Let $G$ be a group of order $260$. For each prime $p$ dividing $|G|$, determine the order of a Sylow $p$-subgroup.

We have $|G| = 260 = 2^2 \cdot 5 \cdot 13$.

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    $\begingroup$ The order of a subgroup is its size. The question you have posted doesn't ask how many subgroups there are, or whether they are normal. $\endgroup$ – Mark Bennet Mar 2 '14 at 21:39
  • $\begingroup$ Then I misinterpreted the question and I'll rework on it. But hypothetically, if the question did ask how many subgroups there are, how would I have done in answering that? $\endgroup$ – Cookie Mar 2 '14 at 22:46
  • $\begingroup$ Note you can always have a single Sylow subgroup for a given prime (the cyclic group of order $260$ has just one subgroup of each order $r|260$) $\endgroup$ – Mark Bennet Mar 2 '14 at 22:54

We have $|G| = 260 = 2^2 \cdot 5 \cdot 13$

So $G$ contains at least one of the following:

  • Sylow $2$-subgroup of order $4$
  • Sylow $5$-subgroup of order $5$
  • Sylow $13$-subgroup of order $13$
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  • $\begingroup$ Yes, but note you might want to say "At least one..." $\endgroup$ – Pedro Tamaroff Mar 23 '14 at 17:09

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