2
$\begingroup$

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

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

$\endgroup$
  • 4
    $\begingroup$ 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
1
$\begingroup$

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 $13$
| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Yes, but note you might want to say "At least one..." $\endgroup$ – Pedro Tamaroff Mar 23 '14 at 17:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.