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Math Help - Algebra, Problems For Fun (11)

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    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.
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    Senior Member TheAbstractionist's Avatar
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    Quote Originally Posted by NonCommAlg View Post
    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.
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