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How to formalize this unary operation?

  1. Mar 2, 2009 #1
    I have two commutative groups [tex](G,\circ, I_\circ)[/tex] and [tex](G,\bullet,I_\bullet)[/tex], and I defined an isomorphism [tex]f[/tex] between them: so we have [tex]f(u \circ v)=f(u) \bullet f(v)[/tex]

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

    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?

    Thanks in advance!
  2. jcsd
  3. Mar 2, 2009 #2


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

    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!)
  4. Mar 3, 2009 #3
    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 [tex]f(u \bullet_A v)=f(u) \bullet_B f(v)[/tex] must be satisfied for every operation of n-arity, we get exactly what I wanted.

    Was that correct?
  5. Mar 3, 2009 #4
    a further question to add to what has been already discussed:

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

    which we call [tex]':G \rightarrow G[/tex]

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

    [tex]\forall u \in S, \\\ u'=u[/tex]

    As far as I understood when I'm defining a universal-algebra I cannot make any statement involving [tex]\forall, \in, \exists [/tex] 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:

    [tex]u' \ast k_1 = u[/tex]
    [tex]u' \ast k_2 = u[/tex]


    [tex]u' \ast\ k_n = u[/tex]

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