A question concerning binary operations and isomorphisms

Question

Consider two nonempty sets $A$ and $B$, such that $A$ is isomorphic to $B$. Now, if $B$ is a group under some binary operation $*$, does it necessarily imply that there exists an operation $*'$, under which $A$ is a group? If this is true, what happens if I strengthen my condition, saying for instance that $B$ is an abelian group under $*$? Would there necessarily be a $*'$ under which $A$ is abelian?

The reason I ask this is because of an interesting question I came across which I've been trying to answer for days (please do not answer it, if you happen to know the answer);

Consider $S = \mathbb N \cup \{0\}$. Is it possible to make $S$ an abelian group under some binary operation?

Answer 1

Let $\phi: A \to B$ be our set isomorphism. Now, since $B$ is a group with operation $*$, we can simply define $*'$ so that

$\begin{align*} *': A \times A &\to A \\ a_1*'a_2 &= \phi^{-1}\left(\phi(a_1)*\phi(a_2)\right). \end{align*}$

The bijection between $A$ and $B$ takes care of the identity element and inverses; $1_A = \phi^{-1}(1_B)$ and for $a \in A$, its inverse will be $a^{-1} = (\phi^{-1}(a))^{-1}$.

I can't see any reason why any property of the group $(B, *)$ wouldn't carry through to the group $(A, *')$, since the bijection between $A$ and $B$ essentially just renames elements in the group $(B, *)$; with the above definitions, the definition of $*'$ turns our bijection $\phi$ of sets into an isomorphism of groups.

Answer 2

If there is a bijection $f\colon A\to B$ and $*$ is a group operation on $B$, you can define, for $x,y\in A$, $$ x*'y=f^{-1}(f(x)*f(y)) $$ This defines an operation on $A$ and $f$ becomes an isomorphism between $(A,*')$ and $(B,*)$, since, by definition, $$ f(x*'y)=f(x)*f(y) $$ and $f$ is bijective. It's just routine to check that $(A,*')$ is a group if $(B,*)$ is a group; abeliannes is preserved.