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How can I prove that there is no simple group of order $448=2^6\cdot 7$? I tried with Sylow's theorems, I proved that (if $G$ is simple) the number of 2-Sylows is 7 and that the number of 7-Sylows is 8 or 64, but I don't know how to continue, could you help me please?

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(There is a monthly maximum for questions... :D ) – Mariano Suárez-Alvarez Jan 3 '12 at 5:36
Hint: If there are 64 Sylow 7-subgroups, consider how may elements of order 7 there are. – Geoff Robinson Jan 3 '12 at 5:45
@Alex: Well, if there are 384 elements of order $7,$ how many Sylow $2$-subgroups can there be? (Actually, I see a more direct approach to the question than this anyway). – Geoff Robinson Jan 3 '12 at 5:53
only 1, right thank you – Alex M Jan 3 '12 at 5:59
@Jack, Jyrki: I do not believe the statement about groups of order $q^{n}p.$ For example, the symmetric group $S_4$ has order $2{3}.3,$ and a Sylow $3$-subgroup normalizes the normal Klein $4$-group, but not a whole Sylow $2$-subgroup (there is noting special about these primes for this question). – Geoff Robinson Jan 3 '12 at 9:12
up vote 6 down vote accepted

Let $n_2$ be the number of $2$-Sylow subgroups of $G.$ Then, $n_2$ is odd and divides $7.$ If $n_2=1$ we're done, but if $n_2=7$, by Sylow theorem, conjugation of these seven $2$-Sylow subgroups defines a homomorphism $G \to S_7.$ The kernel of this homomorphism cannot be trivial so we're done.

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why the kernel cannot be trivial? – Alex M Jan 3 '12 at 7:42
+1: This works for groups of size $2^6\cdot 7$, because that number does not divide $7!$. But curiously it wouldn't work for $2^4\cdot 7$ :-) – Jyrki Lahtonen Jan 3 '12 at 7:45

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