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Let $G$ be a $2$-Sylow subgroup of $S_{17}$. We have to prove that $G$ has a subgroup isomorphic to $\mathbb Z_8\times \mathbb Z_8$.

My progress: It is easy to calculate that $|G|=2^{15}$. Now, any $p$-group has subgroups of order $p^k$ for all $1\le k\le n$. Thus, $G$ has a subgroup of order $64$. I am unable to proceed further and prove that this subgroup is in fact $\mathbb Z_8\times \mathbb Z_8$.

Shaun
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Tabludif
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  • You just have to write down any embedding of $C_8 \times C_8$ into $S_{17}$. – Qiaochu Yuan Jul 23 '24 at 05:49
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    So you want the group generated by an 8-cycle in $S^{17}$ and another 8-cycle disjoint from the first one? – Eric Towers Jul 23 '24 at 05:51
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    All $p$-Sylow subgroups of a group are isomorphic to each other and every $p$-subgroup of a group $G$ is contained in a $p$-Sylow subgroup of $G$. These two observations mean that it's enough to prove that $S_{17}$ has a subgroup isomorphic to $C_8 \times C_8$. (Do you see why that's so?) Can you do that? – Robert Shore Jul 23 '24 at 06:21
  • What other said. If you want to learn more, observe that every Sylow 2-subgroup of $S_{17}$ is conjugate to a Sylow $2$-subgroup of $S_{16}$. And the latter Sylow subgroup has a nice description in terms of binary trees. – Jyrki Lahtonen Jul 23 '24 at 06:24
  • @RobertShore Yes, now I can take the group generated by two disjoint $8$-cycles and conclude that another group isomorphic to it is a subgroup of $G$. Thanks! – Tabludif Jul 23 '24 at 06:33

1 Answers1

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There are many subgroups of order $64$ in $G$, and not all of them are isomorphic to $\mathbb{Z}_8 \times \mathbb{Z}_8$.

So you will not be able to prove that every subgroup of order $64$ in $G$ is $\mathbb{Z}_8 \times \mathbb{Z}_8$, since that is not true.

Instead, you should try to construct some subgroup isomorphic to $\mathbb{Z}_8 \times \mathbb{Z}_8$. As hinted in the comments, try to look at a subgroup generated by two disjoint $8$-cycles.

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