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Math Help - isomorphism group problem

  1. #1
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    Oct 2008
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    isomorphism group problem

    Definition: when H, K are subgroups of G, we define HK to be the set of all elements of G that can be written in the form hk where h is in H and k is in K.
    1) let H be a subgroup of a group G and N be a normal subgroup of G.show that HN is a subgroup of G and N be a normal subgroup of HN.
    2) let H,K and N be a subgroup of a group G, K is normal subgroup of H and N is normal subgroup of G.prove that NK is normal subgroup of NH.
    3) let H1 and H2 be subgroups of a group G and N1 subgroup of H1 and N2 subgroup of H2.then show that
    N1(H1 intersection N2) is normal subgroup of N1(H1 intersection H2)
    and (H1 intersection N2)(H2 intersection N1) normal subgroup of (H1 intersection H2)
    can u help me of these 3 proofs?
    email: captain_mathku@yahoo.com
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  2. #2
    Junior Member
    Joined
    May 2008
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    33
    to prove that HN is a sub group, you need to show that HN is closed under the group operation, it contains the identity and has inverse for each element (associativity is given from G). this is all straight forward with  gng^{-1} \in N, \; \forall g \in G, n \in N . for example - closure  h_1 n_1 h_2 n_2 = h_1 h_2 h_2^{-1} n_1 h_2 n_2 = (h_1 h_2) (n_1' n_2) where  h_2^{-1} n_1 h_2 = n_1'

    2. again, by definition for every nk in NK and n_1 h_1 in NH you have
     (n_1 h_1) nk (n_1 h_1)^{-1} = (n_1 h_1) n (h_1^{-1} h_1) k (h_1^{-1} n_1^{-1}) = n_1 (h_1 n h_1^{-1}) (h_1 k h_1^{-1}) n_1^{-1} = n_1 n' k' n_1^{-1} (k'^{-1} k') = n_1 n' (n_1^{-1})' k' .
    so you get that  (n_1 h_1) (nk) (n_1 h_1)^{-1} \in NK \rightarrow (n_1 h_1) NK (n_1 h_1)^{-1} \subset NK, \; \forall n_1 h_1 \in NH \rightarrow NK \triangleleft NH

    3. I guess you forgot to write the N1,2 are normal in H1,2. anyway the proofs would probably be like the one above
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