Semigroups with no morphisms between them

Question

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?

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

Answer 1

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.

$\square$

Answer 2

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

Answer 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 $n$.