Burnside’s normal complement theorem (1)

Posted: January 20, 2011 in Basic Algebra, Groups
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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 exist \alpha, \beta \in S_n, \ h_i,h_i' \in H, \ 1 \leq i \leq n such that gg_i = g_{\alpha(i)}h_i and gg_i' = g_{\beta(i)}'h_i'. Then there exist \gamma \in S_n and h_i'' \in H such that for all i,

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

ii) h_i' =h_{\beta(i)}''^{-1} h_{\gamma(i)} h_i''.

Proof. i) The same argument as Lemma 1.

ii) By i), there exists \gamma \in S_n and h_i'' \in H such that

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

for all i. Let \delta := \gamma^{-1} \alpha \gamma. Then (1) and \gamma \delta = \alpha \gamma gives g_{\delta(i)}'=g_{\alpha \gamma(i)} h_{\delta(i)}'' and hence

g_{\alpha \gamma(i)}=g_{\delta(i)}'h_{\delta(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 replacing g_{\alpha \gamma(i)} on the right side of (4) with what we got in (2) gives

gg_i' = g_{\delta(i)}' h_{\delta(i)}''^{-1} h_{\gamma(i)} h_i''

and so, by Lemma 1, \delta =\beta and h_i' =h_{\beta(i)}''^{-1} h_{\gamma(i)} h_i''. \ \Box

Lemma 3. 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 exist \alpha, \beta \in S_n and h_i, h_i' \in H such that gg_i=g_{\alpha(i)}h_i and gg_i'=g_{\beta(i)}'h_i' for all i. Then h_1h_2 \cdots h_n = h_1' h_2' \cdots h_n'.

Proof. By Lemma 2, ii), there exist \gamma \in S_n and h_i'' \in H such that 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 a unique \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, \beta \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_{\beta(i)}h_i'. Then

gg_{\beta(i)}=g_{\alpha \beta(i)}h_i

and thus

gg'g_i=gg_{\beta(i)}h_i'=g_{\alpha \beta(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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