"Here's a thought" or "Final Showdown"

[leski]

Okay, so I'm coming back to Ottawa on Monday the 22nd, and leaving on the 1st. Not counting the 1st ('cause I'm leaving then to go back for Gypsy), that makes 10 days. I also hand in my take-home exam on Friday, so that leaves Sat and Sun, two days, where I'm staying in Kingston to get some extra shifts in (read: $$$$). So that makes 12 days of

Comments

[vorpal]

I struggled with parts of the first question for awhile, and deduced from the theorems in my Advanced Modern Algebra book that G must contain a Sylow 5-subgroup, but I couldn't show that, given that G !~= A_5, that it was unique (which would imply that it's normal, obviously).

I believe that this page explains how this works, but I didn't read it thoroughly and it may be a bit over my head: link (I think I do have some catching up to do on group theory...)

I'm still unsure, though... if G has 6 Sylow 5-subgroups, and this implies that it's simple, does this imply that it's isomorphic to A5? I know what A5 is, obviously, but I've never examined the structure of it... I'm assuming that An < A5 for 1 <= n <= 4... are none of them normal subgroups of A5?

Yeah, I used some of the flowchart proof in my paper.

[leski]

Okay, if G has 6 Sylow 5-Subgroups, then the flowchart (which I don't think is a good proof, I didn't use all of it) says that it's simple. There're a few facts that you need now:
1) A5 is simple, and the only simple (and non-abelian) group of order 60, up to isomorphism (this was proved in one of the lectures)
2) G is of order 60 and simple.

So since |G| = |A5| and A5 is the only possible group for G to be isomorphic to is A5. Therefore, G ~= A5. Woo.