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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- October 10th 2008, 10:23 AMdori1123index
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).

- October 10th 2008, 10:57 AMThePerfectHacker
- October 10th 2008, 12:48 PMdori1123
that's the hint given in the book, but I don't get it...

- October 10th 2008, 01:18 PMThePerfectHacker
Let be a left coset where . We know that for some since is a complete collection of cosets. Therefore for some . Now for some since is a complete collection of cosets. Therefore, for some . Thus, . We have shown that contains all cosets of in . Therefore, there are

__at most__cosets of in . The reason why at most and not exactly is because it is possible that different 's and 's can produce the same coset. It remains to show that if then . Can you finish the last step? - October 10th 2008, 05:05 PMdori1123
- October 11th 2008, 04:36 PMThePerfectHacker