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How many homomorphisms are there from $\mathbb Z_3$ to $S_4$?

I got this on a quiz today and calculated 9 but I'm not certain I was right. I sent 1 to id and to the 8 3-cycles... did I do wrong?

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An homomorphism $f : \mathbb{Z}_{3} \to S_{4}$ is uniquely determined by $f(1)$, which must be an element of order that divides $3$, then ord$(f(1)) = 1$ or ord$(f(1)) = 3$.

In $S_{4}$ the only elements of order $3$ are the $3$-cycles, so your solution seems correct.

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but $A_3$is the only subgroup of $S_4 with order 3. By lagrange theorem the order of the image is either1 or 3. so arent there 2 homoms? – tetrr May 7 '14 at 11:50
the homomorphism is determined not by the subgroup image, but by the images of the elements – WLOG May 7 '14 at 11:52
$A_{3}$ is not the only subgroup of order 3 in $S_{4}$, there are 7 more. – shrey Jun 27 '15 at 12:39

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