Let
$G=(C_{p_1} : C_{3}) \times(C_{p_2} : C_{3})$
where $p_1,p_2\equiv{1}\pmod{3}$.
How many elements does the group $G$ have of each order? Furthermore, what is the total number of conjugacy classes?
I assumed that G contains exactly p-1 elements of order p, 2(p-1) elements of order 3p and 2(3p+1) elements of order 3, for each p_{i}. But I could be wrong.
Similarly, can I ask the same question for $G= A_{4} \times(C_{p} : C_{3})$ where $p\equiv{1}\pmod{3}$.
I am trying to adapt a current proof where $G= C_{3} \times(C_{p} : C_{3})$ and $p\equiv{1}\pmod{3}$. The proof is shown above and the authors claim that similar arguments can be used to prove the two cases I have presented above.
Best Answer
For a direct product $G \times H$, the conjugacy classes are of the form $C \times D$, where $C$ and $D$ are conjugacy classes of $G$ and $H$, and the order of the elements in $C \times D$ is the least common multiple of the orders of elements in $C$ and in $D$. This makes it straightforward to answer your questions in direct products provided that we can answer them in the factors.
So lets apply that to the case when both $G$ and $H$ are nonabelian groups with structure $C_p:C_3$.
The factors have one element of order $1$, $p-1$ of order $p$ in $(p-1)/3$ classes, and two classes of elements of order $3$, both of size $p$.
So in $(C_{p_1}:C_3) \times (C_{p_2}:C_3)$ with $p_1 \ne p_2$, we have (if I have counted correctly):
one element of order $1$,
$p_1-1$ of order $p_1$ in $(p_1-1)/3$ classes,
$p_2-1$ of order $p_2$ in $(p_2-1)/3$ classes,
$2p_1 + 2p_2 + 4p_1p_2$ of order $3$ in $2 + 2 + 4 = 8$ classes,
$2p_2(p_1-1)$ of order $3p_1$ in $2(p_1-1)/3$ classes,
$2p_1(p_2-1)$ of order $3p_2$ in $2(p_2-1)/3$ classes, and
$(p_1-1)(p_2-1)$ of order $p_1p_2$ in $(p_1-1)(p_2-1)/9$ classes.
The only significant difference when $p_1=p_2$ is that the elements in the final class have order $p_1$.