Is $\varphi(ab)=\varphi(b)\varphi(a)$ an equally valid definition of homomorphism?

Given the groups $$G$$ and $$\bar G$$, usually a homomorphism is defined as a map $$\varphi \colon G \longrightarrow \bar G$$ such that $$\varphi(ab)=\varphi(a)\varphi(b)$$. Now, if we'd define homomorphism a map such that, instead, $$\varphi(ab)=\varphi(b)\varphi(a)$$, we'd have anyway that $$\varphi(e)=\bar e$$ (based on the fact that the unit is the same left-side and right-side) and $$\varphi(a^{-1})=\varphi(a)^{-1}$$ (based on the fact that the inverse is the same left-side and right-side).

So, up to the point of this basic Lemma, I'd be led to conclude that an equally valid definition of homomorphism is that one of a group map such that $$\varphi(ab)=\varphi(b)\varphi(a)$$.

My questions are:

1. Am I wrong?
2. If the answer to 1 is "no", can we keep simultaneously both definitions of homomorphism or do we have to make an original choice between the two, and keep it along the way to avoid inconcistencies downstream the above basic Lemma?
• Correct me if I'm wrong, but I don't see a fundamental difference between right and left multiplication, since you could define a new "left multiplication" based on the right multiplication, i.e., c*d = dc. – Andrew Yuan Feb 11 at 9:20
• One difference is that your "antihomomorphisms" are not closed under composition. And that is a desirable property. In fact a composition of two "antihomomorphisms" is a normal homomorphism. – freakish Feb 11 at 9:31
• According to this definition, the identity mapping on a noncommutative group would not be a homomorphism, right? That seems like a problem. – littleO Feb 11 at 9:32
• The opposite group is relevant here, and this is basically what @AndrewYuan noticed in their comment. – user1729 Feb 11 at 10:03

A map $$\phi:G\rightarrow H$$ between, say, groups such that $$\phi(ab)=\phi(b)\phi(a)$$ for all $$a$$, $$b\in G$$ is usually called an antihomomorphism.

The simplest instance is that of the inverse map $${\rm inv}_G:G\longrightarrow G,\qquad{\rm inv}_G(g)=g^{-1}.$$ which in fact is an antiautomorphism.

Note that if $$\overline{\rm Hom}(G,H)$$ denotes the set of antihomomorphisms between $$G$$ and $$H$$ there is a bijection $${\rm Hom}(G,H)\longrightarrow\overline{\rm Hom}(G,H)$$ given equivalently by $$f\mapsto f\circ{\rm inv}_G$$ or $${\rm inv}_H\circ f$$.

A problem with antihomomorphisms is that in general $$\overline{\rm End}(G)$$ is not a group, it is not even closed under composition.

Obviously, every distinction becomes pointless if $$G$$ is abelian.

We can define the group $$\overline{G}^{\mathsf{op}}$$ by stating that this group has the same elements as group $$\overline{G}$$ but has a composition $$\circ^{\mathsf{op}}$$ defined by: $$x\circ^{\mathsf{op}}y:=y\circ x$$where $$\circ$$ denotes the composition on group $$\overline{G}$$.

Then a function $$\phi:G\to\overline{G}$$ with the property $$\phi(ab)=\phi(b)\circ\phi(a)$$ induces a grouphomomorphism $$\psi:G\to\overline{G}^{\mathsf{op}}$$ prescribed by $$\psi(a)=\phi(a)$$.

This because: $$\psi(ab)=\phi(ab)=\phi(b)\circ\phi(a)=\phi(a)\circ^{\mathsf{op}}\phi(b)=\psi(a)\circ^{\mathsf{op}}\psi(b)\tag1$$

If $$\overline{e}$$ is the identity element of $$(\overline{G},\circ)$$ then it can easily be verified that is also the identity element of $$(\overline{G},\circ^{\mathsf{op}})$$.

Then on base of $$(1)$$ it can be deduced that $$\psi(e)=\overline{e}$$ (hence also $$\phi(e)=\psi(e)=\overline{e}$$).

This because: $$\psi(e)=\psi(ee)=\psi(e)\circ^{\mathsf{op}}\psi(e)$$ and $$\overline{e}$$ is unique in satisfying $$\overline{e}\circ^{\mathsf{op}}\overline{e}=\overline{e}$$.

(Actually this can be concluded likewise and more directly from $$\phi(e)=\phi(e)\circ\phi(e)$$ also)

On base of this we can deduce that also: $$\phi(a^{-1})=\psi(a^{-1})=\psi(a)^{-1}=\phi(a)^{-1}$$