Homomorphisms from $S_{5}$ to $\mathbb{Z}_{12}$ , what does it mean the ID subgroup? I'm trying to find the number of homomorphisms from $S_{5}$ to $\mathbb{Z}_{12}$ , meaning : 
$S_{5}$ $\longrightarrow$ $\mathbb{Z}_{12}$
I'm using the 1st Isomorphism theorem .
$G/Ker(f)≅Im(f)$
So , we can have the following subgroups for Ker :


*

*Id 

*$A_{5}$

*$S_{5}$
For the first : $120/1≅Im(f)=120$ but by Lagrange theorem $|Im(f)|  | |\mathbb{Z}_{12}| $ , but 120 doesn't divide 12 , then this can't be .
Question : what does it mean the ID subgroup ? why its size is 1 ?  
Regards 

Firstly, I am not sure what your question that comes under the title "Question" is. Let me try my best at it:


*

*The "ID" subgroup you seem to be interested in is called the identity subgroup, which for an abstract group $G$ is the singleton set $I=\{e_G\}$ where $e_G$ is the identity element of $G$. It is routine to check that $I$ is actually a subgroup of $G$. 


For example, in $S_5$, $Id_{S_5}=(1)(2)(3)(4)(5)$ is the identity permutation which fixes all the five symbols whose group of permutations in $S_5$. 

Your approach is right, but you fail to observe that, if $a \mid b$, then $a \le b$. This will get you out of those unnecessary contradictions.   

Let $f$ be a homomorphism from $S_5$ to $\Bbb Z_{12}$. Then, you have following restrictions on $f$: 


*

*The first isomorphism theorem, together with Lagrange's theorem tells you that, $$|S_5/\operatorname {Ker} f|~~ \mbox{$=$} ~~|\operatorname{Im} f| ~~~\mbox{divides}~~ |\Bbb Z_{12}|$$



So, for knowing the cardinality of $\operatorname{Ker} f$,  notice that you  need to find all those $x$ such that $$\dfrac{120}{x} \mid 12$$ Firstly, this in particular means that, $\dfrac {120} x \le 12$. So, you have that, $x \ge 10$. 

As you know that only normal subgroups in $S_5$ are $1=\{\operatorname{Id}_{S_5}\}, A_5$ and $S_5$, you are forced to conclude that $\operatorname{Ker} f \in \{A_5, S_5\}$. 
So,... 
