# Math Help - Is the product of a finite set of an Abelian Group alway the neutral element?

1. ## Is the product of a finite set of an Abelian Group always the neutral element?

Okay so lets say G is my finite set with G = { g1, g2, g3,... gn } and we define an operation on G with * : G x G --> G so that we have an abelian group with (G,*)

So for every gi in G we have an inverse gj in G, so that gi * gj = 0G (neutral element)

So cant I just switch around my elements in my finite set (g1*g2*g3*...*gn) so that I always have one element together with its inverse element? it should be possible because it is assoziative and commutative (because its an abelian group).
So both elements as a pair equal the neutral element 0G. This means that I am multiplying 0G with 0G over and over. The neutral element in multiplication is "1". So I do 1*1*1*1.

So isnt g1 * g2 * g3,..., * gn = 1?

2. ## Re: Is the product of a finite set of an Abelian Group alway the neutral element?

An element can be its own inverse. For example, in $\mathbb{Z}_2$ with addition, the sum of all elements is 0 + 1 = 1.

3. ## Re: Is the product of a finite set of an Abelian Group alway the neutral element?

Thanks for your quick answer. But I dont know if your statement is pro or contra. it still equals 1.
To me the proof still seems correct or am I overlooking something there.

4. ## Re: Is the product of a finite set of an Abelian Group alway the neutral element?

It is customary to view abelian groups with the operation of addition rather than multiplication. You are confusing multiplicative and additive units in my response.

5. ## Re: Is the product of a finite set of an Abelian Group alway the neutral element?

Yea thats really confusing to me. I thought I can use one or the other. Is it wrong to argue with multiplication in this case? You said its customary, but not a rule.

6. ## Re: Is the product of a finite set of an Abelian Group alway the neutral element?

You argued in post #1 that you can divide the product of all elements in the group into pairs of inverse elements. I told you that some elements are their own inverses, so you don't always get a pair. Since every element in the big product is only listed once, you don't get the unit from that element. This is where the proof does not go through.

How to denote the group operation: as * or +, is irrelevant. What is important is to correctly specify the unit of that operation. In $\mathbb{Z}_2$ with addition, the unit is 0 and not 1! Since 0 + 1 = 1, this is a counterexample to your theorem.

To repeat, when you apply a general theorem to a specific group, it is important to instantiate the concepts (operation, unit, inverse) of the general theorem correctly. Suppose the theorem talks about an arbitrary abelian group A with operation + and unit 0. Suppose also that you have proved that a given set G is an abelian group with respect to operation * and unit 1. Then you can instantiate the theorem's + with your concrete * and the theorem's 0 with 1. But if G is also a group with respect to another operation & and unit @, it would be wrong to apply the theorem to G by instantiating theorem's + with * and theorem's 0 with @. In other words, the theorem says, "If (A, +, 0) is an abelian group, then...", and you say, "Since (G, *, 1) is an abelian group, I can apply the theorem and interpret its conclusion by replacing A with my G, + with * and 0 with 1". But you can't apply the theorem to (G, *, @) or (G, &, 1).

Now, it is still customary to denote the unit of * with 1 in general theorems. Denoting it with 0 as in post #1 is strange. Denoting the operation of an abelian group with * is OK, but + is more customary.

I would also add that I am not a specialist in group theory, and while the product (or sum) of all elements in an abelian group is not necessarily the unit, it may possess some other interesting properties.