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How many? questionby Zaare
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#1
Feb1205, 03:06 AM

P: 54

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
Feb1205, 08:21 AM

P: 499

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 nonisomorphic. 


#3
Feb1305, 06:36 AM

P: 54

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?



#4
Feb1305, 10:16 AM

P: 499

How many? question



#5
Feb1305, 11:29 AM

P: 54

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=ab. In this example the operation is associative and right inverse exists. I took 1 as identity: (1)+g=1g=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? 


#6
Feb1305, 01:06 PM

P: 499

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=ab 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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