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, $$\ker \varphi:=\{x\in G\mid \varphi(x)=e\}$$


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?


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)

  • $\begingroup$ I highly doubt this is really the case, or at least as simple as that; for start, it is not true that $\ker \varphi=0$ implies that $\phi$ is injective if $\varphi$ is a morphism of monoids. $\endgroup$
    – Arnaud D.
    Dec 28, 2016 at 15:25
  • $\begingroup$ Your Proposition 2 is true, I think (to be sure, try checking if the standard proof for groups still goes through -- I don't have the time for it). But your Proposition 1 is false; $\ker\varphi$ does not "know" which elements of $G$ get sent to the same images under $\varphi$. The proper analogue of a normal subgroup for monoids would be some sort of "normal congruence relation". See the "Congruences and Quotients" section in Alan J. Cain's Nine Chapters on the Semigroup Art ( www-history.mcs.st-and.ac.uk/~alanc/pub/c_semigroups/index.html ). $\endgroup$ Dec 28, 2016 at 15:27
  • $\begingroup$ @darijgrinberg: Thanks for suggesting the link. $\endgroup$
    – user170039
    Dec 28, 2016 at 15:32
  • $\begingroup$ @darijgrinberg: I am not sure what you mean by the "standard proof for groups" but if you mean the proof of Theorem 9.2 of this book to be "standard proof" then I don't think that the argument goes through. $\endgroup$
    – user170039
    Dec 28, 2016 at 15:37
  • 1
    $\begingroup$ @user170039: Okay, Gallian is somewhat overcomplicating things. You want to prove that if $aH = a'H$ and $bH = b'H$, then $abH = a'b'H$. Argue as follows: From $bH = b'H = Hb'$, you obtain $abH = aHb' = a'Hb'$ (since $aH = a'H$). Thus, $abH = a'Hb' = a'b'H$ (using $Hb' = b'H$ again). $\endgroup$ Dec 28, 2016 at 15:42

2 Answers 2


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.


As Arnaud D. already pointed out, your approach is too naive to work properly, but you can consider the kernel congruence associated to $\varphi$. This approach works for many types of algebras, including monoids. See First Isomorphism Theorem for more details.

Defining the kernel of an homomorphism of monoids is another story and it took a long time to obtain an appropriate definition. The main problem with your definition 3, which has been frequently used, is that it does not contain enough information.

The difference between monoids and groups can be outlined as follows. Let $\pi: H \to G$ be a surjective group morphism and let $K = \pi^{-1}(1)$. Then $\pi(x_1) = \pi(x_2)$ iff $x_1x_2^{-1} \in K$. Let now $\pi: M \to G$ be a surjective monoid morphism and let $K = \pi^{-1}(1)$. Even if $K$ is given, it does not help to decide whether $\pi(x_1) = \pi(x_2)$, a major difference!

In order to avoid this problem, one needs to consider more general algebras than monoids, namely categories. I will just explain the case of a surjective monoid morphism $\pi$ from a monoid $M$ to a group $G$ (the general case, where $G$ is a monoid, is slightly more difficult to handle).

Definition. The kernel category $Ker(\pi)$ of $\pi$ has $G$ as its object set and for all $u,v \in G$ $$ Mor(u, v) = \{ (u, m, v) \in G \times M \times G \mid u\pi(m) = v \} $$ Note that $Mor(u, u)$ is a monoid equal to $\pi^{-1}(1)$ and that $G$ acts naturally (on the left) on $Ker(\pi)$.

I will not give here any detail on this theory, but I would like to give its spirit. In group theory, if $\pi: H \to G$ is a surjective group morphism and $K$ is the kernel of $\pi$, the synthesis problem consists in constructing $H$ given $G$ and $K$. This is in general a very difficult problem and there is a huge literature on this topic. In the same way, if $\pi: M \to G$ is a surjective monoid morphism, one would like to get information on $M$, given $G$ and $Ker(\pi)$. The results obtained in monoid theory are much weaker than the corresponding results for groups, but are nevertheless nontrivial to obtain.

For more information on this topic see these slides and the following references:

[1] S. Margolis, J.-É. Pin, Inverse semigroups and extensions of groups by semilattices, J. of Algebra 110 (1987), 277-297.

[2] J. Rhodes, B. Steinberg, The $q$-theory of finite semigroups. Springer Monographs in Mathematics. Springer, New York, 2009. xxii+666 pp. ISBN: 978-0-387-09780-0

[3] B. Steinberg, B. Tilson, Categories as algebra. II. Internat. J. Algebra Comput. 13 (2003), no. 6, 627--703.

[4] Tilson, Bret. Categories as algebra: an essential ingredient in the theory of monoids. J. Pure Appl. Algebra 48 (1987), no. 1-2, 83--198.

  • $\begingroup$ I'm not sure I follow the part about he synthesis problem. $H=G\times K$ seems to work fine. Or do you mean finding all possible $H$? $\endgroup$
    – tomasz
    Jun 4, 2018 at 11:58
  • $\begingroup$ @tomasz Here $H$, $G$ and $K$ are given, but the problem is to build $H$ from $G$ and $K$: for instance, is it always a semidirect product? $\endgroup$
    – J.-E. Pin
    Jun 4, 2018 at 13:40
  • $\begingroup$ That's how I build it: take a product. That is a solution to the problem. But I understand you mean to find all solutions. $\endgroup$
    – tomasz
    Jun 4, 2018 at 17:00

You must log in to answer this question.