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The question is to work out the number of semi-direct products for from Q to H, where

$$ H = C_{42} , Q = C_{3} $$

I did:

$Aut(C_{42}) = C_2 \times C_6 = C_{12}$

3 divides 12 telling us that they're are some semidirect products.

If we let Q = (1, 2, 3). We can see that they're are 3 elements in Q (1, 2 and 3) that divide 12, thus there are 3 semi direct products.

Is this correct reasoning?

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$C_2\times C_6\not\cong C_{12}$. (I assume that's supposed to be $C_{12}$.) $C_{12}$ has elements of order $4$, $C_2\times C_6$ does not. –  Thomas Andrews Dec 12 '12 at 21:37
Yeah it was, sorry. For some reason I'm not getting a preview of the questions I type up. So I have to say that there are 3 elements in Q that divide elements in $C_2$ and $C_6$ and so there are 3 semi direct products? –  Kaish Dec 12 '12 at 21:42
Not sure what you mean by "divide." You are looking for the number of homomorphisms $C_3\to\operatorname{Aut}(C_{42})\cong C_6\times C_2$, which, since $C_3$ is cyclic, corresponds to the number of elements of $C_6\times C_2$ with order equal to $1$ or $3$. It's not obvious to me that different such elements yield different groups, although clearly the trivia case is distinct since it yields a commutative group. –  Thomas Andrews Dec 12 '12 at 21:58
By divide I mean if we say $C_3 = (1, 2, 3), C_2 = (1, 2), C_6 = (1, 2,..., 6)$. Then there are elements in $C_3$ that divide elements in $C_2$ and $C_6$. Element '1' divides $1, 2 \in C_2$ and also all elements in $C_6$. Element 2 divides 2 in $C_2$ and $2, 4, 6 \in C_6$. Like that. But that doesn't seem proper to me –  Kaish Dec 12 '12 at 22:06
Yeah, that notion of "divides" is very undefined. In particular, we usually write $C_3=\{0,1,2\}$, not $\{1,2,3\}$ and your definition of "division" seems to depend on the labeling. –  Thomas Andrews Dec 13 '12 at 13:40

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