# Thread: Determine if this is a group

1. ## Determine if this is a group

Problem:
Determine if the set with the binary operation forms a group.
Set S = All real numbers except -1.
$a*b = ab+a+b$

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I know I need to check three things: associativity, if there is an identity, and if there is an inverse for every element in S.

I know how to do associativity since it's straightforward. For the identity, I just guess and checked and found that $e=0$ works. (Is there any way to find an identity without guessing/checking?)

For the inverse, I am having difficulty.

I know that this must be satisfied to have an inverse:

$a*a' = a'*a = e$

Since I know the identity element is 0,

$a*a' = a'*a = 0$

Expanding this out with the definition of the binary operation,

$aa'+a+a' = a'a+a'+a = 0$

I said that an inverse does not exist, because from the last equation, it is impossible to get $aa'+a+a'$ equal to 0 for all a. And therefore, this is not a group.

Am I correct?

2. ## Re: Determine if this is a group

let's try to find an identity in a more systematic way:

we want a*x = a.

(it should be obvious that a*b = b*a, so we just need to check for a one-sided identity).

now a*x = ax + a + x. set this equal to a, and try to solve for x:

ax + a + x = a
ax + x = 0
(a+1)x = 0. now, can we divide by a+1? yes we can, because a will not be -1.
x = 0/(a+1) = 0.

are you sure it is impossible to find an inverse?

again, we want b such that a*b = 0. we want to solve for b in terms of a, if we can.

ab + a + b = 0
(a+1)b = -a.....i think you can finish. verify your answer.