# Proving two subgroups of the same order are both normal

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?

• Is $|G|$ even or odd? If it is odd, it's not so difficult. – daruma Jan 23 at 1:01
• @daruma Apologies, it was supposed to say |G| is odd! – Jacob Jan 23 at 1:05

## 2 Answers

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!.

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 $$B$$.