A group $G$ isomorphic to $\mathcal{P}(A)$ [duplicate]

Denote $\mathcal{P}(A)$ as the power set of $A$. Especifically, let $A = \{1,2\}$. $\mathcal{P}(A)$ is an abelian group under symmetric difference $\triangle$. Is there a group $G$ such that $G \cong \mathcal{P}(A)$?

Maybe $G= \langle (12)(34) \rangle \leqslant S_4$, and this is because of Cayley's Theorem. But is there some other group not being $\langle (12)(34) \rangle$? Is it $V_4$, the Klein-four group?

Solution 1:

I think that $\mathcal{P}(A)$ is isomorphic to $\mathbb{Z}/2 \times \mathbb{Z}/2 \times \mathbb{Z}/2=(\mathbb{Z}/2)^3$, with isomorphism given by the indicator function $$ \Phi\colon \mathcal{P}(A) \to (\mathbb{Z}/2)^3, \qquad B \mapsto (1_B(1), 1_B(2), 1_B(3)), $$ where $1_B(x)=1$ if $x \in B$, and zero otherwise. Note that $\Phi$ admits an inverse: given some $(a_1, a_2, a_3)$, we take $B=\{i \mid a_i = 1\}$. In particular, $\Phi$ is bijective, so we just have to show that it is a group homomorphism. This boils down to show that $$ 1_{B \triangle C}(x) \equiv 1_B(x) + 1_C(x) \pmod{x} $$ which is easy to check (for instance, by considering the four options if $x$ is or is not in $B$ and $C$). (Let me know if this is enough, or if you need more details!)