Example of a group of order 80 with more than one Sylow 5-subgroup?

Question

I have noticed that it can be 1 or 16 Sylow-5 subgroups. I have provided an example of group $G$ where there is 1 Sylow-5 subgroup. It is $\mathbb Z_{80}$. But I don't know what to do if there are 16 Sylow-5 subgroups. Is there an example of such a group?

Answer 1

Each subgroup of order $5$ has $4$ elements of order $5$, and any two such subgroups have a trivial intersection. Hence, in this case, there are $4\times 16=64$ elements of order $5$.

Answer 2

Let $H = \mathbb Z_5$ and $K = \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2 \times \mathbb Z_2$.

Since $|\operatorname{Aut}(K)| = (2^4 - 1)(2^4-2)(2^4-2^2)(2^4-2^3)$ is divisible by $5$, there exists a nontrivial homomorphism $\phi : H \to \operatorname{Aut}(K)$, e.g. map a generator of $H$ to a generator of any subgroup of $\operatorname{Aut}(K)$ of order $5$.

Hence there exists a semidirect product $G = H \ltimes_{\phi} K$ where $K \lhd G$. Since $\phi$ is nontrivial, this is not a direct product, hence $H$ is not normal, hence there must be more than one Sylow $5$-subgroup, so by Sylow counting, there must be $16$ such subgroups.