This question was taken from Contemporary Abstract Algebra, Joseph A. Gallian (10th edition), Q51 of Chapter 7.
Prove that a group of order $np^m$ where $p$ is prime and $n<p^m$ has at most one subgroup of order $p^m$.
This question is unusually hard to prove, as opposed to other problems in the same book.
Here is my working on a weaker version of the problem.
Claim: A group of order $np$, where $p$ is a prime and $n<p$, has at most one subgroup of order $p$.
Proof: Let $G$ be a group of order $np$, and suppose $H,K$ are subgroups of G with order $p$.
$H\cap K$ is a subgroup of $H$, by Lagrange's Theorem, $|H\cap K|=1$ or $p$.
$|H\cap K|$ can't be 1, else $|HK|=\frac{|H||K|}{|H\cap K|}=p^2>np=|G|$, a contradiction.
Then we must have $|H\cap K|=p$, which implies $H=K$, so there is only one subgroup of order $p$ in $G$.
Note: Sylow Theorems are not allowed to be used.
Any help will be greatly appreciated!
Best Answer
I don't have access to a copy of Gallian, so I cannot check, but the claim is false as stated.
Consider the group $G=S_3\times C_2$ of order $12=3\cdot2^2$, $3<2^2$. We see easily that $G$ has three subgroups of order four.