# Let $G$ be a $p$-group and let $H$ be a proper subgroup of $G$. Show that $H$ is a proper subgroup of $N_G(H)$

Let $$G$$ be a $$p$$-group and let $$H < G$$. Show that $$H < N_G(H)$$.

If $$H \trianglelefteq G$$, then it should be clear that $$H. So we suppose, $$H$$ isn't normal. So we let $$H$$ act on the set

$$S = \{gHg^{-1}: g \in G, gHg^{-1} \neq H\}$$

via conjugation. If I can show that $$p$$ doesn't divide $$|S|$$, then I can use the fixed point theorem to complete the problem. Any advice on how to show that $$p$$ doesn't divide $$|S|$$?

• $p$ doesn't divide $|S|$ since $p$ divides $|S|+1$ Sep 24, 2022 at 17:44
• @kabenyuk I appreciate your input, but this result isn't intuitively clear to me. Can you expand upon this idea? Sep 24, 2022 at 17:46
• The total number of conjugates of $H$ is the index of its normalizer, which is a power of $p$ because $G$ is a $p$-group. Here you are assuming the normalizer is not all of $G$, so the index is a nontrivial power of $p$. Your set $S$ contains all but one of the conjugates of $H$, so it is one less than a nontrivial power of $p$, hence mot divisible by $p$. Sep 24, 2022 at 19:17
• This is not right. For example a Tarski Monster is a $p$-group, but its subgroups of order $p$ are self-normalizing. Sep 25, 2022 at 7:59
• @DavidC.Huang I do not understand what you are asking, or why you are asking it. How do I arrive at what? And why does it matter if the order of $G$ is a power of $2$? Are you under the impression that $2$ is not a prime? You claim in your post that all you need to do is show that the cardinality of $S$ is not a multiple of $p$. So why is it a problem if you get that the cardinality of $S$ is $1$? Is $1$ a multiple of $p$? Again, what exactly is the problem? Sep 26, 2022 at 1:07

The result is only true if $$G$$ is finite, which you are assuming but never state explicitly. As Derek Holt points out in comment, that is bad practice: if you are considering only finite groups, you should say so (or, in this site, at least tag it as [finite-groups]).

You state that you only need to show that the cardinality of $$S$$ is not a multiple of $$p$$ when $$H$$ is not a normal subgroup of $$G$$, so that is what the argument below will show.

Since you talk about the fixed point theorem, presumably you know about group actions. Let $$G$$ act on its subgroups by conjugation. Let $$H$$ be a proper subgroup of $$G$$.

Let $$T=\{gHg^{-1}\mid g\in G\}$$ be the set of all conjugates of $$H$$. Your set $$S$$ is just $$T\setminus\{H\}$$. The set $$T$$ is the orbit of $$H$$ under the action.

By the Orbit-Stabilizer Theorem, the cardinality of $$T$$ is equal to the index of the stabilizer of $$H$$ under the action. The stabilizer is $$N_G(H) = \{g\in G\mid gHg^{-1}=H\}.$$ Thus, the cardinality of $$T$$ is $$[G:N_G(H)]$$, and the cardinality of $$S$$ is $$|T|-1 = [G:N_G(H)] - 1$$.

Because $$G$$ is a $$p$$-group, every subgroup has order a power of $$p$$ and index a power of $$p$$. So the cardinality of $$T$$ is a power of $$p$$. Say $$p^i$$.

That means that the cardinality of $$S$$ is one less than a power of $$p$$. If $$i\gt 0$$, then $$|T|\equiv 0\pmod{p}$$, so $$|S|\equiv -1\pmod{p}$$, Since $$-1$$ is never a multiple of a prime, then it follows that $$|S|$$ is not a multiple of $$p$$ and we are done.

If $$i=0$$, so $$|T|=1$$, then that means that $$gHg^{-1}=H$$ for all $$g\in G$$, so $$H\triangleleft G$$. Since we are assuming that $$H$$ is a proper subgroup of $$G$$, then this gives $$H\subsetneq N_G(H)=G$$, and we are done without having to worry about $$S$$ at all.

Hint

1. Finite $$p$$- groups are nilpotent.
2. For any proper subgroup $$H$$ of a nilpotent group $$G$$, we have $$N_G(H)\supsetneq H$$.

Both these results are well-known.

The result is false without requiring the $$p$$- group to be finite, as Tarski monsters show.

• The notion of nilpotent groups is a bit advanced for where I am. I advise submitting an answer that uses more elementary methods Sep 25, 2022 at 3:53
• Idk any other way to prove it off the top of my head. Sep 25, 2022 at 4:09
• I think the easiest proof uses the fact that $Z(G) \ne 1$ for a finite nontrivial $p$-group, together with induction on $|G|$. Sep 25, 2022 at 12:31