# Are Sylow subgroups unique in groups with specific conditions?

• AKG
In summary, the conversation discusses the conditions for a subgroup K to be normal in a group G using the example of subgroups H and K. The conditions include H being normal in K, |H||K| = |G|, and the intersection of H and K being the identity. The conversation also explores the possibility of K being the only subgroup of its order in a group of order pnqm and the concept of cosets forming a group isomorphic to H. However, it is pointed out that the set of cosets does not form a group unless K is normal in G.
AKG
Homework Helper
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.

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. what's 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.

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).

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.

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.

## 1. What is a normal subgroup?

A normal subgroup is a subgroup of a group that is invariant under conjugation by all elements of the original group. This means that if an element is in the normal subgroup, and another element from the original group is used to conjugate it, the resulting element will also be in the normal subgroup.

## 2. How can you determine if a subgroup is normal or not?

One way to determine if a subgroup is normal is by checking if it is invariant under conjugation by all elements of the original group. Another way is by using the quotient group and checking if the cosets of the subgroup are also subgroups.

## 3. What is the significance of a normal subgroup?

Normal subgroups are important in group theory because they allow us to define quotient groups, which are essential in understanding the structure of a group. Normal subgroups also help in simplifying the analysis of a group by reducing it to a simpler quotient group.

## 4. Can a subgroup be both normal and not normal at the same time?

No, a subgroup can only be either normal or not normal. It cannot be both at the same time. A subgroup is considered normal if it is invariant under conjugation by all elements of the original group, and not normal if it is not.

## 5. How can normal subgroups be used in real-life applications?

Normal subgroups have various applications in fields such as physics, chemistry, and cryptography. In physics, they are used to understand the symmetries of physical systems. In chemistry, they help in predicting the symmetry of molecules. In cryptography, they are used to construct secure encryption algorithms and protocols.

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