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Prove that any subgroup of order $p^{n-1}$ in a group $G$ of order $p^n$, $p$ a prime number, is normal in $G$.

Remark: I have met the similar problem here but I would like to complete my solution which is different.

Proof: $G$ be a group of order $p^n$ and $H$ it's subgroup of order $p^{n-1} $. Let $S=\{Hx: x\in G\}$ the set of right cosets of $H$ in $G$. Then $o(S)=i_G(H)=p$.

Then it's easy to show that exists $\psi:G\to S_p$ where $\psi$ is homomorphism and $K=\text{Ker} \psi$ is the largest normal subgroup of $G$, contained in $H$.

Let $f=\psi \mid_H$ is the mapping $\psi$ restricted to $H$ then consider $f:H\to S_p$ then $f(H)$ is a subgroup of $S_p$. Thus, $o(f(H))\mid p!$

Also $H/\text{Ker} f \cong f(H) $ but $\text{Ker} f=\text{Ker} \psi \cap H=\text{Ker} \psi=K$ $\Rightarrow$ $\dfrac{o(H)}{o(K)}=o(f(H))$ $\Rightarrow$ $\dfrac{p^{n-1}}{o(K)}=o(f(H))\mid p!$

So we have two cases $o(f(H))=p^{\alpha}$ where $\alpha\in \{0,1\}$

1) If $\alpha=0$ then $o(f(H))=1$ so $o(K)=o(H)$ and $K\subset H$ $\Rightarrow$ $K=H$ where $H$ is normal in $G$ since kernel is always normal subgroup.

2) If $\alpha=1$ then $o(f(H))=p$ and I don't know how to complete this case.

Would be very grateful if anybody can show how to do in the second case.

RFZ
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    This can be viewed as a special case of this: https://math.stackexchange.com/questions/164244/normal-subgroup-of-prime-index – Ben West Feb 15 '18 at 19:38
  • @BenWest, Yes I've already saw this post but some moments of its solution are written quite unclear. Could you clarify it? – RFZ Feb 15 '18 at 19:50
  • What part confuses you? – Ben West Feb 15 '18 at 19:53
  • @BenWest, I was trying to apply it's solution to my problem but as you see I've got two cases. The first one seems good to me however, the second one is not so good. I know that in this case $f(H)$ is cyclic so how to complete? Also is the first case correct? Could we conclude that $o(f(H))=1$? – RFZ Feb 15 '18 at 20:02

2 Answers2

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As you say there is a homomorphism $\psi$ from $G$ to $S_p$ defined by its action on cosets of $H$. The image of such a map is transitive, and the order of the image is a power of $p$. Thus $|\psi(H)|=p$. The image of $\psi$ has order $p$, and is a cyclic group. The kernel of $\psi$ must be $H$, and so $H$ is normal in $G$.

Angina Seng
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  • Sorry, but I do not understand you. Are you claiming that the first case is wrong in my solution? I have never met the notion "transitive". What does it mean? The kernel of $f$ is $K$ not $H$. Could you clarify these questions. I am not able to digest your answer. – RFZ Feb 15 '18 at 19:54
  • Unfortunately I cannot understand why the case $o(f(H))=1$ never occurs? Could you explain it please? – RFZ Feb 15 '18 at 20:05
  • @RFZ I was muddled by your introduction of $f$, which is unnecessary. It's the homomorphism $\psi$ that you need to consider. – Angina Seng Feb 15 '18 at 20:07
  • Anyway, if we consider $\psi$ how it follows that the first case never occurs? – RFZ Feb 15 '18 at 20:09
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    That is explained in the post. The image of an action of a group on the cosets of a subgroup is transitive. Or for a more basic explanation, for $g \in G \setminus P$, $gP \ne P$, and so $\psi(g) \ne 1$. – Derek Holt Feb 15 '18 at 20:55
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Let me clarify a few things.

The map $\Psi$ is defined by $\Psi(g):aH\mapsto gaH$. An element in Ker$\Psi$ has to fix every single coset, in particular $H$ itself: $$g\in \operatorname{Ker}\Psi\Longrightarrow gH=H\Longrightarrow g\in H$$ Therefore $\operatorname{Ker}\Psi\subset H$.

A consequence of this is that $\operatorname{Ker}\Psi$ is not the whole group $G$.

Now look at $G/\operatorname{Ker}\Psi\simeq \operatorname{Im}\Psi$. Since $G$ is a $p$-group, and since $\operatorname{Ker}\Psi\neq G$, we know that $$|\operatorname{Im}\Psi|=p^a$$ for some positive $a$. But we also know that $p^a$ has to divide the order of $\mathfrak{S}_p$, therefore $a=1$.

It follows that $$\dfrac{|G|}{|\!\operatorname{Ker}\Psi|}=p$$ Therefore $$|\!\operatorname{Ker}\Psi|=p^{n-1}=|H|$$

And since $\operatorname{Ker}\Psi\subset H$, we have equality $\operatorname{Ker}\Psi= H$.

Finally, since $H$ is a kernel, $H$ is normal.

Arnaud Mortier
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  • Dear, Arnaud! This is the answer that I wanted to hear! Elegant, detailed, beautiful! Thanks for that! The best answer. – RFZ Feb 17 '18 at 18:18
  • @RFZ You're welcome :) – Arnaud Mortier Feb 17 '18 at 23:33
  • Hi, Arnaud! Could you help me with this question, please? https://math.stackexchange.com/questions/2661532/sylows-first-theorem-proof-by-wielandt – RFZ Feb 22 '18 at 11:30