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:(adsbygoogle = window.adsbygoogle || []).push({});

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

[tex]xyx^{-1}y^{-1} = (xyx^{-1})y^{-1} \in Ky^{-1} = K[/tex] but also

[tex]xyx^{-1}y^{-1} = x(yx^{-1}y^{-1} \in xH = H[/tex] so

[tex]xyx^{-1}y^{-1} \in K \cap H = \{e\}[/tex]

which essentially means for [tex]xy = yx \in HK[/tex]

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!

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# Homework Help: Abstract Algebra: Groups of order 21

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