Question

Determine if the set Z is a group under the binary operation *: x * y =xy +y. If yes, is it abelian? If not, which conditions fail to hold?

Answer 1

Check for associativity.
Does the following hold \(\forall x,y,z\in\mathbb{Z}\)?
\[\begin{align*}(x*y)*z&=x*(y*z)\\
(xy+y)*z&=x*(yz+z)\\
(xy+y)z+z&=x(yz+z)+(yz+z)\\
xyz+yz+z&=xyz+xz+yz+z\end{align*}\]

Answer 2

Do you know what abelian is? I am not sure but this problem has got me confused.

Answer 3

Abelian means commutative. does x*y=y*x for all x and y?

Answer 4

Abelian groups are commutative under our binary operation. Here we have the operator \(\otimes:\mathbb{Z}\to\mathbb{Z}\) s.t. \(x\otimes y=xy+y\). I'll let you determine whether it's a group or not (check our axioms -- closure, identity, inverse, associativity).

For some \(x,\, y\in\mathbb{Z}\) we have \(x\otimes y=xy+y\). Now, we also know \(y\otimes x=yx+x=xy+x\). Notice that \(x\otimes y=y\otimes x\) only when \(xy+y=xy+x\Leftrightarrow y=x\) and therefore this does not hold in general.

Answer 5

-some +all i.e. \(\forall x,y\in\mathbb{Z}\)

Answer 6

So this is not a group because of associativity?