Homomorphisms from $S_{5}$ to $\mathbb{Z}_{12}$ , what does it mean the ID subgroup?

Question

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

Answer 1

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 $G$.

Answer 2

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 $S_5$.

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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.

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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,...