1. ## Cosets of Subgroups

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

2. 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 $\displaystyle G=<g_1, \cdots, g_n>$ and $\displaystyle [G:H]=m.$ let $\displaystyle Hx_1, \cdots , Hx_m$ be the cosets of $\displaystyle H,$ where $\displaystyle x_1=1.$ if $\displaystyle g_i^{-1} \notin \{g_1, \cdots , g_n \},$ then we can add $\displaystyle g_i^{-1}$ to the set of the generators of $\displaystyle G$ and hence we may

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

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

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

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