In this problem we work out step-by-step the procedure for checking if a given set and a given operation together form a group.

Denote by $X$ the set of all integers (including 0 and the negative integers). If $x, y$ are integers then so is
$$x * y := x + y + 11$$
so this defines an operation on the set $X$. We now check if $( X, * )$ is a group.

Recall that this means checking

(1) Associativity
(2) Existence of identity
(3) Existence of inverse

Depending on the situation you can check (1) before or after (2) and (3), but you do need to check (2) before (3) since the definition of inverse involves the identity element.


Part (1): Associativity

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We begin with (1). That means we must check to make sure

$$( x * y ) * z = x * ( y * z ) \textrm{ for ALL } x, y, z \in X .$$
We check this by computing both sides separately and then comparing them.

The left side is: $\phantom{ = }$ $( x * y )$ $* z$ $=$ $* z$ $=$

The right side is: $\phantom{ = }$ $x *$ $( y * z )$ $=$ $x *$ $=$

Does $( X, * )$ satisfy the associativity law? ? yes no


Part (2): Existence of Identity

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Next we check (2). That is, we want to know if there exists an element $e \in X$ such that

$$(x * e) = x = (e * x) \textrm{ for ALL } x .$$

The first equality means: $0$ $=$ $(x * e )$ $- x$   $=$ $- x$   $=$ So therefore $e =$

The second equality means: $0$ $=$ $x -$ $(x * e )$   $=$ $x -$   $=$ So therefore $e =$

Based on these two calculations, it follows that $( X, * )$ has an identity, namely
$e =$


Part (3): Existence of Inverse

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Finally, we check (3). This means for ANY $x \in X$ we want to find $y \in X$ such that

$$( x * y ) = e = ( y * x )$$
where $e$ is the value we have just determined.

The first equality means: $e$ $=$ $x * y$   $=$ Therefore, $y =$ (Use the value of $e$ you calculated earlier)

The second equality means: $e$ $=$ $y * x$   $=$ Therefore, $y =$ (Use the value of $e$ you calculated earlier)

Based on these calculations, do the elements of $( X, * )$ have inverses? ? yes no


Conclusion

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So is $(X, *)$ a group? ? yes no

Additionally, is $( X, * )$ an Abelian group? ? yes no