{3} = 1$ for all $x \in H$). Concretely, $$ \left\{\, \begin{bmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 & 1 \end{bmatrix} : a, b, c \in F \,\right\}, $$ where $F = \mathbb{Z}/ 3 \mathbb{Z}$ is the field with three elements.

$\endgroup$
0
23
$\begingroup$

Let $p$ be an odd prime. Let $G$ be the non-abelian group of matrices of the form $$\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}\in\operatorname{GL}(3,\mathbb F_p).$$ Then $|G|=p^3$ and each element $g\ne 1$ has order $p$; this follows from the fact that $g-1$ is nilpotent, hence $(g-1)^p=(g-1)^3=0$ and finally $g^p=(1+(g-1))^p=1^p+(g-q)^p=1$.

Likewise the abelian group $H=(\mathbb Z/p\mathbb Z)^3$ is also of exponent $p$, i.e., all elements $\ne1$ have order $p$. As $H$ is abelian and $G$ is not, certainly $G\not\cong H$.

$\endgroup$
1
  • 1
    $\begingroup$ Hagen, I changed $H=(\mathbb Z/p\mathbb Z)^2$ into $H=(\mathbb Z/p\mathbb Z)^3$ $\endgroup$ May 24, 2015 at 16:26

Not the answer you're looking for? Browse other questions tagged .