Number of prime order elements of two non-isomorphic groups of same order

Question

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 ?

Answer 1

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 $\mathfrak A_4$.

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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.

Answer 2

There are no examples up to order $700$, by a quick search in Magma.

There are also no examples with only elements of prime order (or $1$). This follows from a result from a paper of Deaconescu et al, which classifies such finite groups. They are one of the following:

1. A $p$-group of exponent $p$;
2. The alternating group $A_5$;
3. A Frobenius group $P\rtimes C_q$ for some $p$-group $P$ of exponent $p$ and some prime $q\neq p$.

In all three cases the number of elements of each order can be deduced purely from the group order, and of course there is no overlap of orders between the three categories.

The general problem seems difficult.