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I need help in understanding following problem.

Let $G$ be a finite group and $H\leq G$. Prove that the number of subgroups of $G$ conjugate to $H$ is a divisor of $|G|$.

I want to understand what does a subgroup conjugate to $H$ mean.

Does it mean that every element of that subgroup is conjugate to every element in $H$.

That is for $K\leq G$ to be conjugate to $H$.

For every $k\in K$ $ \exists x_{i} \in G $ such that $k= x_{i}^{-1}H x_{i}$ where $ i = {1,....,|H|}$

I have been thinking about this for a while. I just want to make sure I am on the right track. Any input will be appreciated.

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2 Answers 2

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The number of conjugates of $H$ is equal to the index of the normaliser of $H$ in $G$, i.e. $\,\dfrac{\lvert G\rvert}{\lvert N_G(H)\rvert}$.

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  • $\begingroup$ Thanks for pointing the typo. It's getting late here…. $\endgroup$
    – Bernard
    Commented Feb 11, 2015 at 22:11
  • $\begingroup$ is that something that needs to be proven or does it follow immediately from the definitions? $\endgroup$
    – user100463
    Commented Jan 18, 2017 at 0:41
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    $\begingroup$ In my opinion, it follows from the definitions; $G$ acts by conjugation on the set of its subgroups and the normaliser is the subgroup of elements which fix $H$. It's just the transcription of the class formula for this specific case. $\endgroup$
    – Bernard
    Commented Jan 18, 2017 at 0:56
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    $\begingroup$ It's a divisor of $n$, hence a divisor of $\lvert G\rvert$ since $\;[G\colon H]=[G\colon N_G(H)]\cdot[ N_G(H)\colon H]$. $\endgroup$
    – Bernard
    Commented Jan 18, 2017 at 1:09
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    $\begingroup$ The equality is valid for any tower of subgroups: $H\subset K\subset G$, not only with the normaliser. $\endgroup$
    – Bernard
    Commented Jan 18, 2017 at 1:18
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Two subgroups $H$ and $K$ of $G$ are conjugate if there exists a $g\in G$ such that $g^{-1}Hg=K$.

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  • $\begingroup$ In particular, this is different to what you wrote. $\endgroup$ Commented Feb 11, 2015 at 21:59

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