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

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

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

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

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