Prove that there does not exist a simple group of order 760. I was trying to solve it by using Sylow's Theorem but I am unable to prove it.
Thanks for any help.
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Well, what did you conclude from Sylow's theorems? – Tobias Kildetoft Apr 23 '13 at 17:45
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@Tobias For k=0 conclusion is fine,but what about for k=1 where 19k+1|40. – Ester Apr 23 '13 at 17:49
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What do you mean by $k$? – Tobias Kildetoft Apr 23 '13 at 17:50
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n_19 congruent to 1(mod p) by Sylow's Theorem – Ester Apr 23 '13 at 17:59
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2Ok, so you can assume that the group has $20$ $19$-Sylow subgroups. How many $5$-Sylow and $2$-Sylow subgroups can it have? – Tobias Kildetoft Apr 23 '13 at 18:01
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This is a bit more subtle than a straightforward application of Sylow's theorem. Here is a brief sketch (omitting the details) of one possible proof.
Suppose $G$ is simple. If $P$ is a Sylow $19$-subgroup, then $[G:N_G(P)]=20$ so that $G\hookrightarrow S_{20}$. In fact we may assume $G\hookrightarrow A_{20}$, else $G\cap A_{20}$ is a nontrivial normal subgroup of $G$.
Let $x\in N_G(P)\setminus C_G(P)$. Then theorem $(R)$ here shows that $x$ has at most $2$ fixed points. We also know that $x$ must have order $2$ or $19$. These two facts together show that either $x$ is a $19$-cycle or $x$ is the product of $9$ disjoint transpositions. We are done in the latter case because $x$ is an odd permutation, a contradiction.
If $x$ is a $19$-cycle, then theorem $(P)$ of the above reference shows that $|N_{A_{20}}(P)|=171$. But we know that $|N_G(P)|=38$, contradicting Lagrange's theorem.
Jared
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If x is not in the centralizer, then x has order 2 (P is abelian). We need to know that such an x exists (that N≠C); this is Burnside's transfer theorem. – Jack Schmidt Apr 24 '13 at 20:33
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If N=C, then G also has a normal subgroup of size 2, which can avoid BTT (which gives a normal subgroup of index 19). – Jack Schmidt Apr 24 '13 at 20:34