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the question is as follow:

1) $H \lt S_{6}$ of order 9. prove that H is not cyclic.

2) $A \lt S_{6}$ of order 16. prove that A is not abelian.

so for 1) I know that $H$ is abelian but can't so therefore it's isomorphic to $\mathbb Z/3 \mathbb Z\times \mathbb Z/3 \mathbb Z$ or $\mathbb Z/9 \mathbb Z$ both are cyclic so what am I missing?

about 2) I have no sense of direction.

Thank you!

Kim Seel
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1 Answers1

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The first question follows from the fact that all Sylow $3$-subgroups of $S_6$ are isomorphic to $C_3\times C_3$, which is not cyclic, see the comment. The order of a $3$-Sylow subgroup in $S_6$ is $9$, because $6!=5\cdot 3^2\cdot 2^4$. Now $C_3\times C_3$ is such a Sylow subgroup, but every other Sylow $3$-subgroup is conjugated to it, hence isomorphic.

The second question goes similarly, with Sylow $2$-subgroups.

Dietrich Burde
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  • Would you care to summarize the proof? Other wise, it seems you are asking the OP to take your word, with no proof, and only a link to a potential proof. – amWhy Mar 20 '17 at 12:48
  • @dietrich-burde in the exam the say that $H=<\sigma \tau>$ and that $<\sigma> \cap <\tau> = {e}$ hence $H = <\sigma> \times <\tau>$ from a theorem I've learned there's a requirement that $gcd(<\sigma>, <\tau>)=1$ for that, and that doesn't apply here. – Kim Seel Mar 20 '17 at 12:55
  • No, there is no such requirement, see this question. – Dietrich Burde Mar 20 '17 at 14:51