If $G$ is a group of order $p^k$, with $p$ prime, then a subgroup $H$ of order $p^{k-1}$ is normal.
I'm trying to prove the existence of only one subgroup of order $p^{k-1}$.
Let $n_p$ be the number of $p$-Sylow subgroups. We know that $n_p = 1 + pq$ and $1 + pq$ divides $p^k$. So, $n_p = 1$, which implies that we have only one $p$-Sylow subgroup.
My doubt is: did I misunderstand what $n_p$ is? Because the Sylow Theorem states that $G$ has a subgroup of order $p^a$ if $p^a$ divides $|G|$, so there are subgroups of order $p^a$ with $a < k$, and this would already imply that $n_p > 1$.
I'm really confused about this $n_p$.
I know it's a very basic doubt; I'm just starting to study Sylow's theorems now. In all the books I've read, I understand the examples but when it comes to this case, these things seem to contradict the theorem.
Thanks.
Best Answer
It is false. Consider $G=C_2\times C_2$, where $C_2$ denotes the cyclic group of order $2$. Then $|G|=2^2$ but $C_2\times\{1\}$ and $\{1\}\times C_2$ are distinct subgroups of $G$ of order $2$.