Counting Non-Isomorphic Binary Operations on G with p Elements

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In summary, In order to find a non-isomorphic group with this operation, you must first find a left and right identity, and then find an inverse for the left identity. However, 1 is an identity for the subset, so the group obtained is just the positive reals.
  • #1
Zaare
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"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.
 
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  • #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.
 
  • #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?
 
  • #4
Zaare said:
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.
 
  • #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:
  • #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.
 

FAQ: Counting Non-Isomorphic Binary Operations on G with p Elements

How do you count non-isomorphic binary operations on G with p elements?

To count non-isomorphic binary operations on G with p elements, you must first understand what is meant by "non-isomorphic" and "binary operations." Non-isomorphic means that the operations are not the same or identical. Binary operations are mathematical operations that take two elements from a set and combine them to produce a third element. To count, you must consider all possible combinations of the elements in G and determine which operations are non-isomorphic.

What is the significance of counting non-isomorphic binary operations on G with p elements?

The significance of counting non-isomorphic binary operations on G with p elements lies in its applications to group theory and abstract algebra. By counting these operations, we can gain a better understanding of the structure and properties of G, and how it relates to other mathematical structures.

Is there a formula for counting non-isomorphic binary operations on G with p elements?

Yes, there is a formula for counting non-isomorphic binary operations on G with p elements. It is given by p^p^(p+1)/2, where p is the number of elements in G. This formula takes into account the number of possible combinations of elements and eliminates any duplicate operations that are isomorphic.

Can non-isomorphic binary operations on G with p elements be represented visually?

Yes, non-isomorphic binary operations on G with p elements can be represented visually using Cayley tables. A Cayley table is a visual representation of a binary operation on a finite set, where each row and column represents an element from the set and the corresponding cell contains the result of the operation on those two elements.

How does the number of non-isomorphic binary operations on G with p elements change as p increases?

The number of non-isomorphic binary operations on G with p elements increases rapidly as p increases. In fact, the number of non-isomorphic binary operations approaches infinity as p approaches infinity. This is because as p increases, the number of possible combinations of elements also increases, leading to a larger number of non-isomorphic operations.

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