Let $G$ be a group of order $p^n$, where $p$ is a prime and $n \in \mathbb{N}$. Show that any subgroup of $G$ of order $p^{n−1}$ is a normal subgroup.

Question

I found a solution for this question which solved it using induction, I would like to ask for an alternate solution for this using Sylow's theorems.

My approach

Using Sylow's theorem we can say that there exists a subgroup of order $p^{n-1}$, say $H$. Also, if other subgroup exist they will be conjugates of $H$. Let's say $K$, with $|K|=p^{n-1}$ and $K\neq H$ exists.

$K = gHg^{-1}, \, H\cap K =e$

$$ |HK|= \frac{|H||K|}{|H\cap K|} = p^{2n-2} > p^n \,\,\forall \,\, n \in \mathbb{N} $$ $\therefore$ H is unique, $H=gHg^{-1}$, H is normal.

Is this solution correct?

Note that the claim holds more generally for subgroups of index equal to the smallest prime dividing the group order — Hagen von Eitzen, Nov 08 '22 at 07:01
You don‘t want to use Sylow here. The Sylow $p$ group of $G$ is $G$ itself — Hagen von Eitzen, Nov 08 '22 at 07:04
Also, Sylow p-subgroups need not intersect trivially. So, $H\cap K=e$ isn't always true. — Umesh Shankar, Nov 08 '22 at 07:05
So …. what else? Hint: Given $H$ of index $p$, look at the action of $G$ on the $p$-element set of cosets of $H$. — Hagen von Eitzen, Nov 08 '22 at 07:09
@HagenvonEitzen got the mistake, I'll try using the action of G — Atom Bomb, Nov 08 '22 at 07:14
See also here https://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index — Nicky Hekster, Nov 08 '22 at 08:15

score 1 · Accepted Answer · answered Nov 08 '22 at 08:18

1

Sketch: let $G$ be a $p$-group, apply induction on $|G|$: if $|G|=p$, then it is clear. Let $H \lt G$ and $|G:H|=p$. Then either $G=HZ(G)$ and hence $H \unlhd G$, or $Z(G) \subseteq H$. In the latter case apply induction, using the fact that $Z(G) \gt 1$.

answered Nov 08 '22 at 08:18

Nicky Hekster

49,281

A nice simpler argument for this particular case (in comparison to the more general one you linked to)! – Jyrki Lahtonen Nov 09 '22 at 07:07
@JyrkiLahtonen Thanks, and yes the link shows (as you know) how to generalise to $|G:H|=p$, smallest prime dividing $|G|$ then $H \lhd G$. – Nicky Hekster Nov 09 '22 at 08:53

Let $G$ be a group of order $p^n$, where $p$ is a prime and $n \in \mathbb{N}$. Show that any subgroup of $G$ of order $p^{n−1}$ is a normal subgroup.

1 Answers1