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Thread: Algebra, Problems For Fun (11)

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    Algebra, Problems For Fun (11)

    This one is a well-known and important result:

    Let $\displaystyle G$ be a group and $\displaystyle H \neq G$ a subgroup of $\displaystyle G.$ Prove that if $\displaystyle [G:H] < \infty,$ then $\displaystyle \bigcup_{g \in G} gHg^{-1} \neq G.$
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    Quote Originally Posted by NonCommAlg View Post
    This one is a well-known and important result:

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

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

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

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

    This proves the result for the finite case.

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