# How many semi-direct products $H \rtimes Q$ can be be constructed for $H \cong C_{42}, Q \cong C_{3}$

I was also given a hint:

Represent $C_{42} \cong C_2 \times C_3 \times C_7$, find its group of automorphisms, then look for elements of order 3.

So, I found the group of automorphisms to be:

$$\operatorname{Aut}(C_{42}) = C_1 \times C_2 \times C_6$$

Then I said as 3 divides ($1 \times 2 \times 6 = 12$), we can work out the Euler function of 3, and this gives me:

$$\varphi (3) = 3 - 1 = 2$$

So there are 2 semi direct products. Is this correct?

• Perhaps we're having here a problem of notation: for me, $\,H\rtimes Q\,$ means the semidirect product that stems from some homomorphism $\,H\to\operatorname{Aut}(Q)\cong C_2\,$ , so you should be interested in elements of order two (involutions) in $\,H\,$ that can be mapped to $\,C_2\,$... – DonAntonio Nov 14 '12 at 13:36
• @DonAntonio: I believe it is standard that in a semidirect product $H \rtimes Q$, $H$ is the normal subgroup and $Q$ is the complement. The directioon of the $\rtimes$ symbol mimics that of the $\lhd$ symbol. – Derek Holt Nov 14 '12 at 15:45
• Perhaps I missed that one, @DerekHolt . Thanks – DonAntonio Nov 14 '12 at 15:47

Hint: We know that: $$\mathbb Z_n\rtimes\mathbb Z_m=\langle a,b|a^n=b^m=1, bab^{-1}=a^l, l^m\equiv1 \;\;(\text{mod}\; n)\rangle$$ So I think we should focus on the condition contained above to find the proper number. This presentation can be proposed by using the hint given to you. Here we assume that $\mathbb Z_m=\langle b\rangle$ and $\mathbb Z_n=\langle a\rangle$.