Is this a group $G=\mathbb{Z}$ with $a*b=4$

Question

Okay! Well I know its not a group but for my own sake I'm having trouble explaining why and seeing why it isn't.

$G=\mathbb{Z}$ with $a*b=4$

I believe that it's not a group because Inverse Elements don't exist i.e.
$a*b=1$ no such $b \in G$ exists that'll, but apparently this isn't a proper explanation.

apparently the real answer is that there does not exist a proper identity, can someone explain why this fails?

Answer 1

The definition of an inverse element depends on the definition of an identity element. We can't assume that that the identity is $1$ in this group. So first we need to check if there exists some $e \in G$ such that for every $g \in G$, we have that $e * g = g = g * e$. And no such $e$ exists. Otherwise, we could take $g = 5$ to obtain: $$ e * 5 = 5 \implies 4 = 5 $$ a contradiction.

Answer 2

The reason the argument is not proper is because it assumes that $1$ is the identity element.

For $a \ne 4$ we have $1*a = 4 \ne a$ so $1$ is not an identity element.

Now we can fix this by saying: If $G$ is a group than there will be an element $e$ that is an identity element. At this time we don't know what $e$ actually is or even if such an $e$ exists. But if $G$ is a group there is one.

Pick any $a,b \in \mathbb Z$. $a*b =4$ so unless $4 = e$ we do not have $b = a^{-1}$. For any $c \ne 4$ we have $4*c =4 \ne c$ so $4$ is not the identity element. And therefore since $a*b = 4\ne e$ for all $b$, we know that $a$ can not have an inverse.

That IS a proper argument.

But it is an unnecessiarily complicated argument.

It'd be easier just to show that $G$ has no identity element. Let $a \ne 4$, then for any element $e$ we have $a*e = 4\ne a$ so no possible $e$ can be an identity element. So $G$ can not be a group.