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So far I have tried the following: By Sylow's theorem there must be 1 or 8 $Syl_7 (G)$. If there is only 1, we are done since this would be the normal subgroup of size 49. So suppose there are 8. Let G act on the set of 8 by conjugation this produces a homomorphism $\phi : G \rightarrow S_8$

Claim: $|Ker(\phi)|$ = 49 or 7

$Im( \phi ) \le S_8$ . So $|Im(\phi)|$ divides 8! Because 392 does not divide 8! the mapping cannot be injective. So $Ker(\phi)$ is not trivial

Size of $Ker( \phi)$, Size of $Im( \phi) $, Possible?

2, 196, not possible

4, 98, not possible

8, 49, not possible

7, 56, possible

49, 8, possible

14, 28, ?

28, 14, ?

56, 7, ?

The first 3 combinations are impossible because the resulting image doesn't divide the 8!. I am having a hard time coming up with reasons for why the last 3 cannot occur.

For $Im( \phi ) = 7$ would it be correct to assert that since this would technically fix one of the $Syl_7(G)$ the fixed one would be normal?

I have no idea for $Im( \phi )$ = 14 and 28 though. Any help would be much appreciated!

Math Lady
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1 Answers1

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Consider the image of $\phi$. As $G$ acts transitively on the Sylow $7$-subgroups of $G$ then the image of $\phi$ is a transitive subgroup of $S_8$ and so its order $m$ is divisible by $8$. But $m\mid 392=8\times 7^2$ so $m\in\{8,56,392\}$. However, $392\nmid|S_8|$, so $m=8$ or $56$, which means the kernel of $\phi$ has order $49$ or $7$.

Angina Seng
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