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Math Help - direct products problem

  1. #1
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    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\}}.
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  2. #2
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    Quote Originally Posted by deniselim17 View Post
    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.
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