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Math Help - Normal group problem

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    Normal group problem

    Let N be normal in G, and let H be a subgroup of G. Prove that NH is a subgroup of G.

    I'm a confused about NH, what kind of element belong to this set?
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    Quote Originally Posted by tttcomrader View Post
    Let N be normal in G, and let H be a subgroup of G. Prove that NH is a subgroup of G.

    I'm a confused about NH, what kind of element belong to this set?
    NH means the set of all products of an element of N times an element of H. So an element of NH is of the form nh, where n is in N and h is in H.
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    Quote Originally Posted by tttcomrader View Post
    Let N be normal in G, and let H be a subgroup of G. Prove that NH is a subgroup of G.

    I'm a confused about NH, what kind of element belong to this set?
    The set NH = \{nh| n\in N \mbox{ and }h\in H\}. If N,H are subgroups of G then not necessarily does it means NH is a subgroup. If however, N is normal, i.e. gng^{-1} \in N for any n\in N and g\in G then NH is a group.

    However, it turns out that N need not be normal in G. It can have a weaker property but yet NH.

    Theorem: If N has the property that hnh^{-1} \in N for any n\in N and h\in H then NH is a group.
    (Do you see why this would prove you case? Meaning, if N is normal then certainly hnh^{-1}\in N. So of course it works).

    Proof: All I will show is closure of NH which is the only part which takes a little work. Let x_1 \in NH and x_2\in NH we want to show x_1x_2\in NH. But this means x_1 = n_1h_1 and x_2 = n_2h_2 for n_i \in N and h_i \in H. Thus, x_1x_2=n_1h_1n_2h_2. But is this product in NH? Not necessarily, so we need to use the additional condition in the theorem. Which says h_1n_2h_1^{-1}\in N, i.e. h_1n_2h_1^{-1} = n_3\implies h_1n_2 = n_3h_1 for some n_3\in N. Thus, n_1h_1n_2h_2 = n_1(h_1n_2)h_2 = n_1(n_3h_1)h_2 = (n_1n_3)(h_1h_2) which certainly is in NH. Q.E.D.
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