# How to formalize this unary operation?

#### mnb96

Hello,
I have two commutative groups $$(G,\circ, I_\circ)$$ and $$(G,\bullet,I_\bullet)$$, and I defined an isomorphism $$f$$ between them: so we have $$f(u \circ v)=f(u) \bullet f(v)$$

How can I formalize the fact that I want also an unary operation $$\ast : G \rightarrow G$$ which is preserved by the isomorphism? namely, an unary operation such that $$f(u^{\ast}) = f(u)^{\ast}$$ ?

Is it possible somehow to embed the unary operation into the group in order to form an already-known algebraic structure? or it is just not possible to formalize it better than I already did?

#### Hurkyl

Staff Emeritus
Gold Member
I don't know if there is a special name for "groups with an additional unary operation", but any homomorphism $(G,\circ,\star,1) \to (H, \bullet, *, i)$ of "groups with an additional unary operation" should indeed satisfy $f(x^\star) = f(x)^*$.

For a theoretical sledgehammer, "groups with an additional unary operation" are an example of a universal algebra, just like groups, rings, and vector spaces over R are. (But not fields!)

#### mnb96

Thanks hurkyl!
your explanation was clear and indeed, if I understood correctly, what I wanted to create is essentially an isomorphism between two (universal) algebras of signature (2,1,1,0).

In fact, from an universal-algebraic point of view, a group has three operation: one binary associative (arity=2), one inverse (arity=1), and the identity element (arity=0); when I include another unary operation of arity=1, we have (2,1,1,0) signature.

And since the homomorphic property $$f(u \bullet_A v)=f(u) \bullet_B f(v)$$ must be satisfied for every operation of n-arity, we get exactly what I wanted.

Was that correct?

#### mnb96

a further question to add to what has been already discussed:

let's say I have this "group with an additional unary operation" $$(G,\ast, I,\\ ^{-1},\\')$$

which we call $$':G \rightarrow G$$

how can I formalize in terms of universal-algebras that for a subset $$S \subseteq G$$ we always have:

$$\forall u \in S, \\\ u'=u$$

As far as I understood when I'm defining a universal-algebra I cannot make any statement involving $$\forall, \in, \exists$$ and stuff like that, because one must use only operators of n-arity.
Do I really need to introduce a nullary-operator for each element in S and do something like this:

$$u' \ast k_1 = u$$
$$u' \ast k_2 = u$$

$$\ldots$$

$$u' \ast\ k_n = u$$

What if those elements are infinite?
what the signature of this algebra would be?

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