Question

Find, with justification, all the Sylow subgroups of Z15.

Answer 1

Answer on Question #44363 - Math - Abstract Algebra

Question Find, with justification, all the Sylow subgroups of $Z_{15}$.

Solution. The group $Z_{15}$ has order 15=3·5, therefore its Sylow subgroups has orders 3 or 5. Since the numbers 3 and 5 are prime, the groups of order 3 or 5 are cyclic. Hence, the desired subgroups are generated by the elements of $Z_{15}$ of order 3 or 5. We know that the order $|n|$ of an element $n \in Z_{15}$ equals $n / gcd(n,15)$. So, $|n| = 3$ if and only if $gcd(n,15) = 5$, while $|n| = 5$ if and only if $gcd(n,15) = 3$. The numbers $n$ between 0 and 14, such that $gcd(n,15) = 5$, are

$$n = 5, 10.$$

Since $10 = 5 \cdot 2$, the element 10 generates the same cyclic subgroup as 5 does. Thus, $Z_{15}$ has a unique subgroup of order 3. It is $\langle 5 \rangle$.

The numbers $n$ between 0 and 14, such that $gcd(n,15) = 3$, are

$$n = 3, 6, 9, 12.$$

All these numbers belong to the same cyclic subgroup $\langle 3 \rangle$ of order 5. Thus, $Z_{15}$ has a unique subgroup of order 5. It is $\langle 3 \rangle$.

Answer: $\langle 5 \rangle = \{0, 5, 10\}$ of order 3 and $\langle 3 \rangle = \{0, 3, 6, 9, 12\}$ of order 5.

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