Confusion about categorical viewpoint of normal subgroups One way to define normal subgroups is as normal monos in $\mathsf{Grp}$. Another uses quotients by internal equivalence relations, which makes use (I think) of the fact $\mathsf{Grp}$ has effective equivalence relations - each equivalence relation is a the kernel pair of its coequalizer.
What's the theory linking these two definitions?
 A: Let me state the two definitions a bit more explicitly.  Let $\mathcal{C}$ be a category with a zero object.  For any objects $A$ and $B$, we let $0:A\to B$ denote the unique map which factors through a zero object.  Then:


*

*A normal monomorphism is a map $i:A\to B$ such that there exists an object $C$ and a map $f:B\to C$ such that $i$ is the equalizer of $f$ and $0:B\to C$.

*An equivalence class of zero (this is a term I just made up) is a map $i:A\to B$ such that there exists an equivalence relation $j:R\to B\times B$ such that $i$ is a pullback of $j$ by the map $(1,0):B\to B\times B$.


If $\mathcal{C}$ has pullbacks, then any normal monomorphism is an equivalence class of zero: suppose $f:B\to C$ is any map and let $j:R\to B\times B$ be the kernel pair of $f$.  Since $j$ is monic, an object $A$ with a map $i:A\to B$ and a map $A\to R$ such that 
$$\require{AMScd}
\begin{CD}
A @>>> R\\
@VV{i}V @VV{j}V  \\
B @>{(1,0)}>> B\times B 
\end{CD}$$
commutes is the same as a map $i:A\to B$ such that $(1,0)\circ i=(i,0)$ factors through $j$.  But by the definition of $j$, $(i,0)$ factors through $j$ iff $f\circ i=f\circ 0=0=0\circ i$.  That is, such an $i$ is the same thing as a map which equalizes $f$ and $0$.  This shows that a map is a kernel of $f$ iff it is an equivalence class of zero for the equivalence relation $j$.
Conversely if all equivalence relations in $\mathcal{C}$ are effective, then every equivalence class of zero is normal.  Indeed, for any equivalence relation $j:R\to B\times B$, there is then some $f:B\to C$ such that $j$ is the kernel pair of $f$.  The argument above then shows that an equivalence class of zero of $j$ is a kernel of $f$.
A: What links kernels and congruences? The Mal'cev property. 
In the setting of categories of algebraic structures with a constant, being a Mal'cev category essentially amounts to the existence of a subtraction operator. With that in mind, it is not surprising that kernels in Mal'cev categories contain the same information as congruences – after all, in $\mathbf{Grp}$, to quotient by a normal subgroup $N$ amounts to quotienting by $x \sim_N y$ where $x \sim_N y$ iff $x y^{-1} \in N$.
