- #1
Bashyboy
- 1,421
- 5
Homework Statement
Let ##\{N_i ~|~ i \in I\}## be family of normal subgroups of G such that
(i) ##G = \left\langle \bigcup_{i \in I} N_i \right\rangle##
(ii) for each ##k \in I##, ##N_k \cap \left\langle \bigcup_{i \neq k} N_i \right\rangle = \{e\}##
Then ##G \simeq \prod_{i \in I}^w N_i##, where ##\prod_{i \in I}^w## denotes the weak product.
Homework Equations
The Attempt at a Solution
Okay. Consider ##\phi : \prod_{i \in I}^w N_i \rightarrow G## defined by ##\phi((a_i)_{i \in I}) = \prod_{i \in I_0} a_i##, where ##I_0 = \{i \in I ~|~ a_i \neq e \}## (define ##J_0## similarly for the element ##(b_i)_{i \in I} \in \prod_{i \in I} N_i##)
Okay. I am trying to show this is a homomorphism; i.e.,
$$\phi \Bigg((a_i)_{i \in I}(b_i)_{i \in I}) \Bigg) = \prod_{i \in I_0} a_i \prod_{i \in J_0} b_i$$
Note that ##\phi \Bigg((a_i)_{i \in I}(b_i)_{i \in I}) \Bigg) = \prod_{i \in I_0 \cup J_0} a_ib_i##. Since ##I_0 \cup J_0## is finite, we may WLOG take it to be ##\{1,...,n\}##. Hence,
$$\phi \Bigg((a_i)_{i \in I}(b_i)_{i \in I}) \Bigg) = a_1b_1...a_nb_n$$.
This is where I am stuck. I am trying to show that I can I group the ##a_i##'s and ##b_i##'s together through induction, but I am having trouble. For ##n=2##, we have ##a_1b_1a_2b_2 = a_1b_1a_2b_1^{-1}b_1b_2 = a_1 a_3b_1b_2##, which is allowed because these group elements come from normal subgroups, and this is close to what we want. However, ##\phi ((a_i)_{i \in I})) \neq a_1a_3##!
If I am not mistaken, condition (ii) implies commutativity among the family of normal subgroups of ##\{N_i \}##, but I am not sure how to utilize this in my proof. Perhaps I am just too tired at the moment...