Does a group \(G\) of order \(12\) with a normal subgroup of order \(4\) have to be isomorphic to \(A_4\)? Give a counterexample.
Solution
No. Take \[ G=C_{12}. \] A cyclic group has a unique subgroup of order \(d\) for every divisor \(d\) of its order. Hence \(C_{12}\) has a unique subgroup \[ H=\langle g^3\rangle\cong C_4 \] of order \(4\), and \(H\) is normal because \(C_{12}\) is abelian.
But \(C_{12}\not\cong A_4\): the first group is abelian and contains an element of order \(12\), whereas \(A_4\) is nonabelian and has element orders only \(1,2,3\).
Thus the existence of a normal subgroup of order \(4\) does not characterize \(A_4\) among groups of order \(12\).