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Let $G$ be a group of order 2012.

I have shown the following:

  • $G$ has a unique normal subgroup $S$ of order 503
  • $S$ is cyclic and the automorphism group $\textrm{Aut}(S)$ contains an unique element of order 2, namely the one sending $x$ to $x^{-1}$.
  • If $H$ is a Sylow 2-subgroup of $G$, then $G/S\simeq H$ and $G=SH$.

I am then asked to show that if $G$ is not abelian, then the Sylow 2-subgroups are isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$. Any suggestions on how to do that?

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  • $\begingroup$ @DanielFischer They would have to be $\mathbb{Z}_4$. But where to I go from there? $\endgroup$ – Student G Jun 5 '14 at 20:35
  • $\begingroup$ @StudentG: you've been asked the impossible. It is perfectly ok for the Sylow 2-subgroups to be cyclic. $\endgroup$ – Jack Schmidt Jun 5 '14 at 20:37
  • $\begingroup$ @JackSchmidt I have? Could you provide a counter example? $\endgroup$ – Student G Jun 5 '14 at 20:41
  • $\begingroup$ Take a semidirect product. $\endgroup$ – Daniel Fischer Jun 5 '14 at 20:44
  • $\begingroup$ OK. Thanks to both of you. $\endgroup$ – Student G Jun 5 '14 at 20:56
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Here is a counterexample: Consider the matrix group defined over the integers mod 503: $$G = \left\langle \left[ \begin{smallmatrix} -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -1 & 0 \\ \end{smallmatrix} \right], \left[ \begin{smallmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{smallmatrix} \right] \right\rangle $$

The first generator has order 4 and generates a cyclic Sylow 2-subgroup, the second has order 503 and generates a normal Sylow 503-subgroup.

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