How many homomorphims from $Z_{12}$ to $S_{5}$

Question

I'm trying to calculate how many homomorphisms exists from $Z_{12}$ $\longrightarrow$ $S_{5}$ . Here are the options :

$(x x x x x)$ order 5

$(xxxx)$ order 4

$(xx) (xx) $ order 2 ???

$(xxx)(xx) $ lcm(2,3)=6 , hence order 6

$(xxx)$ order 3

$(xx)$ order 2

$id$ , meaning every element is mapped to itself , order 1

My questions are :

1. why is the order of $ (xx) (xx) $ is 4 , and not 2 ?

2. We need $Z_{12}$ , so all the above has order 12 in common (lcm) except for $(x x x x x)$ with order 5 . Then this means that I need to find the ones that divide with 5 and subtract it from $|S_{5}|$

Regards

Answer 1

An homomorphism $\phi:\mathbb Z_{12}\to S_5$ is completely specified by the image of $1$, and there is one for each element of $S_5$ whose order is a divisor of $12=2\cdot 2\cdot 3$.

Now, let $\pi$ be an element of $S_5$. The sizes of the cycles in its cycle decomposition are one of

cycle sizes          order
1, 1, 1, 1, 1        1
2, 1, 1, 1           2
2, 2, 1              2
3, 1, 1              3
3, 2                 6
4, 1                 4
5                    5

and this table also shows the order of the corresponding elements.

We see that all elements of $S_5$ have orders which divide $12$, except those of order $5$. There are $4!$ $5$-cycles in $S_5$, so there are $5!-4!$ elements in $S_5$ whose order divide $12$, so there are exactly that many homomorphisms $\mathbb Z_{12}\to S_5$.