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?

  • $\begingroup$ 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 $\,C_2\,$... $\endgroup$
    – DonAntonio
    Commented Nov 14, 2012 at 13:36
  • $\begingroup$ @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. $\endgroup$
    – Derek Holt
    Commented Nov 14, 2012 at 15:45
  • $\begingroup$ Perhaps I missed that one, @DerekHolt . Thanks $\endgroup$
    – DonAntonio
    Commented Nov 14, 2012 at 15:47

1 Answer 1


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 $\mathbb Z_n=\langle a\rangle$.

  • $\begingroup$ What are a and b? $\endgroup$
    – Kaish
    Commented Nov 14, 2012 at 14:28
  • $\begingroup$ I added some above. $\endgroup$
    – Mikasa
    Commented Nov 14, 2012 at 14:39
  • $\begingroup$ This is definitely worth upvotes! At last, we'll start with one! ;-) $\endgroup$
    – amWhy
    Commented Apr 4, 2013 at 0:36

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