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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?

Joshua Ortiz
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    Your question doesn't make sense. If two groups are isomorphic to a third, then they are necessarily isomorphic to each other. If your candidate $G$ is isomorphic to $P(A)$, then "some other group not being" $G$ cannot be isomorphic to $P(A)$. Also, $V_4$ has $4$ elements while $P(A)$ has $8$. – balddraz Dec 15 '21 at 04:47
  • The group is an Abelian group of order $8$. There are exactly three of these: $C_8, C_2\times C_4, C_2\times C_2\times C_2$. Since the symmetric difference of every set with itself is $\emptyset$, I think the third option is the right one. – markvs Dec 15 '21 at 05:08

1 Answers1

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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!)

Nicolás Vilches
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