Answer to Question #147401 in Abstract Algebra for Sourav Mondal

Question

If G is a finite commutative group of order n
and if a prime p divides n, show that the
number of Sylow-p subgroups of G is one.
Find the unique Sylow-3 and Sylow-2
subgroups of the cyclic group Z24.

Answer 1

Let H be a Sylow p-subgroup of G. Then since G is abelian, then H is normal. But any two sylow p-subgroups are conjugate. Hence if K is another, then $gHg^{-1}=K$ for some $g\in H.$ But H being normal, $gHg^{-1}=H.$ Hence $K=H.$ So nly one Sylow p-subgroup.

$|\mathbb{Z}_{24}|=24=2^3.3.$ So Sylow 2-subgroup has order 8. $\mathbb{Z}_{24}$ being cyclic each subgroup is cyclic. Cyclic subgroup of order 8 is generated by 24/8=3. Hence

$H=\{[0],[3],[6],[9],[12],[15],[18],[21]\}$ is the unique Sylow 2 -subgroup.

Similarly, Sylow 3 is $K=\{[0],[8],[16]\}.$ It is a cyclic group generated by the element. Its order is 24. Now sylow 3 subgroup will have order 3. Hence it is also cyclic of order 3 and must be generated by an element of order 3. Since 1 has order 24, so 24/3=8 will have order 3. And hence the result.