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Math Help - Normal Subgroups

  1. #1
    Member elninio's Avatar
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    Normal Subgroups

    Let H and K be normal subgroups of G such that H∩K = {e} and HK = G.

    Prove that G ∼= H K
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  2. #2
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    Quote Originally Posted by elninio View Post
    Let H and K be normal subgroups of G such that H∩K = {e} and HK = G.

    Prove that G ∼= H K
    Hint - Can you define an isomorphism from G->HxK. (Use the fact G=HK and H∩K={e})
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  3. #3
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    Quote Originally Posted by elninio View Post
    Let H and K be normal subgroups of G such that H∩K = {e} and HK = G.

    Prove that G ∼= H K

    First show that if h_1\,,\,h_2\in H\,,\,\,k_1\,,\,k_2\in K\,,\,\,then\,\,h_1k_1=h_2k_2\Longleftrightarrow h_1=h_2\,,\,k_1=k_2 , next define f: G\rightarrow H\times K \,\,\,as\,\,\,f(g=hk)=(h,k) and show f is a isomorphism of groups.

    Tonio
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  4. #4
    Member elninio's Avatar
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    Tonio, I'm not sure if this is the same thing as what you've posted but i've worked on this problem for over an hour and I know now that I can simply show that f is in isomorphism in a map defined f: H x K -> G by f((h,k)) = hk (Where hk is in G).

    Does this seem correct so far? This is where I am having my trouble. I worked on one isomorphism problem similar to this before but I cant seem to apply any of the same methods or logic to this one.

    How do I show f is an isomorphism in this problem?
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  5. #5
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    Quote Originally Posted by elninio View Post
    Tonio, I'm not sure if this is the same thing as what you've posted but i've worked on this problem for over an hour and I know now that I can simply show that f is in isomorphism in a map defined f: H x K -> G by f((h,k)) = hk (Where hk is in G).

    Does this seem correct so far? This is where I am having my trouble. I worked on one isomorphism problem similar to this before but I cant seem to apply any of the same methods or logic to this one.

    How do I show f is an isomorphism in this problem?

    After you show what I proposed in my prior post, you have:
    f\left((h_1,k_1)+(h_2,k_2)\right)=f(h_1h_2,k_1k_2)  =h_1h_2k_1k_2=h_1k_1h_2k_2=f(h_1,k_1)f(h_2,k_2) since by now it must be clear H, K commute POINTWISE.
    Also, f(h,k)=hk=1=1\cdot 1 \Longleftrightarrow h=1=k\Longrightarrow\,Ker(f)=1 , and surjectivity is immediate.

    Tonio
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