Is \(A_4\) a simple group? What are the conjugacy classes in \(S_4\)? What about in \(A_4\)?
Solution
1. The group \(A_4\) is not simple.
The subset \[ V_4=\{e,(12)(34),(13)(24),(14)(23)\} \] is a subgroup of \(A_4\). Conjugation preserves cycle type, so conjugating any nonidentity element of \(V_4\) by an element of \(A_4\) again gives a double transposition. Hence \(V_4\trianglelefteq A_4\). Since \[ 1<|V_4|=4<12=|A_4|, \] this is a proper nontrivial normal subgroup, so \(A_4\) is not simple.
2. The conjugacy classes in \(S_4\) are determined by cycle type.
Two permutations in a symmetric group are conjugate exactly when they have the same cycle type. Thus the classes in \(S_4\) are: \[ \begin{array}{c|c} \text{cycle type} & \text{class size}\\ \hline 1^4 & 1\\ 2\,1^2 & 6\\ 2^2 & 3\\ 3\,1 & 8\\ 4 & 6 \end{array} \] The sizes sum to \(1+6+3+8+6=24\).
3. The conjugacy classes in \(A_4\) are\[ \{e\}, \qquad \{(12)(34),(13)(24),(14)(23)\}, \]and two classes of four \(3\)-cycles each.
Only the even cycle types from 2 occur in \(A_4\): the identity, the three double transpositions, and the eight \(3\)-cycles.
The three double transpositions form one conjugacy class in \(A_4\): for example conjugation by \(3\)-cycles permutes them transitively.
Now let \(\sigma=(123)\). Its centralizer in \(A_4\) is \[ C_{A_4}(\sigma)=\langle\sigma\rangle, \] which has order \(3\). Hence its \(A_4\)-conjugacy class has size \[ [A_4:C_{A_4}(\sigma)]=12/3=4. \] There are eight \(3\)-cycles total, so they split into exactly two conjugacy classes of size \(4\). Equivalently, a \(3\)-cycle and its inverse lie in the two different \(A_4\)-classes.
4. These classes also exhibit the normal subgroup \(V_4\).
A normal subgroup is a union of conjugacy classes containing the identity. The union of the identity class with the double-transposition class is exactly \(V_4\), recovering the normal subgroup from 1.