Determining if a set is a group or not I have a question that I am trying to solve.
The question relates to how to determine if a given set it a group or not. I am not really understanding how to answer the question.
Q: Determine if the set Z/ with binary operation * defined via a * b = ab -b is a group.
Should I be doing something like <1> = {1, 1^2, 1^3....} ?
                                 <2> = {2^1, ..........} ?
Should I be checking to see what values are considered generators?
 A: Research and answer the following:
++ Is the set non-empty?
++ Is the binary operation Actually closed, meaning: $\;a,b\in\Bbb Z\implies a*b\in\Bbb Z\;$ ?
++ Does there exist an unit element in $\;\Bbb Z\;$ wrt that operation? Meaning: does there exist $\;e\in\Bbb Z\;$ s.t. $\;\forall\,m\in\Bbb Z\;,\;\;m*e=e*m=m\;$
++ Is it true that for any $\;m\in\Bbb Z\;$ there exists $\;n_m\in\Bbb Z\;$ s.t. $\;n_m*m=e\;$ ?
++ Is is true that for $\;a,b,c\in\Bbb Z\;,\;\;a*(b*c)=(a*b)*c\;$
A: A group is a non empty set, with an operation, so that the operation upon the elements of a set have very specific characteristics.
So this is a matter of checking that the set, and this operation have those characteristics. 
1).  Is $*$ closed? That is if $a \in \mathbb Z$ and $b \in \mathbb Z$ then is $a*b$ well defined and is $a*b \in \mathbb Z$?
Well, if $a$ is an integer and $b$ is an integer then $a*b = a\cdot b - b$ is indeed an integer.  
So yes $*$ is closed.
2) Is $*$ associative?  That is if for any three $a,b,c \in \mathbb Z$ does $a*(b*c) = (a*b)*c$?
$a*(b*c) = a*(bc - c) = a(bc - c) - (bc -c) = abc - ac -bc + c$.
And $(a*b)*c = (ab - b)*c = (ab -b)c - c = abc - bc -c$.  These two values need not be the same.  For example: $1*(2*3) = 1*(2\cdot 3 - 3) = 1*3 = 1\cdot 3 - 3 =0$ whereas $(1*2)*3 = (1\cdot 2 - 2)*3 = 0*3 = 0*3 - 3 = -3$.  These are not the same.
As $*$ is not associative and this is not a group and we can stop right now.
But to fully answer your question, had it been associative we'd have to ask:
3) Do an identity element exist?  This is, is there an $e\in \mathbb Z$ so that for any $a\in \mathbb Z$ then $a*e = e*a = a$.
So we need $a*e = ae - e = a$.  That means $e(a-1) = a$.  If $a \ne 1$ that means $e = \frac a{a-1}$.  But this must be true for ALL $a\in \mathbb Z$.
So we need $e = \frac 2{2-1} = \frac 3{3-1} = \frac 4{4-1}$ ... etc.  This is simply not possible.
So there is not identity element $e$ so that $a*e = a$ for all $a$.
So this is not a group.
Finally we must ask:
4) For any $a\in\mathbb Z$ is there an inverse of $a$.  That is, is there a $a^{-1} \in \mathbb Z$ so that $a*a^{-1} = a^{-1}*a = e$ where $e$ is the identity element of question 3?
Well, since there ISN'T any identity element in question 3, there can't be an inverse.
So this is not a group.
