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How many? -question

by Zaare
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Zaare
#1
Feb12-05, 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.
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chronon
#2
Feb12-05, 08:21 AM
chronon's Avatar
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 non-isomorphic.
Zaare
#3
Feb13-05, 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?

chronon
#4
Feb13-05, 10:16 AM
chronon's Avatar
P: 499
How many? -question

Quote Quote by Zaare
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?
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.
Zaare
#5
Feb13-05, 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=|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?
chronon
#6
Feb13-05, 01:06 PM
chronon's Avatar
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=|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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