How to find Isomorphism between groups of order 8

Question

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.

Answer 1

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.

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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 $8$.

Answer 2

HINT: In general, if $X$ is a finite set, $P(X)$ is the power set of $X$ and $A \in P(X)$, what is the identity element of $(P(X), \triangle)$? What do you get when you take symmetric difference $A \triangle A$? What does it tell you about the order of $A$ in $(P(X), \triangle)$? Is $\triangle$ a commutative operation?

If you are trying to find a bijection, you can consider the map $\phi: P(X) \to S$ such that $\phi(A) = (a_1,a_2,a_3)$ where $$a_i = \begin{cases} 1, & \text{if $i \in A$} \\[2ex] 0, & \text{if $i \notin A$} \end{cases}$$

where addition operation is under modulo $2$.

Then, you can show that $\phi$ is an isomorphism by taking arbitrary two sets $A,B \in P(A)$ and showing that $\phi(A\triangle B) = \phi(A)+\phi(B)$ by also considering the meaning of symmetric difference operation (for instance if we take $\{1,2\}$ and $\{1\}$, what is their symmetric difference and how is it related to addition modulo $2$ of $(1,1,0)$ and $(1,0,0)$? As an intuition, symmetric difference can be also thought as $A \triangle B = (A \cup B) \backslash (A \cap B)$. So, elements in the intersection do not belong to the resulting set).

Answer 3

Hint: Prove that the group is abelian and that all the nonidentity elements have order two.

There are $|\rm {GL}_3(2)|=168$ isomorphisms.