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  1. #1
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    index

    If H is a subgroup of K, K is a subgroup of G, then |G:H| = |G:K||K:H| (do not assume G is finite).
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    Quote Originally Posted by dori1123 View Post
    If H is a subgroup of K, K is a subgroup of G, then |G:H| = |G:K||K:H| (do not assume G is finite).
    Let \{ a_i K | 1\leq i\leq n\} be cosets of K in G and let \{ b_j H | 1\leq j \leq n \} be cosets of H in K.
    Show that all the cosets of H in G are given by \{ a_i b_j H\}.
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    that's the hint given in the book, but I don't get it...
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    Quote Originally Posted by dori1123 View Post
    that's the hint given in the book, but I don't get it...
    Let aH be a left coset where a\in G. We know that a \in a_i K for some a_i \in G since \{ a_iK\} is a complete collection of cosets. Therefore a = a_i b for some b\in K. Now b\in b_j H for some b_j \in K since \{ b_j H\} is a complete collection of cosets. Therefore, b = b_j h for some h\in H. Thus, a = a_i b = a_i b_j h \implies a \in a_i b_j H. We have shown that \{ a_i b_j H | 1\leq i \leq n , 1\leq j \leq m \} contains all cosets of H in G. Therefore, there are at most n\cdot m cosets of H in G. The reason why at most and not exactly is because it is possible that different a_i's and b_j's can produce the same coset. It remains to show that if a b H = a' b'H then a = a',b=b'. Can you finish the last step?
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    Quote Originally Posted by ThePerfectHacker View Post
    It remains to show that if a b H = a' b'H then a = a',b=b'. Can you finish the last step?
    So if abH = a'b'H, then abh_1 = a'b'h_2 for some h_1, h_2 \in H, and so abh_1h_2^{-1}(b')^{-1}(a')^{-1} = 1 \iff h_1h_2^{-1} = 1, b(b')^{-1} = 1, and a(a')^{-1} = 1. So a = a' and b = b'.
    Is this correct? does it make sense?
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    Quote Originally Posted by dori1123 View Post
    Is this correct? does it make sense?
    This is how to do it. If abH = a'b'H then it means aK = a'K (why?) and therefore a = a'. From here it follows that b=b'.
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