Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 1 / 10G / i / Solution

The alternating group $A_4$ containsThe product of any two different nonidentity elements is the third, and each squares to the identity, so $V$ is a Klein four-group. Conjugation by a permutation replaces the labels in each cycle, preserving the property of being a product of two disjoint transpositions. Thus $V$ is a normal subgroup of $A_4$. Its order is $4$, strictly between $1$ and $|A_4|=12$. $A_4$ is not simple.