Suppose H and K are subgroups of G with H normal in K, |H||K| = |G|, and the intersection of H and K being identity. Then HK = G. Since HK is the union of hK for all h in H and since hK = h'K iff h = h', wouldn't the set of cosets of K be {hK : h in H}? Also, wouldn't this form a group isomorphic to H? I have a theorem which states that if the cosets of a subgroup, in this case K, form a group under the normal multiplication, then K is normal in G. Well the hypothesis seems to be satisfied, so K would indeed be normal in G, no? But we haven't assumed K is normal, so K being normal would have to be a consequence of the assumptions, but it seems like an unlikely consequence. Is my gut feeling off, or have I made a mistake in the reasoning above?(adsbygoogle = window.adsbygoogle || []).push({});

If the reasoning is correct, then if K and H are Sylow subgroups, then since K is normal, K will be the only subgroup of its order. This seems even less likely, since it seems to suggest that a group of order p^{n}q^{m}for primes p and q and natural n and m will only have one Sylow-p subgroup and one Sylow-q subgroup. It seems to me, for some reason, that we should be able to make such a group that has multiple Sylow-p or Sylow-q subgroups.

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# Normal subgroup or not?

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