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Math Help - Cosets of Subgroups

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    Cosets of Subgroups

    If G is a finitely generated group, H a subgroup of finite index. Show that H is finitely generated.
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  2. #2
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    Quote Originally Posted by thomas_donald View Post

    If G is a finitely generated group, H a subgroup of finite index. Show that H is finitely generated.
    let G=<g_1, \cdots, g_n> and [G:H]=m. let Hx_1, \cdots , Hx_m be the cosets of H, where x_1=1. if g_i^{-1} \notin \{g_1, \cdots , g_n \}, then we can add g_i^{-1} to the set of the generators of G and hence we may

    assume that g_i^{-1} \in \{g_1, \cdots , g_n \} for all 1 \leq i \leq n. now for any 1 \leq i \leq m and 1 \leq j \leq n we have x_ig_j \in G = Hx_1 \cup Hx_2 \cup \cdots \cup Hx_n and so x_ig_j=h_{ij}x_k, for some h_{ij} \in H and 1 \leq k \leq m.

    call this result (1). now let K=<h_{ij}: \ 1 \leq i \leq m, \ 1 \leq j \leq n>. the claim is that K=H. obviously we only need to prove that H \subseteq K:

    let h \in H \subseteq G. then h=g_{i_1}g_{i_2} \cdots g_{i_r}, where g_{i_j} are not necessarily distinct. then by (1): \ \ h=x_1h=x_1g_{i_1}g_{i_2} \cdots g_{i_r}=h_{1i_1}x_jg_{i_2} \cdots g_{i_r}, for some j. again applying (1) we get:

    h=h_{1i_1}h_{ji_2}x_{\ell}g_{i_3} \cdots g_{i_r}, for some \ell. continuing this way we'll eventually get h=\tilde{h}x_t, for some t and some \tilde{h} \in K. but then x_t=\tilde{h}^{-1}h \in H and hence x_t=1. so h=\tilde{h}\in K, i.e. H \subseteq K. \ \Box
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