Are the numbers of elements of two distinct prime orders not equal in finite groups?

Question

Are the numbers of elements of two distinct prime orders not equal in finite groups?

Answer 1

The answer is "no":

One can construct a counterexample of the form $G\times H$. Say $p$ and $q$ are distinct primes, and for some integer $m$ one can find two groups $G, H$ with with $q\nmid|G|$ and $m$ elements of order $p$ in $G$, and $p\nmid|H|$ and $m$ elements of order $q$ in $H$. Then $G\times H$ has $m$ elements of order $p$, and $m$ of order $q$.

Now take $p=5$ and $q=3$. Then $G=C_{11}\rtimes C_5$ has $44$ elements of order $5$ (every element not in $C_{11}$), and $H=C_{21}\rtimes C_3$ has $44$ elements of order $3$ ($2$ in $C_{21}$ plus all $42$ outside it). So $G\times H$ has 44 elements of order $3$ and of order $5$.

To check this in Magma:

P:=DirectProduct(SmallGroup(55,1),SmallGroup(63,3));
assert #[g: g in P | Order(g) eq 5] eq 44;
assert #[g: g in P | Order(g) eq 3] eq 44;

Answer 2

Consider $C_6$. It has 1 element of order 1, 1 element of order 2, 2 elements of order 3, and 2 elements of order 6.

Notice here that the numbers of elements of order 2 and of order 3 differ.