If $G$ is a group, $|G|=8p^n$, $n>1$, $p>7$, $p$-Sylow subgroup is cyclic, prove there is a subgroup with every order $m$ that divides $|G|$

Question

If $G$ is a group, $|G|=8p^n$, for $n\ge 1$ and $p>7$, $p$-Sylow subgroup is cyclic.

Prove there is a subgroup with every order of $m$ that divides $|G|$.

I thought to prove that $G$ is Abelian and then it exists.

Here $p^n$ is a cyclic $p$-Sylow group so its Abelian.

My problem is to prove that $2$-Sylow group is Abelian.

Answer 1

Hint: prove that the Sylow $p$-subgroup is normal and use the fact that a cyclic group $P$ has exactly one subgroup for each order dividing $|P|$. And, of course every group of order $8$ has subgroups of order $1,2,4$ and $8$.

Answer 2

All factors of $8p^n$ are of the form $2^kp^m$, $0 \leq k \leq 3$, $0 \leq m \leq n$. As known, a group of order $p^n$ has subgroup of order $p^m$, $0 \leq m \leq n$. So that handles the factors of form $2^k$, and $p^m$. Now, let $n_p$ denote the number Sylow-$p$ subgroups. Then, $n_p \mid 8$ and $n_p \equiv 1 \pmod{p}$. Since $p>7$, $p>8$. Thus, $n_p=1$, making it a normal subgroup. Since a cyclic group contains a unique subgroup of order dividing the order of the cyclic group, the subgroup of order $p^m$, $0 \leq m \leq n$ is unique in $G$, the original group, making it a normal subgroup as well. Let $P_m \leq G$ denote the subgroup of order $p^m$. So, $P_mH \leq G$ if $H \leq G$ because $P_m \triangleleft G$. To obtain a subgroup of order $2^kp^m$, let $H \leq G$ such that $|H|=2^k$ and $P_m \leq G$ such that $|P_m|=p^m$. Then $P_mH \leq G$ such that $|P_mH|=2^kp^m$ because $P_m \cap H \leq P_m, H$ so, $|P_m \cap H|=1 \implies |P_mH|=\frac{|P_m||H|}{|P_m \cap H|}=2^kp^m$.