Exercise 4.3.24 ($G$ is not the union of the conjugates of any proper subgroup)

Question

Assume $H$ is a proper subgroup of the finite group $G$. Prove $G 
e \bigcup_{g\in G} gHg^{-1}$, i.e., $G$ is not the union of the conjugates of any proper subgroup. [Put $H$ in some maximal subgroup and use the preceding exercise.]

Richard Ganaye · Accepted Answer

\begin{proof} 
Let $H$ be a proper subgroup of the finite group $G$. Assume for the sake of contradiction that $G =
e \bigcup_{g\in G} gHg^{-1}$. Since $G$ is finite, there is a maximal subgroup $M$ such that $H\leq M$. Then $G = \bigcup_{g\in G} gHg^{-1} \subseteq \bigcup_{g\in G} gMg^{-1} \subseteq G$, thus
$$\bigcup_{g\in G} gMg^{-1} = G.$$
If $M \unlhd G$, then $M = \bigcup_{g\in G} gMg^{-1} = G$: this is impossible, since $M$ is maximal in $G$, which implies $M
e G$. So $M$ is not normal in $G$. By Exercise 23, 
$$\left | \left (\bigcup_{g \in G}  gMg^{-1} ight) - \{1\} ight | \leq  (|M| - 1) |G : M|.$$
Therefore
\begin{align*}
|G| &= \left | \bigcup_{g\in G} gMg^{-1} ight |\
&\leq 1 +  (|M| - 1) |G : M|.
\end{align*}
But $M$ is a proper subgroup of $G$, so
$$1 < \frac{|G|}{|M|} ,$$
thus
$$\left( 1 - \frac{1}{|M|}ight) |G| < |G| - 1,$$
or equivalenty
$$(|M| - 1) |G : M| < |G| - 1,$$
so
$$|G| < 1 +  (|M| - 1) |G : M| < |G|.$$
The contradiction $|G| < |G|$ shows that $G$ is not the union of the conjugates of any proper subgroup.
\end{proof}

Exercise 4.3.24 ($G$ is not the union of the conjugates of any proper subgroup)

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Exercise 4.3.24 ($G$ is not the union of the conjugates of any proper subgroup)

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