$G=\mathbb{Z}_{>0}$ with $a\circ{}b=a^b$ is a Group. Or not.

Question

We're asked to prove or disprove the above statement, here's my go at it, I'd appreciate it if someone would give me feedback on it.

Associativity Check: For $G$ to be associative, the following must hold $\forall{}a,b,c\in{}G$ \begin{equation} a\circ{}(b\circ{}c)=(a\circ{}b)\circ{}c \end{equation} Which equates to \begin{align*} a^{b^c}&={a^b}^{c}\\ &=a^{bc} \end{align*} And since this is not true in general, $G$ is not associative.

Identity Check: \begin{equation} \exists{}e\in{}G:\forall{}a\in{}G:e\circ{}a=a\circ{}e=a \end{equation} And because $G=\mathbb{Z}_{>0}$ $\exists$ no $e\in{}G$ such that the property holds, and hence $G$ has no identity element.

Invertibility Check: \begin{equation} \exists{}a^*\in{}G \, \, \, \forall{}a\in{}G: a\circ{}a^*=a^*\circ{}a=e \end{equation} For the same reason that $a^*\in{}G=\mathbb{Z}_{>0}$ there is no inverse such that the property holds and thus $G$ is not invertible either.

Commutativity Check: \begin{equation} a^b\neq{}b^a \end{equation} $\forall{}a,b\in{}G$ and thus $G$ is not Abelian either.

I feel like I've messed up the second and third check, how would I improve it? Thanks

Answer 1

To prove $G$ is not a group you only need to check that one axiom fails. Associativity and identity fails, and invertibility doesn't make sense without an identity.

The problem with your checks is that you have merely stated that they don't hold, you should prove that they don't hold. Associativity is easy as you need only find a counter-example; picking random numbers should work. Identity is a little trickier as you need to show all possible $e\in G$ are not an identity, but still not particularly hard.

I don't know why you're checking if the (not actually a) group is abelian.

Answer 2

On 3, you just say there can't be one, you don't say why. Take example $$1^a = 1$$ which in our presumed group is $1\circ a = 1, \forall a$, then clearly $1$ can not have an inverse element.