Number of $p$-Sylow subgroups

Question

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.

What you are trying to prove is false, except for the case of a cyclic group. — Arturo Magidin, Dec 19 '23 at 17:04
The Sylow $p$ subgroup of a group of order $p^k$ is the whole group. — Arturo Magidin, Dec 19 '23 at 17:05
Only in the case where $p$ is prime. "Sylow $p$ subgroup" only makes sense when $p$ is prime. — Arturo Magidin, Dec 19 '23 at 17:07
You do know what a Sylow $p$ subgroup is? For finite groups, if $p^a$ is the largest power of $p$ that divides the order of $G$, then a subgroup of $G$ is a Sylow $p$ subgroup if and only if it is of order $p^a$. — Arturo Magidin, Dec 19 '23 at 17:10
Btw, the number of those subgroups is congruent to $1$ modulo $p$ (and the same holds for the subgroups of order any other power of $p$). — citadel, Dec 19 '23 at 17:11
@ArturoMagidin I questioned my understanding because of this issue I mentioned; it appears on my list as being true. — ends7, Dec 19 '23 at 17:16
What "list"? You do understand we are not allowed to read your mind without a warrant? — Arturo Magidin, Dec 19 '23 at 18:01
My list of problem. It's unrelated to my question. if you really want to know, come back with a warrant. — ends7, Dec 19 '23 at 18:06

score 3 · Answer 1 · answered Dec 19 '23 at 17:24

3

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 $2$.

answered Dec 19 '23 at 17:24

Deif

1,002

Can you give me some hint, to prove the subgroup of order $p^{k-1}$ is normal? – ends7 Dec 19 '23 at 18:55
1

Yes. Note that by Cayley's theorem $|G:\text{Core}_G(H)|$ divides $|G:H|!$ . – Deif Dec 19 '23 at 19:21
1

It may worth noting that such a group must have a third (at least) subgroup of order $2$. – citadel Dec 19 '23 at 20:09

Arturo Magidin · Accepted Answer · 2023-12-19T21:14:30.253

The only groups of order $p^n$ that have a unique subgroup of order $p^{n-1}$ are the cyclic groups of order $p^n$.

This follows because:

Proposition. A group is a union of proper subgroups if and only if it is not cyclic. In particular, a group with a unique maximum proper subgroup is cyclic.

Proof. If $G$ is not cyclic, then for all $x\in G$ we have that $\langle x\rangle$ is a proper subgroup of $G$. Since $$G=\cup_{x\in G}\{x\}\subseteq \cup_{x\in G}\langle x\rangle \subseteq G$$ it follows that $G$ is the union of the proper subgroups $\{\langle x\rangle\mid x\in G\}$.

Conversely, if $G$ is cyclic, $G=\langle x\rangle$, and $G$ is a union of subgroups $G=\cup_{i\in I}H_i$, then $x\in H_i$ for some $i\in I$, and therefore, $G=\langle x\rangle\subseteq H_i\subseteq G$, hence $H_i=G$. So if $G$ is a union of subgroups, at least one of them is not a proper subgroup. $\Box$

So now suppose $G$ has order $p^n$. Subgroups of order $p^{n-1}$ are maximal, and every proper subgroup is contained in a subgroup of order $p^{n-1}$ (this is often proven as a consequence or part of Sylow's First Theorem). So if $G$ has a unique subgroup of order $p^{n-1}$, then the union of all its proper subgroups is the subgroup of order $p^{n-1}$, so $G$ is cyclic. Conversely, if $G$ is cyclic then it has a unique subgroup of any given order, in particular of order $p^{n-1}$.

Now, it is true that if $p^k$ divides the order of $G$, then the number of subgroups of order $p^k$ is congruent to $1$ modulo $p$: this was proven by Frobenius as a generalization of Sylow's Third Theorem, and you can find a proof here. However, in general you do not know that it divides the order $G$ unless $p^k$ is the largest power of $p$ that divides the order of $G$ (in which case, it is part of Sylow's Third Theorem). In fact, it need not divide the order, as the Klein $4$-group (or any group of the form $C_p\times C_p$) shows: the Klein $4$-group has $3$ subgroups of order $2$, and while $3\equiv 1\pmod{2}$, $3$ does not divide $4$. Likewise, the group $C_p\times C_p$ has $p+1$ subgroups of order $p$, which does not divide $p^2$.

The fact that the subgroups of index $p$ are normal follows in any number of ways: it is well known that the normalizer of a proper subgroup of a $p$-group always properly contains the subgroup. Or you can use the also very well known group theory result that if the index of $H$ is the smallest prime that divides the order of $G$, then $H$ is normal.

Number of $p$-Sylow subgroups

2 Answers2