GI 12 SolutionBy sylow 3 therem order of G 12 22 3 then

Solution

By sylow 3 therem order of G = 12 = 2^2 * 3 then G has both normal subgroup of oder 3 and a normal subgroup of oder 4.

so, normal subgroup of order 3:

(1+Pk) / 12, where P is prime factor of G

i.e. (1+ 3k) / 12, where k = 0, 1, 2, 3, 4, ......(1)

put k = 0 in equ (1) then 1/12, so 1 has a normal subgroup of order 3

k = 1 in equ (1) then 4/12, so 4 has a normal subgroup of order 3

   k = 2 in equ (1) then 7/12, it is does not divides 12 so it is not a normal subgroup of order 3

therefore G has 2 normal subgroups of order 3 ( i.e. 1, 4 only)

the normal subgroup of order 4 is

(1+2k) / 12, where k = 0, 1, 2, 3, 4, .....(2)

put k = 0 in equ (2) we have 1 / 12 so, 1 is a normal subgroup of order 4

k = 1 in equ (2) we have 3/12, s0, 3 is normal subgroup of order 4

k = 2 in equ (2) we have 5 does not divides 12 so 5 is not a normal subgroup of order 4

therfore G has a 2 normal subgroups of order 4 (i.e. 1 and 3)