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$\begingroup$

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?

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    Perhaps I missed that one, @DerekHolt . Thanks2012-11-14

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

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    This is definitely worth upvotes! At last, we'll start with one! ;-)2013-04-04