Question

Question: If G is a group of order 60 that has a normal Sylow 3-subgroup, prove that G also has a normal Sylow 5-subgroup.

Short answer

It is proved that G also has a normal Sylow 5-subgroup.

Step-by-step solution

Referring to Corollary 9.16, Third Sylow Theorem and Corollary 9.18

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.

Third Sylow Theorem

The number of Sylow p-subgroups of finite group G divides $\left|G\right|$and is of the form 1+pk for some non-negative integer k.

Corollary 9.18

Let G be a group of order pq, where p and q are primes such that p>q.

If $q$does not divide$\left(p-1\right)$ then,$G\cong {\mathrm{\mathbb{Z}}}_{pq}$ .

Given that G is a group of order 60 that has a normal Sylow 3-subgroup.

Proving that G also has a normal Sylow 5-subgroup

The order of G is$60=2^3.3.5$ .

Therefore, byThird Sylow Theorem, Sylow 5-subgroups should be of the form 1+5p.

Thus, G can have either 1 or 6 Sylow 5-subgroups.

Suppose G has 6 Sylow 5-subgroups.

Then, they must be cyclic of order 5.

This gives a total of$5.6=30$ elements of order 5 in group G.

Suppose H is the Sylow 3-subgroup of G.

As it is given that H is a unique Sylow 3-subgrop, byCorollary 9.18 for any $1\le k\le 6$,$H{P}_k\cong {\mathrm{\mathbb{Z}}}_{35}\cong {\mathrm{\mathbb{Z}}}_{15}$ .

Since the subgroup $H{P}_k$has 8 elements of order 15, which gives a total of$8.6=48$ elements.

Thus, G has total $30+48=78$elements.

Since G is a group of order 60, this contradicts our assumption.

This implies G has only one Sylow 5-subgroup, and byCorollary 9.16, it would be normal.

Hence, it is proved that G also has a normal Sylow 5-subgroup.