# Thread: Algebra, Problems For Fun (11)

1. ## Algebra, Problems For Fun (11)

This one is a well-known and important result:

Let $G$ be a group and $H \neq G$ a subgroup of $G.$ Prove that if $[G:H] < \infty,$ then $\bigcup_{g \in G} gHg^{-1} \neq G.$

2. Originally Posted by NonCommAlg
This one is a well-known and important result:

Let $G$ be a group and $H \neq G$ a subgroup of $G.$ Prove that if $[G:H] < \infty,$ then $\bigcup_{g \in G} gHg^{-1} \neq G.$
I love these problems.

It is well known that for any subgroup $H$ of finite index $n$ in $G,$ there exists a normal subgroup $N$ of $G$ which is contained in $H.$ This is proved by considering the set $S=\{aH:a\in G\}$ of all left cosets of $H$ in $G;$ any element $g\in G$ induces a permutation $\pi_g$ of $S$ by $g\mapsto gaH.$ Then the mapping $g\to\pi_g$ from $G$ to the symmetric group of $S$ is a homomorphism with kernel $N=\bigcap_{a\,\in\,G}aHa^{-1}\subseteq H.$

Consider first the case when $G$ is finite. The number of conjugates of $H$ in $G$ is equal to $|G:N_G(H)|,$ the index of the normalizer of $H$ in $G.$ As $|H|\le|N_G(H)|,$ we have $|G:N_G(H)|\le|G:H|=n$ – and as $H$ is proper, $n>1.$ Hence

$\left|\bigcup_{a\,\in\,G}aHa^{-1}\right|\color{white}.$ $\le\ 1+|G:N_G(H)|\left(|H|-1\right)\ \le\ 1+|G:H|\left(|H|-1\right)\ =\ 1+|G|-n\ <\ |G|$

This proves the result for the finite case.

For $G$ infinite, consider the normal subgroup $N$ above. The quotient group $G/N$ is a finite group since it is isomorphic to a subgroup of the symmetric group of degree $n.$ So is $H/N$ as it is a subgroup of $G/N.$ Thus $H/N$ is a proper subgroup of finite index of the finite group $G/N$ and so the union of its conjugates in $G/N$ is proper subset of $G/N.$ The result follows since the mapping $aHa^{-1}\mapsto aN(H/N)a^{-1}N$ is a bijection from the set of all conjugates of $H$ in $G$ to the set of all conjugates of $H/N$ in $G/N.$