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Let G be a finite group and H be a proper subgroup. Prove that the union of the conjugates of H is not the whole of G.

Thanks for any help

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This is false if $H=G$. –  Chris Eagle Mar 17 '12 at 23:39
Are you sure you mean $H$ to be a normal subgroup? –  Tara B Mar 17 '12 at 23:43
@hmmmm: You've been on the site for a few months; by now you probably know that the way to get the best possible answers (best for you) is to state in what context you encountered this problem, and what your thoughts about the problem are so far. Also, writing in the imperative ("Prove", "Show"), when it is not clear you are quoting, is grating to some (many?) of us. –  Arturo Magidin Mar 17 '12 at 23:48
Just curious, is $G$ assume to be finite? –  Dani Hobbes Mar 17 '12 at 23:50
@ArturoMagidin yes sorry I rushed this question which I should not have, sorry about using the phrase "prove" I can see how this is annoying. I will try not to write a question like this again. –  hmmmm Mar 18 '12 at 0:05

3 Answers 3

up vote 20 down vote accepted

(Note: Finite was not specified when I wrote this answer; I'll keep the more general answer, though)

The result is true if we assume that $H$ is of finite index. It may be false if $H$ is of infinite index.

For an counterexample in the infinite index case, let $F$ be an algebraically closed field, let $G$ be the group of all $n\times n$ invertible matrices with coefficients in $F$, and let $H$ be the subgroup of upper diagonal matrices. Since every matrix over an algebraically closed field is similar to an upper diagonal matrix (e.g., the Jordan canonical form), it follows that the union of conjugates of $H$ equals the whole group, even though $H$ does not equal all of $G$.

For a proof in the finite index case, let $[G:H]=n$. Then the action of $G$ on the cosets $H$ by left multiplication gives a homomorphism $G\to S_n$ with kernel $K\subseteq H$. This reduces to the finite case.

In the finite case, let $|H|=k$; then $|G|=kn$. There are at most $n$ distinct conjugates. Since the identity element is in all of the conjugates, the union of the conjugates of $H$ has at most $$n(k-1)+1 = nk-n+1\text{ element}$$ and since we are assuming $n\gt 1$, it follows that $$\left|\bigcup_{g\in G}gHg^{-1}\right| \leq nk-(n-1) \lt nk = |G|,$$ so the union cannot equal all of $G$.

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Can you explain how that reduces to the finite case ? –  WLOG Mar 18 '12 at 18:00
@stef: $K$ is of finite index, since $G/K$ embeds in $S_n$, and the conjugates of $H$ in $G$ correspond to the conjugates of $H/K$ in $G/K$. So $G$ is a union of conjugates of $H$ in $G$ if and only if $G/K$ is a union of conjugates of $H/K$ in $G/K$. –  Arturo Magidin Mar 18 '12 at 19:59
Could you clarify your action of $G$ on $H$? If you take $H$ as a $G$-set by conjugation, I don't see how you get $S_n$, especially the subscript $n$. Are you taking $G/H$ as $G$-set? –  julypraise Apr 5 '14 at 18:41
For a better bound using representations of finite groups, see the paper On a theorem of Jordan, by J-P. Serre (Bulletin AMS, 2003). This paper relates this algebraic result with the exercise discussed on this other thread: math.stackexchange.com/questions/345272/… –  ACL Apr 13 '14 at 10:24
How are there at most $n$ distinct conjugates? Every time I try to see this, I either come up with a maximum of $k$ or $kn$ distinct conjugates, depending on how I look at it. It's clear that there are $n$ left or right cosets of $H$, but it's not clear to me that this must be a maximum for the number of distinct conjugates. –  Travis Bemrose May 1 '14 at 4:04

I think the Orbit-Stabilizer Theorem can be applied here.

Let $G$ have order $n$, and since $H$ is a proper subgroup, let $[G\colon H]=m>1$. Let $N(H)$ be the normalizer of $H$ in $G$, which contains $H$. As such, $[G\colon N(H)]\leq[G\colon H]$.

Let $G$ act by conjugation, so that the orbit of $H$ is the set of all conjugate subgroups. So the stabilizer of $H$ is precisely the set $N(H)$, so by the Orbit-Stabilizer Theorem, the number of all conjugate subgroups is equal to $[G\colon N(H)]$. Now each of the conjugate subgroups has cardinality equal to that of $H$, and each contains the identity element $e$, so there are most $1+[G\colon N(H)](\vert H\vert-1)$ elements in the union. So $$ 1+[G\colon N(H)](\vert H\vert-1)\leq 1+[G\colon H](\vert H\vert-1)=1+\vert G\vert-m=\vert G\vert+(1-m)<\vert G\vert $$ since $m>1$. So the union of the conjugate subgroups is a proper subset.

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Disclaimer: The question was changed after I wrote this answer. The original question is answered by this answer.

Let $H$ be a proper normal subgroup of $G$.

Then for every conjugate you have $gHg^{-1} = H$. Hence $\bigcup_{g \in G} gHg^{-1} = H \subsetneq G$.

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Alas, the question was poorly written, and has now changed. You may want to add a disclaimer, lest you get downvoted for correctly answering something else. –  Arturo Magidin Mar 18 '12 at 0:11
@ArturoMagidin Thanks for telling me. I think I'll just delete my answer for now. I might edit it and repost it at some later point. –  Rudy the Reindeer Mar 18 '12 at 0:17

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