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- Mar 10, 2012

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Hello MHB.

I am reading a book on Combinatorics in which there is proved the Schreier's Lemma. I don't fully understand the proof.

THEOREM: Let $\{g_1,\ldots,g_m\}$ generate a group $G$. Let $k_1,\ldots,k_s$ be coset representatives of for a subgroup $H$ of $G$ where $k_1=1$. Let \(\overline{g}=k_i\) iff \(gH=k_{i}H\). Then $H$ is generated by the set $$S_H=\{k_ig_j\overline{(k_ig_j)}^{-1}:i=1,\ldots,s;j=1,\ldots,m\}$$

PROOF: All the elements of $S_H$ lie in $H$. Now suppose that $h=g_{i_1}g_{i_2}\ldots g_{i_r}\in H$. For $j=0,\ldots,r,$ let $t_j=g_{i_1}\ldots g_{i_j}$, and let $u_j=\overline{t_j}$. Then, with $u_0=1$, we have $$h=u_0g_{i_1}u_1^{-1}.u_1g_{i_2}u_2^{-1}\ldots u_{r-1}g_{i_r}u_r^{-1},$$

Since $u_0=u_r=1$ and all the other $u_i$ cancel with their inverses. But $u_{j-1}g_{i_j}$ lies in the same coset as $u_j$.

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I can't see how it is true that $u_{j-1}g_{i_j}u_j^{-1}\in S_H$. Can somebody please explain this to me?

I am reading a book on Combinatorics in which there is proved the Schreier's Lemma. I don't fully understand the proof.

THEOREM: Let $\{g_1,\ldots,g_m\}$ generate a group $G$. Let $k_1,\ldots,k_s$ be coset representatives of for a subgroup $H$ of $G$ where $k_1=1$. Let \(\overline{g}=k_i\) iff \(gH=k_{i}H\). Then $H$ is generated by the set $$S_H=\{k_ig_j\overline{(k_ig_j)}^{-1}:i=1,\ldots,s;j=1,\ldots,m\}$$

PROOF: All the elements of $S_H$ lie in $H$. Now suppose that $h=g_{i_1}g_{i_2}\ldots g_{i_r}\in H$. For $j=0,\ldots,r,$ let $t_j=g_{i_1}\ldots g_{i_j}$, and let $u_j=\overline{t_j}$. Then, with $u_0=1$, we have $$h=u_0g_{i_1}u_1^{-1}.u_1g_{i_2}u_2^{-1}\ldots u_{r-1}g_{i_r}u_r^{-1},$$

Since $u_0=u_r=1$ and all the other $u_i$ cancel with their inverses. But $u_{j-1}g_{i_j}$ lies in the same coset as $u_j$.

**Thus**$u_{j-1}g_{i_j}u_j^{-1}\in S_H$, and we have expressed $h$ as a product of elements of $S_H$.__________________________________________________________

I can't see how it is true that $u_{j-1}g_{i_j}u_j^{-1}\in S_H$. Can somebody please explain this to me?

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