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I have absolutely no idea how to go about this. I have started to find how many sylow p subgroups there are.

No simple group of order $300$

I've looked at this but I don't know why they've even started the answer in the way they have.

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ZZS14
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  • I don't understand those arguments though, so was hoping someone could go through it again with me afresh. – ZZS14 Dec 03 '13 at 19:46
  • Do look at the link. But also note that 300 = $10^2$ * 3. Should there be subgroups of those orders? What kind of groups are they? – Betty Mock Dec 03 '13 at 19:46
  • @Emily If you don't understand the answers given there. Then you should rephrase your question in terms of what you are confused about in the solutions given in the link and include the link in your question. That way, a duplicate is avoided and you get the specific help you require. – mathematics2x2life Dec 03 '13 at 19:48
  • cyclic groups? i really have no idea – ZZS14 Dec 03 '13 at 19:48
  • @Emily Well,the 3 subgroup is cyclic. The 100 group is abelian. It was a thought off the top of my head -- perhaps can be pursued farther. However the link is good. Do you know what the index of a group is? Can you see they are referring to theorem about normal subgroups. See if you can ask us a specific question. – Betty Mock Dec 03 '13 at 19:51
  • ive just found a theorem in my notes Theorem 3.18 Suppose that the group G has a cyclic Sylow- 2 -subgroup. Then G is not simple. i dont know if it works for |G|=300 as i havent worked it through yet, but if i were to find it had a slow 2 subgroup is that then enough to say its not simple? – ZZS14 Dec 03 '13 at 19:55
  • aren't all sylow 2 subgroups cyclic? – ZZS14 Dec 03 '13 at 19:55
  • @BettyMock why should there be a subgroup of order $100$? And why would it be abelian? – Tobias Kildetoft Dec 03 '13 at 19:57
  • Emily: I am curious about how your notes prove that, as I am only familiar with a rather advanced proof of that fact. – Tobias Kildetoft Dec 03 '13 at 19:59
  • proof: Let G act on itself by right multiplication, and let g be a generator of a Sylow- 2 -subgroup, of order 2r. Then |G| = b·2r with b odd. Now express g as a product of cycles, viewed as an element g ∈ Sym(G). Then all cycles appearing in g must have length 2r, as no power ̸= 1G of g fixes any element, and there must be b such cycles. But this means that g is an odd permutation, hence not contained in Alt(G). So if we view G as a subgroup of Sym(G) then G ∩ Alt(G) is a subgroup of index 2 in G, and hence is normal in G. – ZZS14 Dec 03 '13 at 20:00
  • Ahh, neat. I had only seen it proven as a special case of Burnside's transfer theorem. – Tobias Kildetoft Dec 03 '13 at 20:04
  • The only theorems i have in my notes about a group being simple or not i cant see how they apply to this – ZZS14 Dec 03 '13 at 20:14
  • You still haven't explained which part of the answers to the other instance of this question are giving you problems. – Tobias Kildetoft Dec 03 '13 at 20:16
  • all of it, i have no idea why they've done what theyve done past saying 300=2.2.3.5 – ZZS14 Dec 03 '13 at 20:18
  • @tobias my error, 10 is not a prime. why I was thinking it is I have no idea ... – Betty Mock Dec 04 '13 at 05:28
  • @Emily they note that 300 = $2^235^2$. So there is a 5-Sylow subgroup. The next step says that if G is simple there are 6 of these 5-Sylow subgroups. What theorem are they using to conclude this? Do you have that in your notes? Can you find it online in a search about Sylow subgroups? If you can get this far, we can tackle the next step. – Betty Mock Dec 04 '13 at 05:35

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