Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|cite|improve this question
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

4 Answers 4

up vote 23 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 $G$.

share|cite|improve this answer
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:… – ACL Apr 13 '14 at 10:24
@Travis: (I know it's over a year late, but I just thought through why there at most $n$ distinct conjugates. The idea is that if $g_2 = g_1 h$ for some $h\in H$, then conjugation of $H$ by $g_2$ and $g_1$ gives the same thing. So, the number of conjugates of $H$ is at most the index of $H$.) – Jason DeVito Nov 10 at 18:46

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.

share|cite|improve this answer
How beautiful the answer is. Do you know more questions which is solved by the same technique? – bfhaha Nov 1 at 11:29

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 $\bigcup_{g \in G} gHg^{-1} = H \subsetneq G$.

share|cite|improve this answer
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

I recently encountered a nice exercise in Isaacs, Finite Group Theory, which allows us to say more.

Theorem: if $G$ is a finite group and $H < G$ is a proper subgroup, then the number of elements of $G$ which do not lie in any conjugate of $H$ is at least $|H|$. Since we always have $|H| \geq 1$, this result implies the one in the OP.

The proof uses the permutation character, which is defined as follows. If $G$ acts on the set $\Omega$, then the permutation character is the integer-valued function $\chi$ which counts the number of elements of $\Omega$ fixed by each $g \in G$: $$\chi(g) = |\{\alpha \in \Omega : g \cdot \alpha = \alpha\}|$$

It is a standard exercise to show the following identity, which I will call [PC] for easy reference. $$\sum_{g \in G}\chi(g) = \sum_{\alpha \in \Omega} |G_{\alpha}| = n|G|$$ where $G_{\alpha}$ is the stabilizer of $\alpha$, and $n$ is the number of orbits. Sketch of proof: define $$\delta(g, \alpha) = \begin{cases} 1 & \text{if }g\text{ fixes }\alpha \\ 0 & \text{otherwise} \end{cases}$$ and sum $\delta$ over $G \times \Omega$ two ways.

To obtain the desired result, we define an appropriate group action and apply [PC]. Let $\Omega$ be the set of all left cosets of $H$. Then $G$ and $H$ both act on $\Omega$ by left multiplication.

The action by $G$ is clearly transitive; there is only one orbit.

I will reserve $n$ to refer to the number of orbits under the action by $H$. Note that $n \geq 2$ because $\{H\}$ is one orbit, but it can't be the only orbit because $H$ is a proper subgroup, so there is more than one coset of $H$.

The permutation character for the action by $G$ is $$\chi(g) = |\{aH \in \Omega : gaH = aH\}|$$ and the permutation character for the action by $H$ is simply the restriction of $\chi$ to $H$.

Notice that $gaH = aH$ iff $ga \in aH$ iff $g \in aHa^{-1}$, so the stabilizer of $aH$ under the action by $G$ is $aHa^{-1}$, and under the action by $H$ it is $H\ \cap\ aHa^{-1}$. For the action by $H$, the identity [PC] becomes $$\sum_{h \in H}\chi(h) = \sum_{aH \in \Omega}|H\ \cap\ aHa^{-1}| = n|H| \geq 2|H|$$ For the action by $G$, [PC] gives us $$\begin{aligned} |G| &= \sum_{g \in G}\chi(g) \\ & = \sum_{h \in H}\chi(h) + \sum_{g \in G \setminus H} \chi(g) \\ &\geq 2|H| + \sum_{g \in G \setminus H} \chi(g) \end{aligned}$$ The number of terms in the rightmost sum is $|G| - |H|$. Note that $\chi(g) \geq 1$ iff $g$ fixes at least one element of $\Omega$ iff $g$ lies in some conjugate of $H$. If we define $Z$ to be the set of elements of $G$ which lie in no conjugate of $H$, then clearly $Z \subseteq G \setminus H$, so in the rightmost sum there are $|G| - |H|$ terms, of which $|G| - |H| - |Z|$ are nonzero. Therefore the rightmost sum is at least $|G| - |H| - |Z|$. This gives us $$|G| \geq 2|H| + |G| - |H| - |Z|$$ so $|Z| \geq |H|$, which proves the theorem.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.