## A group is not the union of conjugates of any proper subgroup

Assume $H$ is a proper subgroup of the finite group $G$. Prove that $G \neq \bigcup_{g \in G} gHg^{-1}$; i.e., $G$ is not the union of the conjugates of any proper subgroup.

There exists a maximal subgroup $M$ containing $H$. If $M$ is normal in $G$, then $\bigcup_{g \in G} gHg^{-1} \subseteq \bigcup_{g \in G} gMg^{-1} = M \neq G$. If $M$ is not normal, we still have $\bigcup_{g \in G} gHg^{-1} \subseteq \bigcup_{g \in G} gMg^{-1}$. By the previous exercise, $\bigcup_{g \in G} gMg^{-1}$ contains at most $(|M|-1)[G : M]$ nonidentity elements. Thus $|\bigcup_{g \in G} gHg^{-1}| \leq |G| - [G : M] + 1 < |G|$, since $[G:M] \geq 2$.

In particular, because $G$ is finite, $G \neq \bigcup_{g \in G} gHg^{-1}$. Thus $G$ is not the union of all conjugates of any proper subgroup.

• C  On August 28, 2011 at 12:48 pm

in the second to last sentence, (the middle term of the inequality), do you mean
lGl – [G:M] +1 instead of [g:M]?

• nbloomf  On September 6, 2011 at 10:04 am

Yes. 🙂 Thanks!

• Gobi Ree  On December 12, 2011 at 12:26 am

The phrase “because G is finite” seems unnecessary. The finiteness was used for the formula in the previous exercise.