On group monomorphisms I read that in the category of groups every monomorphism is an equalizer. The proof I read uses the amalgamation property for groups, but I couldn't understand it. Is there an easier way to prove mono=equalizers in Grp category, or can someone explain me how to use amalgamation to prove that in simple words?
 A: Up to isomorphism, a monomorphism in the category of groups is the inclusion of a subgroup, so to prove that every monomorphism is an equalizer we need to show that for every group $G$ and subgroup $H\leq G$, there is another group $K$ with two homomorphisms $\alpha,\beta:G\to K$ such that $H=\{g\in G\vert \alpha(g)=\beta(g)\}$.
The "amalgamation" proof of this takes $K$ to be the amalgamated free product $G\ast_HG$ of two copies of $G$, amalgamating the copies of $H$. Or in categorical language, $K$ is the pushout in the category of groups of the diagram $G\hookleftarrow H\hookrightarrow G$, which comes equipped with two maps $G\to K$. Clearly the equalizer contains $H$, and there is a normal form for elements of an amalgamated free product (see any group theory book that develops the theory of free products and amalgamated free products) that allows you to check explicitly that the equalizer is precisely $H$.
Even if $G$ is finite, $G\ast_HG$ is infinite unless $H=G$, so this proof doesn't show that in the category of finite groups, monomorphisms are equalizers. There's another proof I like that does.
An action of a group on a set is the same as a homomorphism from the group to the group of permutations of the set, so instead of homomorphisms to a group $K$, it's enough to produce two actions of $G$ on a set $X$ that agree on precisely the elements of $H$.
Let $X=G$, and the first action will be the regular action (by left multiplication).
Let $C$ be a set of right coset representatives of $H$ in $G$, including the identity element $1$. If the index $\vert G:H\vert$ is at least $3$, we can choose a fixed-point free permutation $\sigma$ of $C\setminus\{1\}$, and extend to a permutation of $G$ by $\sigma(h)=h$ and $\sigma(hc)=h\sigma(c)$ for $h\in H$, $c\in C\setminus\{1\}$.
The second action will be the regular action after permuting by $\sigma$: i.e., $g\cdot\sigma(x)=\sigma(gx)$. It's easy to check that elements of $H$ act the same way in both actions, but that if $g\not\in H$ then $g1$ and $g\cdot1$ are not equal, as they are in different cosets.
This leaves the case where $\vert G:H\vert=2$, which is easy since then $H$ is normal, and is the equalizer of the natural epimorphism and the trivial homomorphism $G\to G/H$.
