# Normal subgroup or not?

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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?

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 pnqm 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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sounds fishy to me. in any group, a subgroup of index 2 is normal. so are you asking whether in that case, any element of order 2 not in that subgroup generates a normal subgroup?

take S(5) and the normal subgroup A(5), and take any transposition x. Then it would seem that (x) has order 2, meets A(5) =only in the dientity, but I doubt that (x) is normal in S(5).

you seem to be making a big leap oif faith that the set of cosets of K is a group isomorphic to H, just because they have the same cardinaloity as H. whats your arguemnt? to show two things are isomorphic you have to have an isomorphism, and here you have to define the group structure on the cosets. try defining one.

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Well since HK = G and HK is the union of hK for all h in H, the set {hK : h in H} is a collection of cosets of K which cover G, and each hK is unique since there are |G|/|K| = |H| unique cosets of K and |H| unique elements of H. I would think the mapping:

f : h |--> hK

would give an isomorphism. Clearly, it is homomorphic since:

f(hh') = hh'K = (hK)(h'K) = f(h)f(h')

where multiplication of cosets is defined in the natural way. f is clearly surjective, and since we're dealing with finite groups, it is bijective, so it is an isomorphism. The set {hK : h in H} does form a group, with multiplication given by (hK)(h'K) = (hh')K, the identity being K, the inverse of hK being h-1K, and associativity following from the associativity of multiplication in H.

However, you're right, about a subgroup generated by a transposition not being a normal subgroup of S5 but satisfying the conditions for K (choosing H to be A5).

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don't you have to check that the set of cosets is a group with your operation?

i.e. why is (hK)(h'K) = hh'K a well defined group operation?

suppose that hK = gK, i.e. h-1gK = K. i.e. h-1g belongs to K.

then you need hh'K = (hK)(h'K) = (gK)(h'K) = gh'K.

so you need for h'-1 h-1 gh' to be long to K. you know that h-1g belongs to K, but why should h'-1kh' belong to K when k does? this is exactly the definition of normal.

so your group operation is not even an operation unless you know K is normal.

so the cosets do not even form a monoid, much less a group, i.e. there is no natural binary operation on them at all.

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mathwonk said:
don't you have to check that the set of cosets is a group with your operation?

i.e. why is (hK)(h'K) = hh'K a well defined group operation?

suppose that hK = gK, i.e. h-1gK = K. i.e. h-1g belongs to K.

then you need hh'K = (hK)(h'K) = (gK)(h'K) = gh'K.

so you need for h'-1 h-1 gh' to be long to K. you know that h-1g belongs to K, but why should h'-1kh' belong to K when k does? this is exactly the definition of normal.

so your group operation is not even an operation unless you know K is normal.

so the cosets do not even form a monoid, much less a group, i.e. there is no natural binary operation on them at all.
Great, thanks. I had a feeling it had something to do with checking whether the product was well-defined but I wasn't sure what I actually had to do to check that. Thanks.