Abstract problem.?

Question

Let G= A_4 (the even group of permutations). And let K be the subgroup K = {(1), (12)(34),(13)(24),(14)(23)}. Determine with proof whether or not K is a normal subgroup of G.

Answer 1

Yes. What's more, K is a normal subgroup of S_4 (the group of all permutations).
One has to show that p^(-1)kp is in K for every k in K and every permutation p. Now each k in K is a pair of disjoint transpositions, and K consists of ALL such pairs of swaps (and identity). What happens when p^(-1)kp is applied is that the elements of {1,2,3,4} get "renamed", then arranged in two pairs and swapped within each pair as per k, and finally renamed back. This permutation is just another pair of swaps, so it is in K.
More formally, if k = (ij)(mn) then p^(-1)kp = (p(i)p(j)) (p(m)p(n)). The argument is placed on the left, as usual with permutations, so the 1st to apply in p^(-1)kp is p^(-1), then k then p. So p(i) and p(j) are first mapped to i and j, then swapped, and then mapped back to p(j) and p(i). The result is that p(i) and p(j) get swapped. Same with m and n.

Answer 2

You could always make a Cayley table as the last resort.

Notice that the nonidentity elements in K are disjoint 2-cycles. Every element in A4 has an even number of permutations, you need to take an arbritray element of A4 that is not in K, so like (rs)(rt) and find its inverse and show if k in K then (rs)(rt)k((rs)(rt))^-1 is in K. This should give you a good start.