I know that any abelian group of order $75$ has a $\mathbb Z_{15}$ inside it.I was just wondering if there is a $\mathbb Z_{15}$ inside the non-abelian group of order $75$.
If there is a normal subgroup of order $5$ inside it then I can take direct product with a $\mathbb Z_3$ to get a $\mathbb Z_{15}.$
Also,$Aut(\mathbb Z_5\times \mathbb Z_5)\cong GL_2(\mathbb F_5)$.I was trying to look if there is an order $3$ element inside $GL_2(\mathbb F_5)$,which has an eigenvector inside $ \mathbb Z_5\times \mathbb Z_5$.
Any help would be appreciated. Thanks in advance. If there is really a $\mathbb Z_{15}$ inside it, please give me hints.I want to finish it myself.
This is an ad-hoc answer, that does not require the knowledge of the structure of the (unique, up to isomorphism) non-abelian group of order $75$.
Let $G$ have order $75$. By Sylow's theorems, a Sylow $5$-subgroup $F$ is normal in $G$.
If $F$ is cyclic, since $3$ does not divide $\phi(25) = 20$, we have that $G$ is abelian.
So let $F$ be elementary abelian, so that it has $5 + 1 = 6$ subgroups of order $5$. Suppose $G$ has a subgroup $C$ of order $15$, which is thus cyclic. Then a Sylow $3$-subgroup $T$ acts by conjugacy on the $5$ subgroups of order $5$ of $F$ other than the one in $C$, so that it normalises at least one other, and thus centralises it.
We have obtained that $G$ is abelian if it has a subgroup of order $15$.