= Solution
For a <finite group> over the complex numbers, the sum of the squared dimensions of its inequivalent <irreducible representations> equals the <group order>. There is always a one-dimensional <trivial representation>, and part (b) supplies an <irreducible representation> of dimension three. Thus the remaining squared dimensions sum to
$$
12-1^2-3^2=2.
$$
They must be two further dimensions equal to one. Hence
$$
\boxed{\text{the irreducible dimensions are }1,1,1,3.}
$$
The number of inequivalent <irreducible representations> equals the number of <conjugacy classes>, so $A_4$ has four classes. There are only three <cycle types> in part (a); consequently <cycle type> cannot classify conjugacy within $A_4$.
More explicitly, the identity is a singleton class, and <conjugation> by three-cycles permutes the three double <transpositions> transitively. For a three-cycle, the <centralizer> in $A_4$ consists of its three powers, so its <conjugacy class> has $12/3=4$ elements. The eight three-cycles therefore split into two classes of four:
$$
\boxed{\text{class sizes }1,3,4,4;\quad\text{the three-cycle type splits into two classes}.}
$$
For example, $(123)$ and $(132)$ have the same <cycle type> but are not conjugate in $A_4$. Any <permutation> conjugating the first to its inverse must fix its unique fixed vertex and reverse the orientation of the remaining three labels, so it is odd. They are conjugate in $S_4$, illustrating exactly why restricting the ambient <group> changes the answer.
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