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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    $\begingroup$ This is false if $H=G$. $\endgroup$ Commented Mar 17, 2012 at 23:39
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    $\begingroup$ Are you sure you mean $H$ to be a normal subgroup? $\endgroup$
    – Tara B
    Commented Mar 17, 2012 at 23:43
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    $\begingroup$ @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. $\endgroup$ Commented Mar 17, 2012 at 23:48
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    $\begingroup$ Just curious, is $G$ assume to be finite? $\endgroup$ Commented Mar 17, 2012 at 23:50
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    $\begingroup$ @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. $\endgroup$
    – hmmmm
    Commented Mar 18, 2012 at 0:05

7 Answers 7


(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 a 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 triangular matrices. Since every matrix over an algebraically closed field is similar to an upper triangular 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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    $\begingroup$ Can you explain how that reduces to the finite case ? $\endgroup$
    – WLOG
    Commented Mar 18, 2012 at 18:00
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    $\begingroup$ @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 $G/K$. $\endgroup$ Commented Mar 18, 2012 at 19:59
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    $\begingroup$ 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. $\endgroup$ Commented May 1, 2014 at 4:04
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    $\begingroup$ @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 $H$.) $\endgroup$ Commented Nov 10, 2015 at 18:46
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    $\begingroup$ @TravisBemrose: $(gH)(gH)^{-1} = gHg^{-1}$, so the map from the cosets of $H$ to the conjugates of $H$ given by $gH \mapsto gHg^{-1}$ is well-defined (and surjective). So there are at most $|G:H|$ distinct conjugates. $\endgroup$ Commented Feb 14, 2018 at 9:13

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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    $\begingroup$ How beautiful the answer is. Do you know more questions which is solved by the same technique? $\endgroup$
    – bfhaha
    Commented Nov 1, 2015 at 11:29
  • $\begingroup$ @bfhaha: what about the following question? mathoverflow.net/q/292721/34538 $\endgroup$ Commented Feb 14, 2018 at 7:40

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.


Here is an answer by Burnside's lemma. (This is an exercise in Steinberg's ``Representation Theory of Finite Groups'').

If $H\lhd G$, then $\bigcup\limits_{x\in G}xHx^{-1}=H\neq G$ and we are done.

So we assume that $H\not\lhd G$. Let $X$ be the set of all conjugate of $H$. Then $|X|\geq 2$.

Let $G$ acts on $X$ by conjugation. That is, $g\cdot xHx^{-1}=gxHx^{-1}g^{-1}$. It is easy to show that there is only one orbit. By Burnside's lemma, $$1=\text{the number of orbit}=\frac{1}{|G|}\sum\limits_{g\in G}|\text{Fix}(g)|$$ and $|G|=\sum\limits_{g\in G}|\text{Fix}(g)|$. Since $\text{Fix}(e)=|X|\geq 2$, there exists $g_0\neq e\in G$ such that $|\text{Fix}(g_0)|=0$.

If $G=\bigcup\limits_{x\in G}xHx^{-1}$, then for all $g\in G$, $g=xhx^{-1}$ for some $x\in G$ and $h\in H$ and $$g\cdot xHx^{-1}=xhx^{-1}(xHx^{-1})xh^{-1}x^{-1}=xhHh^{-1}x^{-1}=xHx^{-1}.$$ That is, $g$ has a fixed point on $X$, contrary to that there is an element $g_0\neq e\in G$ such that $|\text{Fix}(g_0)|=0$.


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

  • $\begingroup$ 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. $\endgroup$ Commented Mar 18, 2012 at 0:11
  • $\begingroup$ @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. $\endgroup$ Commented Mar 18, 2012 at 0:17

Lemma: the number of distinct conjugates of $H$ is equal to $|G:N_G(H)|$.

Pf of lemma: $\phi: G/N_G(H)\to \{gHg^{-1}\}$ with $\phi(gN_G(H))=gHg^{-1}$ is a bijection for the following reason:

$gN_G(H)=g'N_G(H)\iff \begin{cases}g'^{-1}g\in N_G(H)\\ g'g^{-1}\in N_G(H)\end{cases}\iff \begin{cases} g'^{-1}ghg^{-1}g'\in H&\forall h\in H\\ g'g^{-1} h'gg'^{-1}\in H&\forall h'\in H\end{cases}\iff $$\begin{cases} ghg^{-1}\in g'Hg'^{-1} &\forall h\in H\\g'h'g'^{-1}\in gHg^{-1}&\forall h'\in H \end{cases}\iff \begin{cases} gHg^{-1}\subset g'Hg'^{-1}\\ g'Hg'^{-1}\subset gHg^{-1} \end{cases} \iff gHg^{-1}= g'Hg'^{-1}$

So $\phi$ is well-defined and injective.

it's surjective for obvious reason.

Hence the number of distinct conjugates of $H$ is equal to $|G:N_G(H)|$.

$|\bigcup_{g\in G} gHg^{-1}|\leqslant \sum_{g\in G/N_G(H)} |gHg^{-1}|-|\bigcap_{g\in G} gHg^{-1}|= |G:N_G(H)| |H|-|\bigcap_{g\in G} gHg^{-1}|<\dfrac{|G|}{|N_G(H)|}|H|\leqslant \dfrac{|G|}{|H|}|H|= |G|$

Notice that
$\text{(1)}\forall g\in G/N_G(H), |gHg^{-1}|=|H| \text{ for obvious reasons}\\ \text{(2)} e\in \bigcap_{g\in G} gHg^{-1}\\ \text{(3)} H\leqslant N_G(H)$


Here's a slight variation on some of the earlier approaches. First note that $K=\bigcap_g gHg^{-1}$ is a normal subgroup of $G$ of finite index, so it is sufficient to consider the case of $H/K\leqslant G/K$ and reduce to the case of a finite group $G$.

Second, it is sufficient to consider the case of a maximal proper subgroup $H$. In this case if $H$ is normal there is only one conjugate, $H$. Otherwise $H$ is self-normalising, $N_G(H)=H$. In this case there are exactly $k=|G:H|$ conjugates of $H$. Hence if $G$ is the union of the conjugates of $H$, $$ |G| \leqslant k(|H|-1) + 1 = k|H| + 1-k = |G| + 1-k < |G|, $$ giving a contradiction.


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