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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?

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    $\begingroup$ If I'm not missing anything, the definition of antihomomorphism is the same with homomorphism, so basically they are the same thing. $\endgroup$ – onurcanbektas Jul 16 '17 at 16:01
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    $\begingroup$ For abelian groups they are equivalent. Otherwise not. $\endgroup$ – user370967 Jul 16 '17 at 16:02
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    $\begingroup$ $f(gh) = f(h)f(g) \neq f(g)f(h)$ as $Im(f)$ is not necessarily abelian $\endgroup$ – user370967 Jul 16 '17 at 16:19
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    $\begingroup$ Everything is antıïsomorphic with itself, I believe. Use $f(g)=g^{-1}$. In fact, this should imply that two groups are antıïsomorphic iff they are isomorphic. $\endgroup$ – Akiva Weinberger Jul 16 '17 at 16:47
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    $\begingroup$ @Math_QED I didn't notice the places of the elements. $\endgroup$ – onurcanbektas Jul 16 '17 at 17:58
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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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An antihomomorphism can be used, for example, to turn a left group action into a right group action.

It is dual to the notion of homomorphism: if we see groups as categories with one object, then homomorphisms are covariant functors, antihomomorphisms are contravariant functors. Therefore, it is very natural that the two notions are very much similar.

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As Daniel mentioned in his answer, under the identification of a group with a groupoid with one object, group homomorphisms are functors, and antihomomorphisms are contravariant functors.

Under this identification, given group $G$ (a groupoid with one object), a left action of $G$ on an object $X$ in some category $\mathsf{C}$ is a functor $\lambda: G \to \mathsf{C}$ with $\lambda(1_G)=1_X$. Dually, a right action of $G$ on $X$ is a contravariant functor $\rho: G^{\mathrm{op}} \to \mathsf{C}$ with $\rho(1_G)=1_X$.

Thus, group homomorphisms appear when you want to consider left actions, and antihomomorphisms appear whenever you want to consider right actions. So if you want to see where antihomomorphisms turn up, just think about where right actions turn up. In Klein geometry, for example, the Cartan principal bundle $G \to G/H$ has a canonical right action by $H$.

Fortunately, the group inverse gives an anti-isomorphism $G^{\mathrm{op}} \xrightarrow{\cong} G$ of a group with its opposite group, so we can always convert right actions into left actions and vice-versa by precomposing with the group inverse.

For more details, see Aluffi's Algebra: Chapter 0.

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Sometimes anti-isomorphisms arise naturally: the map $g\to g^t$ on $GL_n(\mathbb{R})$ or $g\to g^\dagger$ on $GL_n(\mathbb{C})$, for example. You can interpret an anti-homomorphism $G \to H$ as a homomorphism $G \to H^{\text{op}}$, though, where $H^{\text{op}}$ is the group with the same underlying set as $H$ and group operation $x.y = yx$. The subject is not particularly interesting, though, since the map $H \to H^{\text{op}}$ given by $x \to x^{-1}$ is an isomorphism. For (noncommutative) rings, though, the situation is more interesting: the rings $A$ and $A^{\text{op}}$ are not necessarily isomorphic.

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Sorry if this has already been written in someone's answer, but it's worth pointing out that you don't really need anti-homomorphisms between groups. Given an anti-homomorphism $f : G \rightarrow H$, the function $g \mapsto (f(g))^{-1} = f(g^{-1})$ is automatically an ordinary homomorphism $G \rightarrow H$, and this process is bijective, and sets up a bijection between anti-isomorphisms and ordinary isomorphisms. So basically, as long as we take the "fundamental" anti-isomorphism $$G \rightarrow G$$ $$g \mapsto g^{-1}$$ seriously, we don't need to worry about any of the others (in some sense.)

The situation changes if we're dealing with monoids, in which case anti-homomorphisms become a bit more useful. In this context, they can also be referred to as "contravariant functors" in the sense of category theory, since a monoid is just a category with one object. And, just like how in category theory we can replace each contravariant functor $\mathbf{C} \rightarrow \mathbf{D}$ with a covariant functor $\mathbf{C}^{op} \rightarrow \mathbf{D}$ and vice versa, so too can we do this in the world of monoids. Nonetheless, I personally think that anti-homomorphisms of categories/monoids are pretty useful; it's just easier to think clearly when you're not introducing ops everywhere in order to artificially coerce everything in sight to be covariant.

That's my two cents.

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