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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 $K$.

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  • $\begingroup$ $H$ is normal in its normalizer. The quotient is nontrivial $p$ group, so it has a nontrivial central element of order $p$. Lift back to $G$. $\endgroup$ Oct 12, 2023 at 17:55
  • $\begingroup$ 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. $\endgroup$ Oct 12, 2023 at 18:32

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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 $K$).

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  • $\begingroup$ 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? $\endgroup$
    – H.Y Duan
    Oct 12, 2023 at 19:43
  • $\begingroup$ @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. $\endgroup$ Oct 12, 2023 at 19:46

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