How to find Isomorphism between groups of order 8 Let $A=\{1,2,3\}$ and $2^A$ is all the subsets of A. Then $(2^A,\triangle)$ is a group. How can I show that the group $\big\{\{1\},\{2\},\{3\},\{1,2\},\{1,3\},\{2,3\},\{1,2,3\},\emptyset\big\}$
with operation $\triangle$, where $\triangle$ is the symmetric diffrence $C△B=(C \cup B) \backslash (C \cap B)$
is isomorphic to the group
$S= \{(0,0,0),(1,0,0),(0,1,0),(0,0,1),(1,1,0),(0,1,1),(1,0,1),(1,1,1)\}= \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$ with operation $+$.
I did not manage to find a bijection, but both groups have order $8$ and the order of the elements correspond.
I did do the cayley tables for both groups but I still did not manage.

Note that a group morphism $\psi:  (\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2,+)\to (2^A ,\triangle)$ is fully determined by the values
$\psi(1,0,0), \psi(0,1,0)$ and $\psi(0,0,1)$. A reasonable guess for our isomorphism would be
$$\psi(1,0,0) = \{1\}, \ \psi(0,1,0) = \{2\}, \ \psi(0,0,1) = \{3\}.$$
I leave the explicit verification that this works as an exercise for you.

Alternatively, you can proceed as follows, once you observe that $\triangle$ is a commutative operation:
By the fundamental theorem of finite abelian groups, you know that $(2^A, \triangle)$ must either be isomorphic to $\mathbb{Z}_8, \mathbb{Z}_2 \times \mathbb{Z}_4$ or $\mathbb{Z}_{2}\times\mathbb{Z}_{2}\times\mathbb{Z}_{2}$. The first two options are impossible because $(2^A, \triangle)$ does not contain elements of order $4$ and $8$.
