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Math Help - normal subgroup proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    normal subgroup proof

    Q: Prove that if H and K are normal subgroups of a group G, then HK\leq\\G.

    A: Since H\leq\\G and k\leq\\G, H and K have the same identity element e. Thus, ee=e\in{HK} which shows HK is not empty.

    Now, let x,y\in{HK}. Then x=ab, y=cd and y^{-1}=(cd)^{-1}=d^{-1}c^{-1}. So,

    xy^{-1}=(ab)(cd)^{-1}=(ab)c^{-1}d^{-1}=a(bc^{-1})d^{-1}.

    If we let x=y, then xy^{-1}=e\in{HK}. So, HK is a subgroup of H.

    Now, to show it is a normal subgroup, do I just compute xHK and show it is equal to HKx?

    I am not sure if I am showing enough in the proof I wrote down. Some help would be great.

    Thanks
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  2. #2
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Danneedshelp View Post
    Q: Prove that if H and K are normal subgroups of a group G, then HK\leq\\G.

    A: Since H\leq\\G and k\leq\\G, H and K have the same identity element e. Thus, ee=e\in{HK} which shows HK is not empty.

    Now, let x,y\in{HK}. Then x=ab, y=cd and y^{-1}=(cd)^{-1}=d^{-1}c^{-1}. So,

    xy^{-1}=(ab)(cd)^{-1}=(ab)c^{-1}d^{-1}=a(bc^{-1})d^{-1}. Mistake here. Read your line above! In any case, what is the conclusion of this calculation? You want to show that xy^{-1} \in HK.

    If we let x=y, then xy^{-1}=e\in{HK}. So, HK is a subgroup of H.

    Now, to show it is a normal subgroup, do I just compute xHK and show it is equal to HKx? Yes! You can just use the fact that xHKx^{-1}=(xHx^{-1})(xKx^{-1})...

    I am not sure if I am showing enough in the proof I wrote down. Some help would be great.

    Thanks
    .
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Quote Originally Posted by Bruno J. View Post
    .
    xy^{-1}=abd^{-1}c^{-1}=(ab)(d^{-1}c^{-1}). Since K is a subgroup, d^{-1}c^{-1}\in{K} whenever c,d\in{K}. So, xy^{-1}\in{K}
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Danneedshelp View Post
    xy^{-1}=abd^{-1}c^{-1}=(ab)(d^{-1}c^{-1}). Since K is a subgroup, d^{-1}c^{-1}\in{K} whenever c,d\in{K}. So, xy^{-1}\in{K}
    But didn't you write y=cd so that c \in H, d \in K? Same as x=ab with a \in H, b\in K...

    You'll certainly need to use the fact that the two subgroups in question are normal. It's not true in general that HK is a subgroup (though it's true if either is normal).
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