Edit:The things that look like `xx` should be xxSo, where it says `xx^(-1)`I meant x times x inverseAnd where it says `xx`I meant x times x

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Edit:

The things that look like `xx` should be xx

So, where it says `xx^(-1)`

I meant x times x inverse

And where it says `xx`

I meant x times x

To be a group, you have a binary operation operating on a set (satisfying some conditions). Here you haven't stated what the set is.

So, I'm going to answer this assuming that the set is `ZZ`, or `QQ`, or `RR`, or `CC`, or something where we have "familiar numbers" and "familiar addition" (no weird definition of "+")

The criteria for a group are:

1) There is an identity element e such that e*x = x*e = x

2) For each x, there is an element, `x^(-1)` such that `x**x^(-1) = x^(-1)**x = e`

3) `(x**y)**z = x**(y**z)`

And for an abelian group:

4) `x**y=y**x`

One at a time:

1)

e*x = e+ex

x*e = x+xe

Assuming by "xe" you mean regular (familiar) multiplication, then ex=xe and these can only be equal if e=x

So this can only happen if there is only 1 element in the group, the identity element. So, unless we have a set with one element, we will fail this condition.

2)

`x**x^(-1) = x + xx^(-1)`

`x^(-1)**x = x^(-1) + x^(-1)x`

This can only happen if, for every x, `x=x^(-1)` and, moreover, `x+xx=e`

Without knowing what set this operator acts on, it is hard to describe what is wrong with this, but this one is likely to fail. For example, if our set contains the natural numbers (for example, any of `ZZ`, `QQ`, etc, but not `ZZ_4` or `ZZ_5`) then we have:

`1+1*1=2`

`2+2*2=6`

These are not the same, so they can't both be e

So, in the case that our set contains as a subset the natural numbers, this condition fails

3)

`x**(y**z) = x**(y+yz) = x+xy+xyz`

`(x**y)**z = (x+xy)**z = x+xy+xz+xyz`

Unless xz=0, this will be false.

So condition 3 fails

4)

`x**y=x+xy`

`y**x = y+yx`

Assuming regular multiplication (so xy=yx), then this will only hold true if x=y (for every x and y) which will only hold true in a one-element group. So condition 4 fails

So this is not a group, and it is not an abelian group