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Suppose $H$ and $H'$ are two finite subgroups of $G$. Do there exist theorems that give conditions so that $H$ and $H'$ are conjugate, that is, there exists $g\in G$ such that $H = gH'g^{-1}$?

Edit: Maybe this question is too general. What about if they are both finite groups of invertible matrices?

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Well, for example if they both are Sylow subgroups for the very same prime... –  DonAntonio Feb 7 '13 at 20:15
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Every finite group is isomorphic to a finite group of invertible matrices, so your edit does not limit the generality! –  Mariano Suárez-Alvarez Feb 7 '13 at 20:28
    
Haha good point! But maybe it helps for the intuition... –  user61408 Feb 7 '13 at 20:34
    
There are many conditions to establish that two subgroups are not conjugate. Any function defined on subgroups that is invariant under conjugation will, of course, work (size of the subgroup, number of elements of given orders, etc.). –  Sammy Black Feb 7 '13 at 20:50
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The subgroups must, first of all, be isomorphic. If they are isomorphic, then your original group $G$ can be embedded in a group in which they are conjugate. If $G$ is finite, then the embedding group can be taken to be finite as well. –  James Feb 7 '13 at 22:42

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up vote 4 down vote accepted

I'll use $K$ instead of $H'$, and write $H^g$ for $g^{-1}Hg$.

One condition is that $H$ must be isomorphic to $K$. The function $\theta_g:x\mapsto x^g$ is a homorphism which maps $H$ bijectively to $K$, so $H\cong K$. (In fact, such functions are referred to as inner automorphisms of $G$, and form a group themselves.)

The number of groups $K$ such that $H$ can be related by an inner automorphism to $K$ is easy to calculate. If we let $G$ act on its set of subgroups by $g\cdot H = H^g$, we see that the stabilizer of $H$ under this action is exactly those $g$ for which $H^g=H$ - that is, the normalizer $N_G(H)$. The orbit of $H$, then, is all subgroups $K$ for which there exists a $g\in G$ so that $H^g=K$. By the Orbit-Stabilizer theorem, there are then $[G:N_G(H)]$ such groups.

Notice that since these groups are all in an orbit together, it must be true that $H\sim K \Leftrightarrow \exists g\in G :H^g=K$ is an equivalence relation. The orbits are referred to as subgroup conjugacy classes.

Determining the subgroup conjugacy classes of groups of invertible matrices is, actually, a rather difficult problem, even in when all subgroups are finite (i.e. in $GL_n(\mathbb{F}_q)$, which you can read about here). One standard result on finite subgroups of matrices in general is that any finite subgroup of $GL_n(F)$ must have a conjugate contained in $O_n(F)$, which is provable using invariant symmetric bilinear forms.

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