# Proof of the Index Property for a Proper Subgroup in Finite $p$-Groups

Based on this question: proper subgroups of finite p-groups are properly contained in the normalizer.

Let $$G$$ be a finite $$p$$-group and let $$H$$ be a proper subgroup. Then there exists a subgroup $$H'$$ such that $$H\lneq H'\leq G$$ and $$H\triangleleft H'$$.

It's known that this $$H'$$ is the normalizer $$N(H)$$ of $$H$$.

Question: Let $$G$$ be a $$p$$-group. I would like to show that there exists a subgroup $$K$$ of $$G$$ containing $$H$$ so that $$H$$ is normal of $$K$$ with index $$p$$ in $$K$$.

My proof:

Consider the action of the group $$G$$ on the set $$S$$ of all right cosets by left translation: $$G\times S\to S$$ by $$g\cdot(Hx)=gHx$$.

This action partitions the right cosets into: $$|S|=|N(H):H|+\sum_i[H:H_i]$$ where $$i$$ represents the conjugacy classes with more than one element. Given that $$|G|$$ is a power of $$p$$, it follows that $$|H|$$ is also a power of $$p$$, and $$|H|\neq |H_i|$$. Therefore, $$[H:H_i]$$ is divisible by $$p$$. Furthermore, since $$[G:H]=|S|$$ is divisible by $$p$$, we can deduce that $$p$$ divides $$|N(H):H|$$. Consequently, $$|N(H):H|=p^k$$ for some $$k$$.

But we cannot choose such subgroup $$K=N(H)$$, the index of $$H$$ in $$N(H)$$ could not be $$p$$ but some power of $$p$$.

How to choose a subgroup $$K\subset N(H)$$ so that $$[K:H]=p$$?

We have the following Lemma in Normal subgroup of prime index:

Let $$p$$ the smallest prime dividing the order of $$G$$. If $$H$$ is a subgroup of $$G$$ with index $$p$$ then $$H$$ is normal.

If we can find such $$K$$ so that $$[K:H]=p$$, then $$H$$ is normal subgroup of $$K$$.

• $H$ is normal in its normalizer. The quotient is nontrivial $p$ group, so it has a nontrivial central element of order $p$. Lift back to $G$. Oct 12, 2023 at 17:55
• No, what you write is complete nonsense. What map from $N(H)/H$ to $G$? There is no such map. You look at the standard quotient map $N(H)\to N(H)/H$ and use the Isomorphism Theorems. These are basic facts. Oct 12, 2023 at 18:32

Let $$H$$ be a proper subgroup of $$G$$. We know that $$N_G(H)$$ properly contains $$H$$, so $$N_G(H)/H$$ is a nontrivial $$p$$-group. Thus, it has a subgroup of order $$p$$, which is of the form $$K/H$$ for some $$K$$, $$H\lt K\lt N_G(H)$$. By the Isomorphism Theorems, $$[K:H]=[K/H:H/H] = p$$. In addition, since $$H/H\triangleleft K/H$$, then $$H\triangleleft K$$ (though that also follows because it is of index $$p$$ in $$K$$).
• Thank you! I just found one Lemma could be another proof. Lemma: Let $G$ be a group of order $p^n$. Then for each $0\le r\le n$, there exists a subgroup of order $p^r$. Assume $|G|=p^n$ for $n\ge 1$ and $|H|=p^k$ for $1\le k<n$. So by this Lemma, there exists a subgroup $K$ of $G$ with order $p^{k+1}$. By another Lemma in my question, since the index $[K:H]=|K|/|H|=p$, then $H$ is a normal subgroup of $K$. It looks good? Oct 12, 2023 at 19:43
• @H.YDuan Looks wrong. $K$ has order $p^{k+1}$. But you have absolutely no way to tell if it contains $H$, so that $[K:H]$ is nonsense. Oct 12, 2023 at 19:46