Cosets of Subgroups

**thomas_donald** · September 18th 2009, 09:40 AM

If G is a finitely generated group, H a subgroup of finite index. Show that H is finitely generated.

**NonCommAlg** · September 18th 2009, 01:29 PM

Originally Posted by thomas_donald

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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