# Determine order of a Sylow p-subgroup

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

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

• 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. – Mark Bennet Mar 2 '14 at 21:39
• 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? – Cookie Mar 2 '14 at 22:46
• 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 $r|260$) – 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 $13$