Question

Question: If every Sylow subgroup of G is normal, prove that G is the direct product of its Sylow subgroups (one for each prime that divides$\left|G\right|$ ). A group with this property is said to be nilpotent.

Short answer

It is proved that G is the direct product of its Sylow subgroups.

Step-by-step solution

Referring to Corollary 9.16

Corollary 9.16

Let G be a finite group and K is a Sylow p-subgroup for some prime p. Then, K is normal in G if and only if K is the only Sylow p-subgroup.

Given that every Sylow subgroup of G is normal.

Proving that G is the direct product of its Sylow subgroups

Let G be the group of order pq, where p and q are both primes.

Therefore, the Order of $G=p.q$.

Consider Sylow p-subgroups of G.

Since every Sylow subgroup of G is normal, byCorollary 9.16, there should be only one Sylow subgroup of order p.

Now, consider the Sylow q-subgroup.

Again, it is normal, so byCorollary 9.16, it is the only normal subgroup of order q.

Now, we know that there is only one p-subgroup and q-subgroup.

Thus, the product of the Sylow subgroup is$pq$ .

Hence, it is proved that G is the direct product of its Sylow subgroups.