Let
[MATH]G=(C_{p_1} : C_{3}) \times(C_{p_2} : C_{3})[/MATH]
where [MATH]p_1,p_2\equiv{1}\pmod{3}[/MATH].
How many elements does the group [MATH]G[/MATH] 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 [MATH]p_{i}[/MATH]. But I could be wrong.
I am trying to adapt a current proof where [MATH]G= C_{3} \times(C_{p} : C_{3})[/MATH] and [MATH]p\equiv{1}\pmod{3}[/MATH].
[MATH]G=(C_{p_1} : C_{3}) \times(C_{p_2} : C_{3})[/MATH]
where [MATH]p_1,p_2\equiv{1}\pmod{3}[/MATH].
How many elements does the group [MATH]G[/MATH] 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 [MATH]p_{i}[/MATH]. But I could be wrong.
I am trying to adapt a current proof where [MATH]G= C_{3} \times(C_{p} : C_{3})[/MATH] and [MATH]p\equiv{1}\pmod{3}[/MATH].