For a given finite group $G$, let $\pi (G)$ denote the set of all prime divisors of $\vert G\vert$.

Let $G_1$, $G_2$ be non-isomorphic finite groups of same order. Then obviously $\pi(G_1)=\pi(G_2)$.

Suppose $G_1$, $G_2$ have the same number of elements of each non-prime order.

Then is it true that $G_1$, $G_2$ have the same number of elements of order $p$, for each $p \in \pi(G_1)$ ?

I guess that it is true; but I couldn't prove it. Can any one give an example for which this is not true ?

  • $\begingroup$ I have spent hours looking for a counter-example since I don't get why such a result would be true. Does anyone know where can be find tables where groups are sorted by the order of their elements? I am looking foward to see someone provide a proof/disproof of this result ! $\endgroup$ – E. Joseph Jan 22 '17 at 14:54
  • $\begingroup$ @E.Joseph I misread the problem. $\endgroup$ – Stella Biderman Jan 22 '17 at 18:01
  • $\begingroup$ I suspect that this is true, and have an idea for a proof: Given a group $G$ and an element of prime order, $g$, prove that replacing $g$ with an element of larger prime order necessitates increasing the size of $G$ by an amount boundable in terms of the old and new order of $g$. A tight enough bound can be used to show that increasing the order of one element and decreasing the order of another can't ever cancel out in terms of its effect on the group size. $\endgroup$ – Stella Biderman Jan 22 '17 at 18:06
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    $\begingroup$ @E. Joseph Such a list, for groups of order up to 512 is available here: drive.google.com/file/d/0B3X5C_7tdfR2VDhTRW9xRndVMkU/view and for the non-solvable groups of those orders is available here: drive.google.com/file/d/0B3X5C_7tdfR2Q0xwZ2FSZUUwcHc/view , which were provided by Alexander Gruber in this mathoverflow.net/questions/107298/… post. In my search in this list, I didn't get such pair of groups ! $\endgroup$ – RKR Jan 22 '17 at 18:26
  • $\begingroup$ @RKR Thank you for the list, that is exactly what I was looking for ! I also searched a lot in this list, but didn't find such a pair. Some pair came really close to give a counter-example though ! $\endgroup$ – E. Joseph Jan 22 '17 at 18:58

Like I said in my comment, I do not see any reason for this result to hold, so I tried to find a counter-example.

I didn't find an explicit one, but I did find what would not work, so I am sharing it here to prevent someone else to lose time in this direction.

I tried to find two groups where they would not be any elements of order $>3$. Then we would have two non isomorphic groups, with not the same number of elements of order $2$ and $3$ respectively and we would have won.

But such a groupe is of one of the following types (which both works as acceptable groups regarding to our conditions):

  • Type $T$. The group is isomorphic to the set

$$(\mathbb Z/3\mathbb Z)^\Gamma\times \{\pm 1\}$$

with the product:

$$(h,a)\cdot (k,b)=(hk^a,ab)$$

where $(\mathbb Z/3\mathbb Z)^\Gamma$ is the set of maps from a given set $\Gamma$ (of cardinality $n$) to $\mathbb Z/3\mathbb Z$.

*We can note that if $n=1$, then a group of type $T$ is isomorphic to $\mathfrak S_3$.

  • Type $S$. The group is isomorphic to the set

$$\left((\mathbb Z/2\mathbb Z)^2\right)^\Gamma\times \mathbb Z/3\mathbb Z$$

with the product:

$$(h,a)\cdot (k,b)=(h\cdot\alpha^a(k),a+b)$$

where we think of $\mathbb Z/3\mathbb 2$ as $\{0,1,2\}$ under addition modulo $3$, and $\alpha$ is a cyclic permutation of three nonidentity elements of $(\mathbb Z/2\mathbb Z)^2$.

*We can note that if $n=1$, then a group of type $S$ is isomorphic to $\mathfrak A_4$.

We can notice that a group of type $S$ and a group of type $T$ won't have the same order unless maybe if $n=1$, which won't work either.

Conclusion: there is no counter-example with of group with only elements of order $2$ and $3$.

I wish someone find a better result.

Sources : one article, another article and a final article.


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