Let G be a finite group of order n For each n below prove th

Let G be a finite group of order n. For each n below, prove that G is not simple. (a) n = 30 (b) n = 182 (c) n = 595

Solution

(a) Let G be a group of order 30.=2.3.5

By Sylow theorems, there exist 3 Sylow subgroups of G and their number M is 1 (mod 3) .As M also divides 30, the only possibilities are M =1 or M =10

Again, by Sylow theorems, there exist 5 Sylow subgroups of G and their number N is 1 (mod 5) .As N also divides 30, the only possibilities are N =1 or N=6

If either M or N is 1, it follows that the corresponding Sylow subgroups are unique. As Sylow subgroups for a given prime are all conjugate, this will imply the existence of normal subgroups. So G cant be simple.

So we need to discuss the case M =10 and N =6.

Each of the 3-Sylow subgroups and 5-Sylow subgroups are isomorphic respectively to the cyclic groups of order 3 /order 5.

Every non-trivial element in each of these groups is a generator of that subgroup.

Thus every non-identity element of G must belong to precisely one of these subgroups.

The total number of such elements is 6.4+ 10.2= 54, which contradicts the number of elements in G.

Hence G must contain a normal subgroup, which proves that a group of order 30 cant be simple.

(b) Let the order of G be 182

182=2 .7.13

The number N of Sylow 7-subgroups is 1 (mod 7) and N should also divide 182 (in fact 26).

As the only possibilities for N are 1,8,15,22,

we conclude that N =1.

So the 7-Sylow subgroup is normal and hence G cant be simple.

(c) Let G have order 595= 5. 191 (191 is a prime)

The number N of 191-Sylow subgroups is 1 (m0d 191) and N should also divide 595.

This forces N =1, Hence , the corresponding 191-Sylow subgroup is unique, hence normal.

So G cant be simple