I am asked:

Does a group $G$ of order 14 have to be cyclic?

Using Sylow's third theorem (work omitted), I conclude that $$|\textrm{Syl}_2(G)|=1 \textrm{ or }7$$ and $$|\textrm{Syl}_7(G)|=1$$ I understand that if $|\textrm{Syl}_2(G)|=1$ then the unique Sylow $2$-subgroup is normal (along with the unique $7$-subgroup) which implies $G\cong \Bbb{Z}/14\Bbb{Z}$ and $G$ is cyclic.

But I'm trying to understand the case when $|\textrm{Syl}_2(G)|=7$. Here are some questions I have:

  • I understand that there are 7 Sylow 2-subgroups. What do they look like? Do they all have order 2? Or do they all/some of them have some different order $2^r$? Are they disjoint except for identity? (our book does a poor job explaining Sylow theorems)
  • We still have a unique Sylow 7-subgroup of order 7, correct? This must mean there are 7 remaining elements to be used in the Sylow 2-subgroups. I am imagining that this implies the 7 Sylow 2-subgroups are of the form $\{e,a_1\},\{e,a_2\},\cdots$. This would let the group have 14 distinct elements. If this is the case, I have no problem disproving "cyclicness".



All sylow-2 subgroups are conjugate, so they have the same order. The largest power of 2 dividing 14 is 2, so all 7 sylow-2 subgroups have order 2.

Yes, you do still have a unique sylow 7 subgroup, so you've accounted for all the elements in your proposed group. To find a concrete representation, consider the symmetries of the regular heptagon.

  • $\begingroup$ Oh, I forgot that the power of the Sylow subgroups must divide the order of the group. Thank you! $\endgroup$ – Patrick Shambayati Mar 6 '15 at 1:22

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