# How many? -question

1. Feb 12, 2005

### Zaare

"How many?"-question

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

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

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

I know that there are 2 such operations if $$\mid G \mid =p^2$$, and 3 such operations if $$\mid G \mid =p*q$$, if $$p$$ and $$q$$ 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. Feb 12, 2005

### chronon

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. Feb 13, 2005

### 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?

4. Feb 13, 2005

### chronon

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. Feb 13, 2005

### Zaare

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
6. Feb 13, 2005

### chronon

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.