1. ## direct products problem

I'm stuck at this question for more than 24 hours. This question is from "THE THEORY OF GROUPS: An Introduction" by Joseph J. ROTMAN, Exercise 4.11.

Here is the question:

If $G$ is a group with normal subgroups $H_{1},H_{2}, ... ,H_{m}$, then $G=\prod_{i=1}^{m} H_{i}$(internal) if and only if $G=\langle \, \bigcup_{i=1}^{m} H_{i} \rangle$ and, for all $j$, $H_{j} \bigcap \, \langle \, \bigcup_{i \ne j} H_{i} \rangle = {\{1\}}$.

2. Originally Posted by deniselim17
I'm stuck at this question for more than 24 hours. This question is from "THE THEORY OF GROUPS: An Introduction" by Joseph J. ROTMAN, Exercise 4.11.

Here is the question:

If $G$ is a group with normal subgroups $H_{1},H_{2}, ... ,H_{m}$, then $G=\prod_{i=1}^{m} H_{i}$(internal) if and only if $G=\langle \, \bigcup_{i=1}^{m} H_{i} \rangle$ and, for all $j$, $H_{j} \bigcap \, \langle \, \bigcup_{i \ne j} H_{i} \rangle = {\{1\}}$.
i don't have the book right now but i think Rotman's definition is this: we say $G$ is the internal direct product of some normal subgroups $H_i, \ i=1, \cdots , m,$ if $G=H_1 H_2 \cdots H_m,$ and

for all $j: \ \ H_j \cap (H_1 \cdots H_{j-1}H_{j+1} \cdots H_m)=\{1\}.$ so the only thing you need to prove is that if we have a bunch of normal subgroups $H_i, \ 1 \leq i \leq m,$ of a group G, then we will have:

$H_1H_2 \cdots H_m = \langle \, \bigcup_{i=1}^{m} H_{i} \rangle,$ which is very easy to prove: since $H_j$ are normal, $H_1H_2 \cdots H_m$ is a subgroup, which contains every $H_j.$ hence it contains the union of them. but by definition

$\langle \, \bigcup_{i=1}^{m} H_{i} \rangle$ is the smallest subgroup which contains the union of $H_j.$ hence $\langle \, \bigcup_{i=1}^{m} H_{i} \rangle \subseteq H_1H_2 \cdots H_m.$ on the other hand, clearly every $H_j$ is contained in $\bigcup_{i=1}^m H_i \subseteq \langle \, \bigcup_{i=1}^{m} H_{i} \rangle.$ since $\langle \, \bigcup_{i=1}^{m} H_{i} \rangle$ is

a subgroup, it must also contain the product of $H_j$, i.e. $H_1H_2 \cdots H_m \subseteq \langle \, \bigcup_{i=1}^{m} H_{i} \rangle.$ this completes the proof.