[Math] First Isomorphism Theorem for Monoids

abstract-algebramonoid

Background

The First Isomorphism Theorem states,

If $G$ and $H$ are two groups and $\varphi:G\to H$ be a group homomorphism, then $\varphi(G)$ is isomorphic to $G/\ker \varphi$.

I was wondering that whether we can generalize this theorem to weaker algebraic structures and I observed the following,

  • The definition of group homomorphism can be easily generalized to what we may call monoid homomorphism as follows,

    $\color{crimson}{\text{Definition 1.}}$ If $(G,\bullet)$ and $(H,\circ)$ be two monoids then a map $\varphi:G\to H$ is said to be a monoid homomorphism if $$\varphi(x\bullet y)=\varphi(x)\circ\varphi(y)$$for all $x,y\in G$.

  • The definition of Normal Subgroup of a group can be suitably generalized too,

    $\color{crimson}{\text{Definition 2.}}$ Let $G$ be a monoid and $H$ be a submonoid of $G$. We will say $H$ to be a normal submonoid of $G$ if $aH=Ha$ for all $a\in G$.

  • The definition of kernel can be given in the following way if $H$ is a monoid,

    $\color{crimson}{\text{Definition 3.}}$ Let $H$ be a monoid and $\varphi: G\to H$ be a monoid homorphism. Let $e$ be the identity element of $G$. Then we can define, $$\ker \varphi:=\{x\in G\mid \varphi(x)=e\}$$

Questions

From this discussion, we can obtain the following more general version of the First Isomorphism Theorem,

$\color{blue}{\text{Proposition 1.}}$ Let $G$ and $H$ are monoids and $\varphi:G\to H$ be a monoid homomorphism. Then prove that $\varphi(G)$ is isomorphic to $G/\ker \varphi$.

I wanted to the argument of Theorem 10.3 of this book. However, a crucial theorem used in proving Theorem 10.3 is Theorem 9.2 and to prove Theorem 9.2 we need to prove that for a monoid $M$ if $N$ be any submonoid of $M$ then, $aN=N$ iff $a\in N$. But this I can't prove. More specifically, I can't prove the following proposition,

$\color{blue}{\text{Proposition 2.}}$ Let $G$ be a monoid and $H$ be a normal submonoid of $G$. Then the set $G/H:=\{aH\mid a\in H\}$ is a monoid under the operation $(aH)(bH)=abH$ where $a,b\in G$.

So, my questions are,

  1. Are the above propositions true?

  2. If so, then can anyone give some hint as to how I should proceed to a proof of both of the propositions?

Remark

So far I have been able to prove the following result,

Theorem. If $G$ be a monoid and $H$ be a submonoid of $G$ then $H=\displaystyle\bigcup_{a\in H} aH$.

Proof Sketch. Let $a\in H$. Then $aH\subseteq H$ by closure of $H$. Since $a$ is arbitrary we have $\displaystyle\bigcup_{a\in H} aH\subseteq H$. To prove the converse observe that, $$a\in H\implies a\in aH\implies \displaystyle\bigcup_{a\in H} aH$$ and since the above statement holds for all $a\in H$, we are done.

but don't know how this helps (if at all)

Best Answer

Proposition $2$ is certainly true as soon as the operation on $G/H$ is well-defined, i.e. as soon as it doesn't depend on the choice of representatives $a$ and $b$ for the cosets $aH$ and $bH$. This is the case if $H$ is a "normal submonoid" as in your Definition 2, and can be shown more or less like in groups.

But it is not true that $aN=N$ is equivalent to $a\in N$ if $N$ is a submonoid: for example if you take the monoid of natural numbers with operation given by multiplication, then $N=\{0,1\}$ is a submonoid, and $0\in N$; but $0N=\{0\}\neq N$.

As I mentionned in a comment, the true problem is that the kernel of a homomorphism can be trivial without the homomorphism being injective. For example consider the homomorphism $$\varphi : (\Bbb N,\cdot ,1)\to (\Bbb N,\cdot ,1): x\mapsto 0 \text{ iff } x\neq 1;$$only $1$ is sent to $1$, so the kernel is trivial; but it is not injective, since every natural other than $1$ is mapped to $0$. Thus the quotient monoid (if it is well-defined) should be isomorphic to $\Bbb N$, while the image of $\varphi$ is the submonoid $N$ defined above. The problem with monoids is that unlike in groups, you can't take inverses, so the kernel doesn't give you enough information.

There is however a form of isomorphism theorem that holds; instead of quotienting by a subobject, you need to quotient by the equivalence relation defined by $$x\sim y\Leftrightarrow \varphi(x)=\varphi(y).$$ Then you can define an operation on the equivalence classes by putting $\overline{x}\cdot \overline{y}=\overline{x\cdot y}$, and show that $\varphi$ factorise through the quotient; and this factorisation will be injective because of the choice of the relation.

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