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Given: Let $G$ be a group, and let $\mathcal{S}$ be the set of subgroups of $G$. For $g\in G$ and $H\in S$, let $g\cdot H=gHg^{-1}$

Question: Deduce that if $G$ is a finite $p$-group, for some prime $p$, the number of subgroups of $G$ that are not normal is divisible by $p$.

Comments:

  • The subgroups of $G$ that are normal have the property that $g\cdot H=gHg^{-1}=H$

  • The question is equivalent to showing $p$ divides $\left|G\right|-\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$

  • $p$-group: $\forall g\in G,|g|=p^k$ for $k\in\mathbb{N}^+$

  • I don't know where to start with this problem, been thinking about it for a while to no avail, I feel that there are too many definitions for me to consider when finding a solution.

  • My problems I have considered and not been able to answer are: what is the order of $|G|$? I think it must be of order that is divisible by $p$, hence the question becomes show $p$ divides the order of $\left|\{H\in S\vert gHg^{-1}\neq H\}\right|$. How do you compute the order of this set?

  • It seems like quite a standard problem, so I would really appreciate it if I could be directed to more information about whatever problem it is.

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Consider the action of $G$ on the set $S$ by conjugation. The singleton elements are the normal subgroups, the others are the "abnormal" subgroups. Then by the orbit stabilizer theorem all orbits have size divisible by $p$, i.e. denoting the orbit of a given subgroup, $H$, by $H^G$ we see $p|\big|H^G\big|$. But then $\{H\in S : H^G\ne \{H\}\}$ is another way to write your set. If $H_1^G,\ldots, H_n^G$ are representatives of the distinct, disjoint orbits we see that

$$\{H\in S : H^G\ne \{H\}\}=\coprod_{i=1}^n H_i^G$$

Therefore

$$\big|\{H\in S : H^G\ne \{H\}\}\big|=\sum_{i=1}^n\big|H_i^G|$$

but $p|\big|H_i^G\big|$ for every $i$, so

$$p|\big|\{H\in S : H^G\ne \{H\}\}\big|.$$

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  • $\begingroup$ Ah, thanks I was wondering what theorem I should use, also the orbit stabiliser theorem makes sense in the context of the question. $\endgroup$ – Sam Houston Jan 6 '15 at 20:28
  • $\begingroup$ @Dansmith The Orbit-stabilizer theorem is the main theorem in this solution. $\endgroup$ – Adam Hughes Jan 6 '15 at 20:35
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The group $G$ acts on its set of subgroups by conjugation, you wish to count the elements that do not have trivial orbits. the size of an orbit is equal to the the order of the group $G$ divided by the stabilizer of $H$ where $H$ is any subgroup belonging to that orbit. Since $G$ is a $p$-group $\frac{|G|}{|stab(H)}$ is a multiple of $p$ since the stabilizer is a proper subgroup of $G$, so all non-trivial orbits have size multiple of $p$

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Let $X$ be the set of not-normal subgroups of $G$. A conjugate of a not-normal subgroup is not-normal again, so $G$ acts on $X$ by conjugation and (by construction) there are no trivial orbits, so the size of every orbit is a positive power of $p$.

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