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Let $X$ be the set of all subgroups of $G$, and let the group action be defined by conjugation, that is for $a\in X$: $$g.A = gAg^{-1}$$ Now let $A$ and $B$ be the only subgroups of $G$ of order $n$ (N.B. that $G$ may have other subgroups, but they will all have order $\neq n$). Suppose $|G|$ is odd, prove that $A$ and $B$ are both normal in $G$. We are given the hint to use the second form of the orbit-stabiliser theorem given to us by: $$|\rm{Orb}_G(x)| = {|G|\over{|\rm{Stab}_G(x)|}}$$ In a previous part of the question we have proven that if we have $K\in \rm{Orb}_G(A)$ that $|K| = |A|$.

My immediate thoughts for this is that if $A,B$ are normal then $\rm{Stab}_G(A)=G$ but have no idea how to use the fact that $|G|$ is odd or how that could factor in, is there something really obvious that I'm missing?

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  • $\begingroup$ Is $|G|$ even or odd? If it is odd, it's not so difficult. $\endgroup$ – daruma Jan 23 at 1:01
  • $\begingroup$ @daruma Apologies, it was supposed to say |G| is odd! $\endgroup$ – Jacob Jan 23 at 1:05
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Suppose $A$ is not normal in $G$. Since conjugate subgroups have the same order, $B$ must be in the orbit of $A$ when $G$ acts on it by conjugation. So $

As there are no other subgroups of that order, $2=|Orb_G(A)|$.

But then, $2$ divides $|G|$ but this is odd!.

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The orbit of $A$ can't have size greater than $2$, because its points are subgroups conjugate to $A$ (and by assumption there are just $2$ subgroups of order $|A|$). But it can't have size $2$ either, by the orbit-stabilizer theorem and the assumption on $|G|$. Therefore, $O(A)$ is a singleton, and hence $gAg^{-1}=A$ for every $g\in G$. Same conclusion holds for $B$.

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