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

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