# 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 :

1. Id

2. $A_{5}$

3. $S_{5}$

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

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It means the subgroup consisting of just the neutral element in the group (also known as the identity element). –  Tobias Kildetoft Mar 22 '12 at 11:48
Please forgive me for asking (if my question is a little bit silly) : when we ask for the Identity subgroup of a symmetric group , does it mean to a group with only one element ? –  ron Mar 22 '12 at 11:52
@ron - Yes. Take a look at my answer below. –  user5137 Mar 22 '12 at 11:54

I suspect what is meant by "the ID subgroup" (which is terrible notation, by the way) is the identity subgroup, ie $\{e\}$, where $e$ represents the identity element of $S_5$.

In general, if $G$ is any group with identity element $e_G$, then $\{e_G\}$ is a normal subgroup of $G$.

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JackManey : So, what is the the ID group of $S_{5}$ ? how much are there ? only one subgroup of the Identity ? –  ron Mar 22 '12 at 11:57
@ron - As I said, the identity group of $S_5$ consists of one element--namely the identity element of $S_5$. Every group has a (necessarily unique) identity element. The singleton set consisting of this element forms a normal subgroup of the given group. –  user5137 Mar 22 '12 at 12:03
Thank you! I think I got it . –  ron Mar 22 '12 at 12:10
@ron You really need to spend time with the definition of what a group is, what a subgroup is! Please go over these definitions to see for yourself that the identity subgroup is necessarily unique and is trivially a normal subgroup of a group. Further, understand that this is a terrible notation. A nice way of calling this subgroup is just with $1$. –  user21436 Mar 22 '12 at 12:11
Understood . thanks . –  ron Mar 22 '12 at 12:45

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,...

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Actually, $\vert G/ker f \vert = \vert Im f \vert$ since $G/ker f \cong Im f$. –  user5137 Mar 22 '12 at 11:51
Thank You for the pointer. I am sorry. –  user21436 Mar 22 '12 at 11:53
Oh, no worries. :) –  user5137 Mar 22 '12 at 11:54