[Math] Anti-isomorphisms

abstract-algebragroup-homomorphismgroup-isomorphismgroup-theory

Definition: Let $G,H$ be groups and $f: G \to H$ a function such that $f(gh) =f(h)f(g)$ for any $g,h \in G$. Then we call $f$ an antihomomorphism. (Note the swapped order of $f(h)$ and $f(g)$.)

I was deriving some properties of antihomomorphisms and I found that there were a lot of similarities with the usual homomorphisms:

For example, for any antihomomorphism $f: G \to H$, we have:

  1. $f(e_G) = e_H$

  2. $f(g^{-1}) = f(g)^{-1}$

If we define $ker f$ in the usual way (i.e. $ker f = \{g \in G|f(g) = e_H\}$), we have:

  1. $f$ injective $\iff ker f = \{e_G\}$

  2. $ker f \unlhd G$

If we denote the existence of a bijective antihomomorphism between $2$ groups with $\asymp$, we have:

  1. $G/ker f \asymp Im(f)$

Also interesting:

  1. $G \asymp H$ and $H \asymp F \Rightarrow G \cong F$

I know that the existence of an isomorphism between $2$ groups means that both groups have exactly the same structure.

So my question is:

From a group theoretic point of view, what is the use of bijective
antihomomorphisms (= anti-isomorphisms) between 2 groups. Can we give
it an interpretation like we have for regular isomorphisms?

Best Answer

These questions relate to the following construction:

Let $(G,\cdot)$ be a group. Define its opposite $G^{op}$ to be the group $(G,*)$ where we define $$x*y=y\cdot x.$$

One may check that this indeed forms a group. Then, an antihomomorphism $f:G\rightarrow H$ is exactly a homomorphisms $G\rightarrow H^{op}$ or, equivalently, a homomorphism $G^{op}\rightarrow H$.$^1$ Essentially all of your results follow from the fact that antihomomorphisms are homomorphisms - their domain is just the opposite of what its marked as.

In some sense, an anti-isomorphism is the same as a normal isomorphism. This is because $G$ and $G^{op}$ are always isomorphic - the map $f:G\rightarrow G^{op}$ defined by $f(g)=g^{-1}$ is an isomorphism.$^2$ Thus, if $G$ and $H$ are anti-isomorphic, they are actually isomorphic (and vice versa).

$^1$ One might observe that a homomorphism $G^{op}\rightarrow H^{op}$ is also a homomorphism $G\rightarrow H$ and vice versa.

$^2$ This follows easily since $f(g\cdot h)=(g\cdot h)^{-1}=h^{-1}\cdot g^{-1}=f(h)\cdot f(g)=f(g)*f(h)$. Obviously, this is bijective since inversion is an involution.

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