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Math Help - Monoid homomorphisms

  1. #1
    Forum Admin topsquark's Avatar
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    Monoid homomorphisms

    Here is the question:
    If f: G \to H is a homomorphism of groups then f(e_G) = e_H and f \left (a^{-1} \right ) = f(a)^{-1} for all a \in G. Show by example that the first conclusion may be false if G, H are monoids that are not groups.
    Just in case:
    A monoid is a set G with a binary operation and contains a two-sided identity element e \in G such that ea = ae = a ~\forall a \in G.
    I have constructed a couple of monoids with homomorphisms, but I'm not getting anywhere. In addition I have a proof of f(e_G) = e_H that does not use the inverses relationship f \left ( a^{-1} \right ) = f(a)^{-1}, so I don't see where this would enter the problem. Additionally the proof does not seem to give me any clues as to how f(e_G) might not be equal to e_H.

    The proof is short so I will include it. Perhaps someone may be able to see a loophole that is escaping me.

    Proof:
    Let G be a set with a binary relation * and let there be an element of G, e_G such that e_G*g =  g*e_G = g for all g in G. Let H be another set with a binary relation \times and let there be an element e_H such that e_H \times h = h \times e_H = h for all h in H. Further let f be a homeomorphism f: G \to H.

    Then
    f(a) \times f(e_G) = f(a*e_G) = f(a). Since by definition h \times e_H = h for all h, f(e_G) must be e_H.

    -Dan
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    Forum Admin topsquark's Avatar
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    I had a thought while making my lunch. I'm reconsidering the line in the proof above which states:
    f(a) \times f(e_G) = f(a) and interpreting f(e_G) as e_H because we can compare it with the equation h \times e_H = h. My thought was that we cannot simply compare the equations and say the inference is correct. So if we can't do the comparison then we can't prove that f(e_G) = e_H. (In other words we cannot solve the equation h \times f(e_G) = h for f(e_G) as the inverse of h need not exist in a monoid.)

    But this raises a different problem for me. Take a look at the problem of finding the inverse of an element ab where a and b are in a group G. We can do this: consider the element b^{-1}a^{-1} \in G. I say this is equal to the inverse of the element of ab in G. For (ab)(b^{-1}a^{-1}) = a (b b^{-1}) a^{-1} = aea^{-1} = aa^{-1} = e. Since we have
    (ab)(ab)^{-1} = e and
    (ab)(b^{-1}a^{-1}) = e

    then (ab)^{-1} = b^{-1}a^{-1} by comparison.

    So is the comparison a valid method or not?

    -Dan
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    the answer depends on how you define a homomorphism between monoids. some definitions REQUIRE that a monoid homomorphism preserve the identity, and in that case, you can't provide a counter-example.

    if, on the other hand, you are only requiring the definition of a homomorphism as given for a group, that is:

    f(ab) = f(a)f(b), then it is easy to construct such a map which does not preserve the identity. consider the following two-element monoid N = {e,a}, where a^2 = a, and e is the identity.

    for any other monoid M define f:M-->N by f(x) = a, for all x in M. then f preserves products but does not send the identity of M to e.
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    Forum Admin topsquark's Avatar
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    Thanks, Deveno, that was a nice counterexample.

    I came up with a solution to my other quetion in my last post.

    If we have
    (ab)(ab)^{-1} = e and
    (ab)(b^{-1}a^{-1}) = e

    Then we know that
    (ab)(ab)^{-1} = e = (ab)(b^{-1}a^{-1})

    Now just multiply both sides by b^{-1}a^{-1} on the right...
    b^{-1}a^{-1}(ab)(ab)^{-1} = b^{-1}a^{-1}(ab)(b^{-1}a^{-1})

    And we are left with (ab)^{-1} = b^{-1}a^{-1}

    -Dan
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    the humorous way of saying this is: if you want to take off your socks and shoes, take off the shoes first, and then the socks.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by topsquark View Post
    Here is the question:
    Just in case:
    I have constructed a couple of monoids with homomorphisms, but I'm not getting anywhere. In addition I have a proof of f(e_G) = e_H that does not use the inverses relationship f \left ( a^{-1} \right ) = f(a)^{-1}, so I don't see where this would enter the problem. Additionally the proof does not seem to give me any clues as to how f(e_G) might not be equal to e_H.

    The proof is short so I will include it. Perhaps someone may be able to see a loophole that is escaping me.

    Proof:
    Let G be a set with a binary relation * and let there be an element of G, e_G such that e_G*g =  g*e_G = g for all g in G. Let H be another set with a binary relation \times and let there be an element e_H such that e_H \times h = h \times e_H = h for all h in H. Further let f be a homeomorphism f: G \to H.

    Then
    f(a) \times f(e_G) = f(a*e_G) = f(a). Since by definition h \times e_H = h for all h, f(e_G) must be e_H.

    -Dan
    Perhaps a more interesting example consider \mathbb{Z}\times\mathbb{Z} as a monoid with coordinate-wise multiplication, the identity obviously being (1,1). Consider then \mathbb{Z}\times\{0\} with coordinate-wise multiplication. This is a monoid with identity (1,0). This serves to show that a subset of a monoid which itself is a monoid with the same operation need not be a submonoid. Moreover, consider then the embedding \iota:\mathbb{Z}\times\{0\}\hookrightarrow\mathbb{  Z}\times\mathbb{Z} this is clearly a homomorphism of semigroups but not a monoid homomorphism since (1,0)\mapsto (1,0)
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