Understanding what the Sylow theorems say about $p$-groups

Question

I have a simple question.

If we consider a group $G$ with order $p^k$ for a prime $p$. For example $125=5^3$.

What we can obtain from sylows theorem? (I already understood it for the other cases, where we have something like $p^km$).

Thanks.

Answer 1

The classical statements of the Sylow theorems don't really tell you anything about about a group $G$ with $|G|=p^k$.

The first Sylow theorem says that $G$ contains a subgroup of order $|G|$, which is clear anyhow.

Since $G$ is the only $p$-Sylow subgroup, second theorem simply tells you that $G$ is conjugate to itself as a subgroup, which is clear (consider conjugation by $e$).

Since there is only $1$ Sylow $p$-subgroup and $1 \equiv 1(\mod p)$, the last theorem is equally unhelpful.

Sometimes a stronger result is proved instead of the first Sylow theorem: given any finite group $G$ and any prime $p$, if $p^m$ divides $|G|$ then $G$ contains a subgroup of order $p^m$. This does tell you something about $p$-groups: they contain subgroups of every order allowed by Lagrange's theorem.

Answer 2

I think Morgan has answered your question completely. So even though Sylow Theorems don't give new information about $p$-groups, there is a lemma in the proof of Sylow Theorems that you might find interesting:

One way to prove Sylow Theorems (at least the first Sylow theorem) is to use the fact that $[G:H]\equiv [N_{G}(H):H] \textrm{ (mod } p)$ whenever $H$ is a $p$-subgroup of $G$. To prove this lemma, consider the canonical action of $H$ on the set of left-cosets $G/H$, and look at the fixed points.

When $G=p^{m}$ itself a $p$-group, you can apply this lemma to conclude that $G$ has a normal subgroup of order $p^{m-1}$. The congruence $[G:H]\equiv [N_{G}(H):H] \textrm{ (mod } p)$ in general proves that normalizer of a proper subgroup $H$ of $p$-group is strictly larger than $H$. Hence, $p$-groups are nilpotent.