Abstract Algebra: Groups of order 21

[vsage]

I was given a problem to prove there are at most 3 groups of order 21, with extra credit for proving there are at most 2. I am pretty stuck on this one but here is what I have so far:

Suppose G is a group of order 21
Let K be a sylow 3-subgroup of G and let H be a sylow 7-subgroup of G.

By Sylow's third theorem, H is a normal subgroup because it can only be expressed one way such that x (mod p) = 1 where x divides the order of G (x = 1 only). By Sylow's Third Theorem, G has either 1 or 7 sylow 3-subgroups.

2 cases now-

For K being a unique sylow 3-subgroup (G has only 1 3-subgroup), H and K are both normal in G. Being prime ordered, they are also both cylic, so let H = <x> and K <y>, then
$$xyx^{-1}y^{-1} = (xyx^{-1})y^{-1} \in Ky^{-1} = K$$ but also
$$xyx^{-1}y^{-1} = x(yx^{-1}y^{-1} \in xH = H$$ so
$$xyx^{-1}y^{-1} \in K \cap H = \{e\}$$
which essentially means for $$xy = yx \in HK$$

The last conclusion coming from the fact that H and K are relatively prime so no elements besides the identity overlap. That being established, HK is an abelian group

-I have 2 problems now. First of all, I don't know how to tie HK back to G, and second of all I'm not sure how to proceed if K is not a normal subgroup (ie 7 subgroups of order 3 exist). Any advice would be much appreciated!

Edit: I proved that G cannot have 7 subgroups of order 3, but the first question still remains. Thanks!

 

[thepatientmental]

First of all, great job on using Sylow's third theorem to establish the normality of H and K in G. This is an important step in proving the uniqueness of groups of order 21.

To address your first problem, let's look at the case where K is a unique sylow 3-subgroup. In this case, as you have correctly shown, HK is an abelian group. Now, since G has order 21, we know that there must be a unique subgroup of order 7 as well. Let's call this subgroup M. Since M is also a sylow 7-subgroup, it must also be normal in G. Therefore, we can write G as the direct product of HK and M, i.e. G = HK × M.

Now, let's consider the elements of G. Since HK is abelian, we know that every element of G can be written as a product of an element of HK and an element of M. So, if we take any two elements of G, say g1 and g2, we can write them as g1 = h1m1 and g2 = h2m2, where h1, h2 ∈ HK and m1, m2 ∈ M. Now, if we multiply g1 and g2, we get g1g2 = h1m1h2m2. Since M is normal in G, we know that m1h2 and h1m2 are also in M. Therefore, g1g2 can be written as g1g2 = h1h2'm, where h2' ∈ HK and m ∈ M. This shows that G is actually abelian, since any two elements in G commute.

To address your second problem, let's consider the case where there are 7 sylow 3-subgroups in G. In this case, let's call these subgroups K1, K2, ..., K7. Now, since these subgroups have prime order, we know that they are cyclic. Let's choose a generator for each of these subgroups, say x1, x2, ..., x7. Now, we can write each element of G as a product of an element of K1 and an element of K2, and so on. So, if we take any two elements of G, say g1 and g2, we can write them as g1 =