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

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

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$\mathbb Z_{12}$ is a cyclic group so you only need to know where to send $\bar 1$. Let $\sigma$ be that image. Then the order of $\sigma$ must divide 12. So what are the possibilities for $\sigma$? – lhf Mar 20 '12 at 1:04
Regarding your question 1: why do you think the order of an element which is a product of two disjoint transpositions is $4$? Regarding your question 2: what you wrote is in fact not a question! – Mariano Suárez-Alvarez Mar 20 '12 at 1:07
I wasn't totally sure regarding this solution , just wanted to make sure . Furthermore , as I can see , I was wrong regarding the order of $(xx)(xx)$ being 4 , since it's indeed 2 . thanks :) – ron Mar 20 '12 at 1:15

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 $\mathbb Z_{12}\to S_5$.

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Great , I understand now my mistake ! thank you! – ron Mar 20 '12 at 1:13
Can you please explain what do you mean by saying "completely specified by the image of 1" ? – ron Mar 20 '12 at 1:17
What I mean is, two homomorphims $\mathbb Z_{12}\to S_5$ which take the same value at $1$ are equal. – Mariano Suárez-Alvarez Mar 20 '12 at 1:50