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Pick the correct statement(s) below:

  • $(a)$ There exists a group of order $44$ with a subgroup isomorphic to $\Bbb Z_2 \oplus \Bbb Z_2 $.
  • $(b)$ There exists a group of order $44$ with a subgroup isomorphic to $\Bbb Z_4 $.
  • $(c)$ There exists a group of order $44$ with a subgroup isomorphic to $\Bbb Z_2 \oplus \Bbb Z_2 $ and a subgroup isomorphic to $\Bbb Z_4 $.
  • $(d)$ There exists a group of order $44$ without any subgroup isomorphic to $ \Bbb Z_2 \oplus \Bbb Z_2 $ or to $ \Bbb Z_4 $.
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    $\begingroup$ For $(1)$ take $G=\mathbb{Z}/2\oplus \mathbb{Z}/2\oplus \mathbb{Z}/11$, for $(2)$ take $G=\mathbb{Z}/4\oplus \mathbb{Z}/11$. $\endgroup$ – Dietrich Burde Apr 18 '14 at 17:33
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They key part is recognising part (c).

This is false. We know that there is only $1$ Sylow 11-subgroup of $G$. And so this must be a normal subgroup of $G$. Call it $H$. Then consider $G/H$. This has order 4. As such is must be isomorphic to either $\mathbb{Z}_2\times \mathbb{Z}_2$ or $\mathbb{Z}_4$. Hence $G \cong H \times \mathbb{Z}_4$ or $G \cong H \times \mathbb{Z}_2\times \mathbb{Z}_2$ or $G \cong H \rtimes \mathbb{Z}_4$ or $G \cong H \rtimes (\mathbb{Z}_2\times \mathbb{Z}_2)$.

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  • $\begingroup$ NOt clear to me the statements a b true then how it differs from c? $\endgroup$ – user124717 Apr 19 '14 at 3:20
  • $\begingroup$ If you tell me what you are struggling on, and what main theorems you have covered so far in you course, then I may be able to provide more help. @DietrichBirdle has given the reason why (a) and (b) are true above. This relies on the classification of abelian groups. $\endgroup$ – harry dunlop Apr 19 '14 at 12:43
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You can pick abelian groups to satisfy (a) and (b). For (c) and (d), are you familiar with the Sylow theorems?

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As i have understood, Sylow's second theorem eliminates option (c). Sylow p subgroups are conjugate to each other, implying they are isomorphic too. The two given subgroups are not isomorphic and hence cannot exist simultaneously in a group of order 44.

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