Showing there's no $2$-group with specific number of elements of each order

Question

Are there any theorems that I can use to show that there are no $2$-groups with a specific number of elements of each order aside from the fact that the number of elements of order $2$ is always odd, Frobenius' theorem and things like if you have an element of order $2^n$ then you have an element of order $2^{n-1}$ etc. ?

For a concrete example that I couldn't rule out using the things mentioned before (but I know for a fact that there's no such group thanks to groupprops) is a group of order $32$ with $17$ elements of order $2$, $10$ elements of order $4$ and $4$ elements of order $8$. If you know of anything else I could use that doesn't work on the specific example let me know anyway because there are a lot more possibilities I haven't been able to rule out for order $64$ that maybe could use whatever you have in mind!

Answer 1

You are basically asking about the number of cyclic subgroups $c_n(G)$ of each possible order when $G$ is a $2$-group. As far as I know, there isn't too much you can say. Here's something I found in Berkovich's book "Groups of prime power order, Vol. I". On p. $32$ he writes:

As we saw (see Exercises $11$–$13$), $c_1(G) \equiv 1 \pmod 4$ if $G$ is either cyclic or a $2$-group of maximal class. The following theorem shows that all other $2$-groups $G$ satisfy $c_1(G) \equiv 3 \pmod 4$.

Theorem $\mathbf{1.17}$. Suppose that a $2$-group $G$ is neither cyclic nor of maximal class.

1. ([Isa$1$, Theorem $4.9$] and [Ber$1$, $§5$].) $c_1(G) \equiv 3 \pmod 4$.
2. ([Ber$1$, $§5$].) If $n > 1$, then $c_n(G)$ is even.

So your $2$-group, if it is to have $c_1(G)=17$ or $c_2(G)=5$, must be of maximal class, hence one in a very short and specific list.