Order of a group with given subgroups

Question

In particular i have the groups $\mathbb{Z}_p^n$ and $\mathbb{Z}_{p^n}$ and i want to prove that if G is a group that contain both groups as subgroups, then $|G|$ is divisible by $p^{2n-1}$. I don't know if in general there is way to know the minimum size of a group with 2 given subgroups, but i couldn't find anything. If G is abelian, then in this case the "minimum" group containing both would be $\mathbb{Z}_p^{n-1}\times\mathbb{Z}_{p^{n}}$, but if G is not abelian i am not sure.

Answer 1

Claim. If $G$ is a finite group that contains subgroups $H\cong \mathbb Z_p^m$ and $K\cong Z_{p^n}$, then $|G|$ is a multiple of $p^{m+n-1}$.

Justification:

Let $P$ be a Sylow $p$-subgroup of $G$. Since $H$ and $K$ are $p$-groups and all $p$ subgroups of $G$ are conjugate to subgroups of $P$, there exist $H', K'\leq P$ that are conjugate to $H$ and $K$. Thus $P$ also contains subgroups isomorphic to $Z_p^m$ and $Z_{p^n}$. If we prove that $p^{m+n-1}$ divides $|P|$ we will also get that $p^{m+n-1}$ divides $|P|[G:P] = |G|$.

Thus it suffices to consider only the case where $G=P$ is a $p$-group.

Choose and fix a group $P$ with subgroups $H$ and $K$ as above. Then

$$|P| \geq |HK| = \frac{|H| |K|}{|H\cap K|}\geq \frac{p^m\cdot p^n}{p} = p^{m+n-1}.$$

Since $|P|$ is a power of $p$ that is greater or equal to $p^{m+n-1}$, $|P|$ is divisible by $p^{m+n-1}$.

(The calculation depends on the fact that $|H\cap K|=1$ or $p$, which follows from the fact that $H\cap K$ is cyclic of exponent $p$.)