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I am revising for my group theory exam and am trying to prove there are no simple groups of order 560. I have actually seen a solution to this problem and it goes as follows.

Assume G is simple. $|G| = 560 = 2^4 \times 5 \times 7. \ \ $ Let $P = \langle x \rangle \in Syl_7(G), \ $then for $k_7 = |Syl_7(G)|, $ we have $k_{7} = 8 \ $. So $\ |N_G(P)| = 70$. Since |Aut(P)|= $6$ and $N_G(P)/C_G(P)$ is isomorphic to a subgroup of Aut$(P) \ $, $C_G(P)$ has an element $y$ of order $5$, and then $xy$ has order $35$. But G $\leq A_8$ and $A_8$ has no element of order $35$ so we have a contradiction.

This all makes sense apart from when it says $C_G(P)$ has an element $y$ of order $5$. Why is this the case?

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$540 \neq 560$. – user21436 Apr 21 '12 at 0:48
Thanks for pointing that out! – Alex Kite Apr 21 '12 at 0:50
Are we trying to prove there is no simple group of order 540 or 560? – user21436 Apr 21 '12 at 0:53
560, sorry for the confusion. – Alex Kite Apr 21 '12 at 0:55
$|N_G(P)| = 2 \cdot 5 \cdot 7$ and $|N_G(P)|/|C_G(P)|$ divides $2 \cdot 3$ (so it must be $2$ or $1$), so $5 \cdot 7$ divides $|C_G(P)|$. Then use Cauchy's theorem. – Dylan Moreland Apr 21 '12 at 0:56
up vote 1 down vote accepted

I'll write an answer based on the comment by Dylan, and we shall have one suffering unanswered question more out of its misery:

Let $\,P\in Syl_7(G)\,\Longrightarrow P=\langle\,x\,\rangle\,\,,\,|P|=7\,$ . Suppose $\,G\,$ is simple and thus $\,P\,$ is not normal in $\,G\,\Longrightarrow n_7=8\,\,,\,\text{with}\,\,n_p=\,$ number of Sylow $p$-subgroups in $\,G\,$ , since $\,8\,$ is the only number both congruent to $\,1\pmod 7\,$ and also dividing $\,|G|=560\,$.

We thus have $$n_7=8=[G:N_G(P)]=\frac{|G|}{|N_G(P)|}\Longrightarrow |N_G(P)|=70=2\cdot 5\cdot 7$$Since $\,|Aut(P)|=6\,$ and $\,N_G(P)/C_G(P)\,$ isomorphic to a subgroup of $\,Aut(P)\,$ , we get $$\left|N_G(P)/C_G(P)\right|=\frac{|N_G(P)|}{|C_G(P)|}=\frac{70}{|C_G(P)|}\mid 6\Longrightarrow |C_G(P)|\in\{35,70\}\Longrightarrow 5\mid |C_G(P)|$$so $\,\exists\,y\in C_G(P)\,\,s.t.\,\,\operatorname{ord}(y)=5\,$ , by Cauchy's Theorem, and since $\,xy=yx\,\,\,and\,\,\,\operatorname{ord}(x)=7\,$ , we get that $\,\operatorname{ord}(xy)=35$ . $\;\;\;\;(**)$

As $\,G\,$ has a subgroup of index $\,8\,$ , as noted above, the action of $\,G\,$ on the left cosets of this subgroup (left multiplication) determines a monomorphism of $\,G\,$ into $\,S_8\,$ (as $\,G\,$ is simple), and thus it must be that in fact $\,G\leq A_8\,$ (otherwise, within $\,S_8\,\,,\,\,G\cap A_8\,$ has index $\,2\,$ in $\,G\,$).

But a simple check with the possible lengths of disjoint cycles reveals at once that there is no element of order $\,35\,$ in $\,S_8\,$ , let alone in $\,A_8\,$, contradicting $\;(**)$ .

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