# Relation between kernel of groups and Semigroups

Let $$G$$ and $$H$$ are groups and $$f:G\to H$$ is group homomorphism. The kernel $$f$$ is defined by \begin{align*} \ker f =\{ g\in G \ : \ f(g)=1_{H}\}. \end{align*} And let $$S$$ and $$T$$ are Semigroups and $$\varphi:S\to T$$ is semigroup homomorphism. The kernel $$\varphi$$ is defined by \begin{align*} \ker \varphi=\{(s, s') \in S\times S \ : \ \varphi(s)=\varphi(s')\}. \end{align*} I know category of groups has a zero morphism, then the equalizer of zero morphism $$0_{GH}:G\to H$$ and $$f:G\to H$$ is kernel of $$f$$. But category of Semigroups doesn't has zero morphism.

I have two questions:

1. What is the relation between these kernels.

2. What is the definition of kernels in the category has no zero object or zero morphism.

Thank you

• Well, if $\varphi$ is a group homomorphism, and writing $\ker_G(\varphi)$ for the group definition of kernel and $\ker_S(\varphi)$ for the semigroup definition, we have that $(s,s')\in\ker_S(\varphi)\Leftrightarrow s^{-1}s'\in\ker_G(\varphi)$. (In semigroups, $s^{-1}$ does not necessarily make sense.) Does that help with your question (1)? – user1729 Sep 24 '19 at 10:07

In general algebraic structures, the kernel of a homomorphism $$f:X\to Y$$ is defined just as for semigroups: $$\ker f:=\{(x, x') : f(x) =f(x')\}$$ which is always a congruence relation on $$X$$, i.e. an equivalence relation closed under the operations (meaning that it's a subalgebra of $$X\times X$$).
For groups [or rings or vector spaces, Boolean algebras, etc.], there is a one-to-one correspondence between congruence relations and normal subgroups [ideals, subspaces, Boolean ideals, etc], namely the equivalence class of the identity element [or, of $$0$$] already determines the whole relation.
In category theory, we can reflect this general notion of kernel by a kernel pair: a pair of arrows $$k_1,k_2:K\to X$$ (where $$K$$ plays the role of the congruence relation by the induced arrow $$K\to X\times X$$), which satisfy $$f\circ k_1=f\circ k_2$$, and whenever $$f\circ u=f\circ v$$ with $$u,v:A\to X$$, there's a unique $$s:A\to K$$ satisfying $$u=k_1\circ s$$ and $$v=k_2\circ s$$.
Working in a category of general algebras, $$s$$ is simply $$A\ni\ a\mapsto (u(a),\, v(a))\ \in\ker f$$.