Given two monoids we always have a morphism from one to the other thanks to the presence of the identity element.

Are there examples of non-empty semigroups that have no morphisms from one to the other? The destination semigroup can't be finite, because if we have an idempotent present, we just map everything to it and get a constant morphism. Apparently, it can't contain an idempotent, period.

Put another way, the subcategory of finite semigroups is clearly strongly connected. Is the same true of the category of all semigroups?

Are there two such non-empty semigroups that don't have a morphism in either direction?

  • $\begingroup$ @DanielSchepler ah yes, good catch, I am interested in the non-empty case. $\endgroup$
    – AlvinL
    Commented Apr 19, 2019 at 18:28
  • $\begingroup$ Note that since you can't have idempotents on either side, your semigroups cannot have a right-compatible topology that makes them compact $\endgroup$ Commented Apr 19, 2019 at 21:02
  • $\begingroup$ @Max this took an interesting turn. Would you formalise your string of thought as an answer, please? $\endgroup$
    – AlvinL
    Commented Apr 20, 2019 at 10:39
  • 1
    $\begingroup$ Well it's not worth an answer, it's just the fact that a compact topological space with a semigroup structure such that right multiplication is continuous has an idempotent, a very famous theorem $\endgroup$ Commented Apr 20, 2019 at 10:48
  • $\begingroup$ @Max ah I see, thanks for the reference, in that case! $\endgroup$
    – AlvinL
    Commented Apr 20, 2019 at 10:51

3 Answers 3


Let $\mathbb N^+$ be the natural numbers without 0, and consider it as a semigroup under addition. Then there can be no morphism $f: A \to \mathbb N^+$ where $A$ is finite, because then the image $f$ will be bounded by some $n \in \mathbb N^+$. Now trying to add an element $a \in A$ to itself $n+1$ times cannot be respected by the map: $$ f(\underbrace{a + \ldots + a}_{n+1 \text{ times}}) = \underbrace{f(a) + \ldots + f(a)}_{n+1 \text{ times}} \geq n+1, $$ which contradicts that the range of $f$ is bounded by $n$.

  • 2
    $\begingroup$ But as stated in the question, $A$ being finite means there is a morphism in the other direction $\endgroup$ Commented Apr 19, 2019 at 19:19
  • $\begingroup$ Yes, the original question asked just for two semigroups such that there is no morphism from one to the other. For the new question I wouldn't know an answer directly. If I think of something I will edit. $\endgroup$ Commented Apr 19, 2019 at 20:08
  • $\begingroup$ Oh sorry I hadn't seen the original question ! $\endgroup$ Commented Apr 19, 2019 at 20:18

No morphism in either direction

Choose two distinct primes $p,q\in\Bbb N_+$ and consider the additive semi-groups

$$\begin{align} P&:=\Big\{\frac n{p^m}\mid n,m\in\Bbb N_+\Big\}\subseteq\Bbb Q_+,\\ Q&:=\Big\{\frac n{q^m}\mid n,m\in\Bbb N_+\Big\}\subseteq\Bbb Q_+. \end{align}$$

Assume there is a morphism $\phi:P\to Q$ and let $a/q^b:=\phi(1)\in Q$ for some $a,b\in\Bbb N_+$. Further, for every $m\in\Bbb N_+$ let

$$\frac{a_m}{q^{b_m}}:=\phi\Big(\frac1{p^m}\Big)\in Q, \qquad\text{for some $a_m,b_m\in\Bbb N_+$}.$$

This means

$$\begin{align} p^m\cdot \frac{a_m}{q^{b_m}}&=\phi\Big(p^m\cdot \frac1{p^m}\Big)\\ &=\phi(1)\\ &=\frac a{q^b}, \end{align}$$

which implies $$p^mq^b\cdot a_m=q^{b_m}\cdot a.$$

Since the left side is divisible by $p^m$, so must be the right side. Since $p$ and $q$ are distinct primes, $a$ must be divisible by $p^m$ for all $m\in\Bbb N_+$, which is a contradiction. Hence there cannot be such a morphism, and since the argument is symmetric, there is no such morphism in either direction.


  • $\begingroup$ I am afraid that $1$ is not in $P$. And if you allow $m=0$ in your definition to include $1$, then $P$ would contain an idempotent, which is not possible. $\endgroup$
    – J.-E. Pin
    Commented Apr 26, 2019 at 6:23
  • $\begingroup$ @J.-E.Pin I dont see what you mean. $1=p/p^{1}$ is in $P$, as $p$ and $1$ are in $\Bbb N_+$. $\endgroup$
    – M. Winter
    Commented Apr 26, 2019 at 6:24
  • $\begingroup$ But then, isn't $\phi(x) = 1$ for all $x$ a morphism? $\endgroup$
    – J.-E. Pin
    Commented Apr 26, 2019 at 6:30
  • $\begingroup$ @J.-E.Pin I consider these semi-groups with the operation of addition. Zero would be an idempotent, but one is okay. $\endgroup$
    – M. Winter
    Commented Apr 26, 2019 at 6:31
  • $\begingroup$ What about changing "semigroups" to "additive semigroups" in the first sentence? $\endgroup$
    – J.-E. Pin
    Commented Apr 26, 2019 at 6:45

There's no morphism from $\{0\}$ (or from any semigroup with idempotent) to the additive semigroup on $\{2,4,6,\dots\}$

  • $\begingroup$ Why did you pick $\{2,4,6,\dots\}$ instead of $\{1,2,3,\dots\}$? (Just curious) $\endgroup$ Commented Apr 19, 2019 at 18:17
  • $\begingroup$ @AlexKruckman doesn't matter I guess. I think the same argument passes as long as there is no idempotent to map to $\endgroup$
    – AlvinL
    Commented Apr 19, 2019 at 18:25
  • $\begingroup$ Yes. Well, my first thought was that the ring $2\Bbb Z$ has unusual ring properties.. $\endgroup$
    – Berci
    Commented Apr 19, 2019 at 19:02
  • $\begingroup$ I read the question as asking for two semigroups such that there is no morphism in either direction. $\endgroup$ Commented Apr 24, 2019 at 22:34
  • $\begingroup$ @Acccumulation There are two questions, he answered one of them. $\endgroup$
    – AlvinL
    Commented Apr 25, 2019 at 7:45

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