Group Isomorphic to Weak Product of Normal

[Bashyboy]

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...

 

[fresh_42]

Can you define a weak product? Also, why do you have to introduce $I_0$ and $J_0$? $a_i=e$ shouldn't change the argument. And you have nowhere mentioned that $I$ is countable. So how would an induction help? And if, then I wouldn't omit the automorphisms involved in semi-direct products, i.e. $a_3=\sigma_1(b_1)(a_2)$. Not sure whether this helps or is more of a burden, if all of them are used in the definition of $\phi$. My guess is, that a categorial proof could be easier, and that's why I asked for a proper definition of the weak product, preferably in terms of commuting diagrams.

 

[Bashyboy]

I introduce (actually, my book does) $I_0$ and $J_0$ so I know which components of $(a_i)_{i \in I}$ and $(b_i)_{i \in I}$ are nonidentity; and because I thought that I needed to to induction on the size of $I_0 \cup J_0$. The set $I$ is not necessarily countable, but I am not doing induction on $I$ but rather on the finite set $I_0 \cup J_0$.

Here is the definition of weak product: The weak direct product of a family of groups $\{G_i ~|~ i \in I\}$, denoted by $\prod_{i \in I}^w G_i$, is the set of all $f \in \prod_{i \in I} G_i$ such that $f(i) = e_i$, the identity of $G_i$, for all but a finite number of $i \in I$.

 

[zwierz]

such that f(i)=eif(i) = e_i, the identity of GiG_i, for all but a finite number of i∈Ii \in I.

perhaps $f(i)\ne e_i$ for finite number of $i$-s?

 

[Bashyboy]

perhaps $f(i)\ne e_i$ for finite number of $i$-s?

I just checked my book and it is accurate as I have quoted it.

 

[zwierz]

I just checked my book and it is accurate as I have quoted it.

well then perhaps your book explains also what an infinite composition of elements from $G$ is

Regarding your initial question
Let $G= \langle N_1\cup N_2\rangle$ and $a_1,b_1\in N_1,\quad a_2,b_2\in N_2$
Consider the following expression $u=a_1a_2b_1b_2$ By definition of a normal subgroup there exist an element $\xi\in N_1$ such that $a_2b_1=\xi a_2$

 

[fresh_42]

I introduce (actually, my book does) $I_0$ and $J_0$ so I know which components of $(a_i)_{i \in I}$ and $(b_i)_{i \in I}$ are nonidentity; and because I thought that I needed to to induction on the size of $I_0 \cup J_0$. The set $I$ is not necessarily countable, but I am not doing induction on $I$ but rather on the finite set $I_0 \cup J_0$.

The basic problem of your attempt is, that you consider the weak product of the $N_\iota$ as a direct product, i.e. you try to define the multiplication componentwise, whereas the multiplication in $G$ is not. So in order to establish a homomorphism, you have to use the same multiplication rules on both sides, i.e. define an appropriate group structure on $\Pi^\omega_\iota N_\iota$. Maybe it's easier the other way around, but I'm not sure, since you also have to deal with well-definition. In any case I still think you have to keep track of transpositions in $\Pi^\omega_\iota N_\iota$ during multiplication.

 

[Bashyboy]

The basic problem of your attempt is, that you consider the weak product of the $N_\iota$ as a direct product, i.e. you try to define the multiplication componentwise, whereas the multiplication in $G$ is not. So in order to establish a homomorphism, you have to use the same multiplication rules on both sides, i.e. define an appropriate group structure on $\Pi^\omega_\iota N_\iota$. Maybe it's easier the other way around, but I'm not sure, since you also have to deal with well-definition. In any case I still think you have to keep track of transpositions in $\Pi^\omega_\iota N_\iota$ during multiplication.

I am not sure I follow this. I was under the impression that the weak product was a special case of the direct product, so of course multiplication would be defined as it is in the general direct product. My book (Hungerford) actually proves this theorem, but sadly they refrain from proving that the map the use is a homomorphism. I have included in this post a screenshot of Hungerford's proof.