## Burnside’s normal complement theorem (1)

Posted: January 20, 2011 in Elementary Algebra; Problems & Solutions, Groups and Fields
Tags: , ,

Throughout this section $G$ is a group, $H$ is an abelian subgroup of $G$ and $[G:H]=n < \infty.$

Definition. Let $G = \bigcup_{i=1}^n g_i H$ be the partition of $G$ into left cosets of $H.$ The set $T=\{g_1, \cdots , g_n \}$ is called a transversal of $H$ in $G.$

Lemma 1. Let $T = \{g_1, \cdots , g_n \}$ be a transversal of $H$ in $G$ and $g \in G.$ There exists a unique permutation $\alpha \in S_n$ and $h_i \in H$ such that $gg_i=g_{\alpha(i)}h_i$ for all $i.$

Proof. Clearly for every $1 \leq i \leq n$ there exists a unique $1 \leq j \leq n$ such that $gg_i \in g_j H.$ Define the map $\alpha : \{1,2, \cdots , n \} \longrightarrow \{1,2, \cdots , n \}$ by $\alpha(i)=j.$ So we just need to prove that $\alpha$ is a bijection. Since $\alpha$ is defined on a finite set, we only need to prove that it is one-to-one. So suppose that $\alpha(i)=\alpha(k)=j.$ Then $gg_i \in g_j H$ and $gg_k \in g_jH$ and thus $g_i$ and $g_k$ are both in the left coset $g^{-1}g_j H.$ Hence $g_i H = g_k H$ and so $i = k.$ To prove that $\alpha$ is unique, suppose that $\beta \neq \alpha$ is another permutation and $gg_i \in g_{\beta(i)}H$ for all $1 \leq i \leq n.$ Then $\beta(i) \neq \alpha(i)$ for some $i$ and $gg_i \in g_{\alpha(i)}H \cap g_{\beta(i)}H = \emptyset,$ which is nonsense. $\Box$

Lemma 2. Let $T = \{g_1, \cdots , g_n \}$ and $T' = \{g'_1, \cdots , g'_n \}$ be two transversals of $H$ in $G$ and let $g \in G.$ By Lemma 1 there exists $\alpha \in S_n$ and $h_i \in H, \ 1 \leq i \leq n$ such that $gg_i = g_{\alpha(i)}h_i.$ Prove that there exist permutations $\beta, \gamma \in S_n$ and $h''_i \in H, \ 1 \leq i \leq n$ such that $gg_i' = g_{\beta(i)}' h_{\beta(i)}''^{-1} h_{\gamma(i)} h_i''$ for all $i.$

Proof. Applying Lemma 1 to $g=1_G$ and $T'$ we find $\gamma \in S_n$ and $h_i'' \in H$ such that

$g_i' = g_{\gamma(i)}h_i'' \ \ \ \ \ \ \ \ \ \ (1)$

for all $i.$ Let $\beta = \gamma^{-1} \alpha \gamma.$ Then (1) and the fact that $\gamma \beta = \alpha \gamma$ gives us $g_{\beta(i)}'=g_{\alpha \gamma(i)} h_{\beta(i)}''$ and hence

$g_{\alpha \gamma(i)}=g_{\beta(i)}'h_{\beta(i)}''^{-1} \ \ \ \ \ \ \ \ \ \ \ \ (2)$

for all $i.$ Also, since $gg_i = g_{\alpha(i)} h_i,$ we have

$gg_{\gamma(i)}=g_{\alpha \gamma(i)}h_{\gamma(i)} \ \ \ \ \ \ \ \ \ \ \ \ (3)$

for all $i.$ Therefore multiplying (1) by $g$ from the left and then applying (3) to the right side we get

$gg_i' = g_{\alpha \gamma(i)}h_{\gamma(i)}h_i'' \ \ \ \ \ \ \ \ \ \ \ \ \ (4)$

for all $i.$ Finally replcaing $g_{\alpha \gamma(i)}$ on the right side of (4) with what we got in (2) completes the proof. $\Box$

Remark. Let’s take another look at the situation in Lemma 2. Well, by Lemma 1 there exist a unique $\alpha' \in S_n$ and $h_i' \in H$ such that $gg_i' = g_{\alpha'(i)}'h_i'$ for all $i.$ The uniqueness of $\alpha'$ and Lemma 2 gives us $\alpha' = \beta$ and $h_i' =h_{\beta(i)}''^{-1} h_{\gamma(i)} h_i''$ for all $i.$

Lemma 3. Let $T_1 = \{u_1, \cdots , u_n \}$ and $T_2 = \{v_1, \cdots , v_n \}$ be two transversals of $H$ in $G$ and let $g \in G.$ By Lemma 1 there exist $\alpha, \beta \in S_n$ and $x_i, \ y_i \in H$ such that $gu_i=u_{\alpha(i)}h_i$ and $gv_i=u_{\beta(i)}h_i'$ for all $i.$ Prove that $x_1y_2 \cdots h_n = h_1' h_2' \cdots h_n'.$

Proof. By Remark 1 we know that there exist $\beta, \gamma \in S_n$ and $h_i'' \in H$ such that $\alpha' = \beta$ and $h_i' =h_{\beta(i)}''^{-1} h_{\gamma(i)} h_i''$ for all $i.$ Thus $\prod h_i' = \prod h_{\beta(i)}''^{-1} \prod h_{\gamma(i)} \prod h_i'',$ where products are over $1 \leq i \leq n.$ Now, since $\beta, \gamma \in S_n,$ we have $\prod h_{\beta(i)}''^{-1} = \prod h_i''^{-1}$ and $\prod h_{\gamma(i)} = \prod h_i.$ Therefore, since $H$ is abelian, we get $\prod h_i' = \prod h_i. \ \Box$

Now the main result of this post:

Theorem. Let $T = \{g_1, \cdots , g_n \}$ be a transversal of $H$ in $G.$ By Lemma 1, for every $g \in G$ there exists $\alpha \in S_n$ and $h_i \in H$ such that $gg_i=g_{\alpha(i)}h_i$ for all $i.$ Define the map $\lambda : G \longrightarrow H$ by $\lambda(g)=h_1h_2 \cdots h_n.$ Then $\lambda$ is a well-defined group homomorphism.

Proof. Lemma 3 shows that $\lambda$ is well-defined. Now let $g,g' \in G.$ Let $\alpha, \alpha' \in S_n$ and $h_i, h_i' \in H$ be such that $gg_i=g_{\alpha(i)}h_i$ and $g'g_i=g_{\alpha'(i)}h_i'.$ Then $gg_{\alpha'(i)}=g_{\alpha \alpha'(i)}h_i$ and thus $gg'g_i=gg_{\alpha'(i)}h_i'=g_{\alpha \alpha'(i)} h_i h_i'.$ Thus, since $H$ is abelian, we have $\lambda (gg') = \prod h_i h_i' = \prod h_i \prod h_i'=\lambda(g) \lambda(g'). \ \Box$