## Epimorphisms in the category of groups

A morphism ${f:X\rightarrow Y}$ in a category ${\mathcal{C}}$ is said to be an epimorphism or epic if the relation ${\alpha\circ f=\beta\circ f}$, for any two morphisms ${\alpha,\beta:Y\rightarrow Z}$ and an arbitrary object ${Z}$, implies ${\alpha=\beta}$. It is not too difficult to see that the epimorphisms of the category of sets are exactly the surjective functions.

Informally, we said that a category is set-based, if its objects are sets with additional structures (e.g., semigroups, groups, vector spaces, rings, topological spaces or topological groups), and its morphisms are functions preserving structure.

Proposition 1. In any set-based category, surjective morphisms are epimorphisms.

Proof: Let ${f:X\rightarrow Y}$ be a surjective morphism in a set-based category ${\mathcal{C}}$. Now consider morphisms ${\alpha,\beta:Y\rightarrow Z}$ in ${\mathcal{C}}$ with ${\alpha\circ f=\beta\circ f}$. Let ${y\in Y}$. Since ${f}$ is surjective, there is an ${x\in X}$ with ${y=f(x)}$. Thus ${\alpha(y)=\alpha(f(x))=\beta(f(x))=\beta(y)}$. Hence ${\alpha=\beta}$. $\Box$

In general, the converse of Proposition 1 is false. It is certainly true in the category of sets or in the category of topological spaces, but it is not in the category of semigroups; indeed, the inclusion ${\mathbb{N}\rightarrow\mathbb{Z}}$ is an epic. We also find non-surjective epimorphisms in the category of Hausdorff spaces and continuous maps.

In this post, we deal with the category of groups, and provide two proofs of the following result.

Theorem 2. In the category of groups, every epimorphism is surjective.

First Proof.

Let ${f:H\rightarrow G}$ be an epimorphism in the category of groups. We recall that, in each category, if the composition ${g\circ h}$ is epic, so is ${g}$. In view of the Theorem of canonical decomposition of group morphisms, there is thus no loss in generality if we suppose that ${H}$ is a subgroup of ${G}$, and that ${f}$ is the inclusion morphism. First, we assume that the index ${(G:H)}$ of ${H}$ in ${G}$ is ${2}$. Then ${H}$ is a normal subgroup of ${G}$. Let ${\alpha,\beta:G\rightarrow G/H}$ be the quotient and constant morphisms, respectively. Then ${\alpha f=\beta f}$. Since ${f}$ is an epimorphism, it follows that ${\alpha=\beta}$. Thus the factor group is trivial. This contradicts ${(G:H)=2}$. The assumption ${(G:H)=2}$ is so false. Now we suppose that there are at least three cosets of ${H}$ in ${G}$. Then we find two elements ${u,v\in G}$ such that the cosets ${H}$, ${uH}$ and ${vH}$ are pairwise disjoint. Since ${uh\mapsto vh:\ uH\rightarrow vH}$ is a bijection, the assignments ${\sigma(uh)=vh}$, ${\sigma(vh)=uh}$ for ${h\in H}$, and ${\sigma(g)=g}$ for ${g\in G\setminus(uH\cup vH)}$, yield a member ${\sigma:G\rightarrow G}$ of the group ${\mathrm{Perm}\,G}$ of all permutations of the set ${G}$. For ${g\in G}$, let ${\alpha_{g}}$ denote the right translation ${x\mapsto xg:\ G\rightarrow G}$. Then ${\alpha:G\rightarrow\mathrm{Perm}\,G, \ g\mapsto\alpha_{g}}$ and ${\beta:G\rightarrow\mathrm{Perm}\,G:\ g\mapsto\sigma^{-1}\alpha_{g}\sigma}$ are group morphisms. For ${x\in H}$ and ${g\in G}$, we compute

$\displaystyle (\beta f)(x)(g)=\beta(x)(g)=\sigma^{-1}\alpha_{x}\sigma(g)=\sigma^{-1}\bigl(\sigma(g)x\bigr).$

Thus

$\displaystyle (\beta f)(x)(g)=\sigma^{-1}(vhx)=uhx=gx \ \text{ if }\ g=uh\text{ with }h\in H,$

$\displaystyle (\beta f)(x)(g)=\sigma^{-1}(uhx)=vhx=gx \ \text{ if }\ g=vh\text{ with }h\in H,$

and

$\displaystyle \sigma^{-1}(gx)=gx \ \text{ if }\ g\in G\setminus\left(uH\cup vH\right).$

Thus ${\alpha f=\beta f}$. However

$\displaystyle \beta(x)(u)=\sigma^{-1}\alpha_{x}\sigma(u)=\sigma^{-1}\bigl(\sigma(u)x\bigr)=\sigma^{-1}(vx)=vx\neq ux=\alpha(x)(u)$

for ${x\in G\setminus(H\cup v^{-1}uH)}$, since ${vx\in G\setminus(uH\cup vH)}$. That is, ${\alpha\neq\beta}$. This contradicts the fact that ${f}$ is an epimorphism. Consequently, ${(G:H)=1}$, that is, ${H=G}$. The surjectivity of ${f}$ is so proved. ${\Box}$

Second Proof.

As stressed in the preceding proof, it is sufficient to show that for a group ${G}$ and an epimorphically embedded subgroup ${H}$, the equality ${H=G}$ holds. We consider a ${G}$-module ${E}$ and form the semidirect product ${E\rtimes G}$ with multiplication ${(x,g)(y,h)=(x+g\cdot y,gh)}$. Let ${f:G\rightarrow E}$ denote a function. If the set ${\{\left.(f(g),g)\ \right|\ g\in G\}}$ is a subgroup of ${E\rtimes G}$, then, for all ${g,g^{\prime}\in G}$, we have

$\displaystyle \bigl(f(g),g\bigr)\bigl(f(g^{\prime}),g^{\prime}\bigr) =\bigl(f(g)+g^{\prime}\cdot f(g^{\prime}),gg^{\prime}\bigr)\in\bigl\{\left.(f\left(h\right),h)\ \right|\ h\in G\bigr\},$

that is, ${f(gg^{\prime})=f(g)+g^{\prime}\cdot f(g^{\prime})}$. Conversely, if ${f(gg^{\prime})=f(g)+g^{\prime}\cdot f(g^{\prime})}$ for all ${g,g^{\prime}\in G}$, then ${(f(g),g)(f(g^{\prime}),g^{\prime})=(f(gg^{\prime}),gg^{\prime})}$, ${f(\mathbf{1})=0}$ and

$\displaystyle (f(g),g)^{-1}=(-g^{-1}\cdot f(g),g^{-1})=(f(g^{-1}),g^{-1}).$

Thus ${\bigl\{\left.(f(g),g)\ \right|\ g\in G\bigr\}}$ is a subgroup of ${E\rtimes G}$ if and only if the equality ${f(gg^{\prime})=f(g)+g^{\prime}\cdot f(g^{\prime})}$ holds for all ${g,g^{\prime}\in G}$. Such functions are called ${1}$-cocycles.

Assume that ${H}$ is a proper subgroup of ${G}$. If there exists a non-constant ${1}$-cocycle ${f:G\rightarrow E}$ with ${f(H)=\{0\}}$, then the morphisms ${G\rightarrow E\rtimes G}$ given by ${g\mapsto(0,g)}$ and ${g\mapsto(f(g),g)}$ agrees on ${H}$, but are different. This contradicts the fact that ${H}$ is an epimorphically embedded subgroup. To conclude the proof, it is sufficient to find such a cocycle.

One source of cocycles are the so-called coboundaries ${f:G\rightarrow E}$ defined by ${f(g)=-m+g\cdot m}$ for some ${m\in E}$. Indeed,

$\displaystyle f(gg^{\prime})=-m+g\cdot(g^{\prime}\cdot m)=-m+g\cdot m-g\cdot m+g\cdot(g^{\prime}\cdot m),$

and therefore

$\displaystyle f(gg^{\prime})=-m+g\cdot m+g\cdot(-m+(g^{\prime}\cdot m))=f(g)+g\cdot f(g^{\prime}).$

We are thus done if we find an element ${m\in E}$ such that ${H\cdot m=\{m\}}$ and ${g\cdot m\neq m}$ for some ${g\in G}$. We take an arbitrary abelian group ${A}$ which contains at least two elements (for instance, ${A=\mathbb{Z}/p\mathbb{Z}}$ for some prime number ${p}$) and set ${E=A^{G}}$. Then ${E}$ is a ${G}$-module via the action

$\displaystyle (g,\varphi)\mapsto g\cdot\varphi:\ G\times E\rightarrow E$

given by ${(g\cdot\varphi)(g^{\prime})=\varphi(g^{\prime}g)}$. If ${H\neq G}$, then there is an element ${g_{0}\in G\setminus H}$. Consider a function ${\varphi:G/H\rightarrow A}$ such that ${\varphi(g_{0}H)\neq\varphi(H)}$. Then the composition of the quotient map ${G\rightarrow G/H}$ after ${\varphi}$ yields a function ${m:G\rightarrow A}$ such that ${m(gH)=\{m(g)\}}$ for all ${g\in G}$ and ${m(g_{0})\neq m(\mathbf{1})}$. Thus ${(h\cdot m)(g)=m(gh)=m(g)}$ for all ${h\in H}$ and ${g\in G}$, but ${(g_{0}\cdot m)(\mathbf{1})=m(g_{0})\neq m(\mathbf{1})}$. Hence ${H\cdot m=\{m\}}$ and ${g_{0}\cdot m\neq m}$. ${\Box}$

### 7 thoughts on “Epimorphisms in the category of groups”

1. Todd Trimble March 8, 2015 / 10:39 pm

Hi; this is a nice post. I think however there is a little detail in the second proof that needs fixing. Namely, the left action of $G$ on $A^G$ isn’t actually a left action; I think you probably want instead $(g \cdot \varphi)(g') = \varphi(g' g^{-1})$.

• Christian Nguembou Tagne April 4, 2018 / 11:24 pm

Thank you for your comment. But, I think it is really a left action. Indeed, for $g,h,g^{\prime}\in G$ and $\varphi\in A^{G}$, we compute

$\displaystyle [g\cdot (h\cdot\varphi)](g^{\prime})=(h\cdot\varphi)(g^{\prime}g)=\varphi\Bigl((g^{\prime}g)h\Bigr)=\varphi\Bigl(g^{\prime}(gh)\Bigr)=[(gh)\cdot\varphi](g^{\prime})$.

Hence $g\cdot (h\cdot\varphi)=(gh)\cdot\varphi$.

2. Michael Weiss July 21, 2017 / 10:28 pm

There’s another permutation-based proof, which I prefer. I think it was first published by Carl Linderholm (AMM 77(2) Feb. 1970 176–177). As already noted, it’s enough to show that if H is a proper subgroup of G, then there are two distinct homomorphisms from G to some other group that agree on the subgroup H. Elements of G act on the set of left cosets G/H by left translation; this gives a homomorphism from G to Perm(G/H), say g–>L_g. Now add a new element to G/H, call it A; let S be the new set, i.e., S is G/H union {A}. Extend L_g to S by declaring that L_g(A)=A. So L is a homomorphism G–>Perm(S).

Now let s be the permutation switching A and H (both elements of S). Let M_g be L_g conjugated by s, i.e., M_g(X)=s(L_g(s^{-1}(X))). (Since s^2=1, I don’t really have to write s^{-1}, but it really doesn’t matter.) Obviously M_g is also a homomorphism G–>Perm(S).

For h in H, L_h and s commute— obvious, since L_h leaves A and H fixed,
while s leaves everything but A and H fixed. So M_h and L_h are the same permutation. On the other hand, if g is not in H (so gH =/= H) then it’s immediate that L_g and s don’t commute: just compute sL_g(A) and compare it with L_gs(A) (or do the same with H instead of A). So L_g and M_g differ when g is not in H. QED

One reason I prefer this: you don’t have to treat G:H=2 as a special case.

3. Cyrille Nganteu(from cameroon) August 30, 2018 / 2:57 pm

Always a good mathematician. Short results but very usefull. Bravo.

4. evelyn November 10, 2018 / 3:57 am

In the second proof, why use 1 and not general, {m(g_{0})\neq m(\mathbf{1})}. Or simply, why was it necessary to use 1 and not general?

• Christian Nguembou Tagne November 10, 2018 / 3:00 pm

In the last part of the second proof, we want to find a function $m$ from $G$ to $A=\mathbb{Z}/p\mathbb{Z}$, where $p$ is a prime number, such that $H\cdot m=\{m\}$ and $g_{0}\cdot m\neq m$. Note that $g_{0}\cdot m$, as well as $m$, is a function from $G$ to $A$. Therefore, to show that $g_{0}\cdot m\neq m$, it suffices to establish that $(g_{0}\cdot m)(x)\neq m(x)$ for some $x\in G$. However, by definition, $(g_{0}\cdot m)(x)=m(xg_{0})$. For example, if $x$ is the identity $\mathbf{1}$, we compute $(g_{0}\cdot m)(\mathbf{1})=m(g_{0})$. Thus, if we find that $m(g_{0})\neq m(\mathbf{1})$, we can conclude that $g_{0}\cdot m\neq m$. That was our approach here. It is also not a necessity to use the identity $\mathbf{1}$ of $G$. One could theoretically use any element $x$ of the group $G$. However, in this context, the identity is the optimal choice.