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How many? -question

  1. Feb 12, 2005 #1
    "How many?"-question

    The problem:
    Let [tex]G[/tex] be a set with an associative binary operation and [tex]e \in G[/tex] an element satisfying the following conditions:

    1) [tex]eg=g[/tex] for any [tex]g \in G[/tex].
    2) For any [tex]g[/tex] there is [tex]h[/tex] such that [tex]gh=e[/tex].

    Assume that [tex]p[/tex] is a prime number and [tex]G[/tex] has [tex]p[/tex]-elements. How many non isomorphic such binary operations are on [tex]G[/tex] which are not groups?

    I know that there are 2 such operations if [tex]\mid G \mid =p^2[/tex], and 3 such operations if [tex]\mid G \mid =p*q[/tex], if [tex]p[/tex] and [tex]q[/tex] are not equal.
    So I'm guessing there answer to the problem is 1. What I've been trying to do for the past week has been to show that 1 such operations exists and to find a contradiction by assuming that a second operation also exists.
    But I haven't even been able to prove the existence.
    Any help would be greatly appreciated.
  2. jcsd
  3. Feb 12, 2005 #2
    I'll use + for the operation, but still use e for the left identity

    A little work shows you that the set of elements of G of the form g+e form a group. So G consists of this group of elements with extra elements g for which g+e is in the group.

    Suppose G has three elements, e,a and x. e and a are a two element group, and x is an 'extra' element. So we first have to decide what x+e is.

    Try x+e=e. Then x+g=x+e+g=e+g=g for each g in G, and we know e+x=x, so this just leaves a+x, which it seems can be either a or x.

    If we try x+e=a then if y=(a+x) then y+e=e, so y is not a or x. Hence a+x=e.
    However, I think we are still free to choose either x+x=a or x+x=x.

    That gives 4 possibilities, which I think are non-isomorphic.
  4. Feb 13, 2005 #3
    I don't quite understand what you mean by "the set of elements of G of the form g+e form a group". For the elements g+e to form a group doesn't g+e have to be g+e=e?
  5. Feb 13, 2005 #4
    Suppose n=f+e and m=g+e are elements of this set. Then n+m=f+e+g+e=f+g+e is also an element. Also n+e=f+e+e=f+e=n. So e is a right identity for this set. As for inverses, n+(-f)+e=f+e+(-f)+e=f+(-f)+e=e+e=e, so (-f)+e is a right inverse for n and is in the set. Hence this set of elements has a right inverse and right identity and so it can be shown that it forms a group.
  6. Feb 13, 2005 #5
    Hmm... In the first part of the problem (which I didn't write here) it was asked if G with this operation was a group. I thought I had found an example which showed that it was not a group.
    I took the set of all real numbers except 0, and defined the operation + as: a+b=|a|b. In this example the operation is associative and right inverse exists.
    I took -1 as identity:
    (-1)+g=|-1|g=1g=g, so a left identity exists, but
    g+(-1)=|g|(-1)=-g, the left identity is not a right identity.
    I thought this showed that the operation with the set was not a group. Where am I thinking wrong?
    Last edited: Feb 13, 2005
  7. Feb 13, 2005 #6
    You're not thinking wrong. I didn't say that G was a group. I said that the subset of elements of the form g+e formed a group.

    Since your example is multiplicative, I'll use . as the operator, so a.b=|a|b
    Now g.e=-|g|, so the subset I mentioned is made up of all negative real numbers. Note that for x<0, x.(-1)=|x|(-1)=x, so -1 is a right identity for this subset.

    Note that 1 is also an identity for your example, and in this case the group obtained is just the positive reals with the normal multiplication, while the negative reals are the 'extra' elements.
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