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    I am given that G1, G2 are finite and K is a subgroup of G1*G2

    Also H1={g in G1 s.t. (g,e) in K) and H2={g in G2 s.t. (g,e) in K)

    I also have been given that |G1| and |G2| are coprime

    I have already shown that H1 is a subgroup of G1 and H2 is a subgroup of G2 and H1*H2 is a subgroup of K.

    I need to show that K=H1*H2.

    I tried to turn this into a pure number theory problem, lettting |H1|=h1 etc so

    h1|g1, h2|g2, h1*h2|k, g1 and g2 are coprime

    but this was not enough to show h1*h2=k (which is what I need).
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    You need to show K is a subgroup of H_1 \times H_2 which combined with the fact H_1 \times H_2 is a subgroup of K will give you that K = H_1 \times H_2
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    (Original post by Noble.)
    You need to show K is a subgroup of H_1 \times H_2 which combined with the fact H_1 \times H_2 is a subgroup of K will give you that K = H_1 \times H_2
    How can I show that K is a subgroup of H1*H2?
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    Let (g_1,g_2) \in K be arbitrary, you want to show that g_1 \in H_1 and g_2 \in H_2

    Let o(g_2) = m then (g_1, g_2)^m = (g_1^m, g_2^m) = (g_1^m, e) \in K as K is closed under multiplication

    Also (g_1^m,e) gives you that g_1^m \in H_1

    | \langle g_1 \rangle | = o(g_1) and | \langle g_2 \rangle | = o(g_2)

    o(g_1) | G_1 and o(g_2) | G_2

    However, since \gcd \{G_1,G_2\} = 1 this implies \gcd \{o(g_1), o(g_2)\} = 1

    Then if you let o(g_1) = p by Bezout's there exists integers u,v such that up+vm=1

    Now, you want to use Bezout's to show that (g_1,e) \in K implies g_1 \in H_1 and (e,g_2) \in K implies g_2 \in H_2
 
 
 
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