# 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 $$8$$.

• I think it should be noted that those are called generators of the group, and a homeomorphism in general is determined by its values on the generators of the domain. Commented Jan 3, 2021 at 7:50

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).

• The order of A is two because I get the identity element and △ is a commutative operation. But I don´t understand how that helps me to find the bijection. The identity element is ∅ for the first group and (0,0,0) for the other. Commented Jan 2, 2021 at 21:22
• So, every element in $P(X)$ has order $2$ and $P(X)$ is an abelian group of order $8$. Here, if you know the characterization of abelian groups, then there are only three abelian groups of order $8$, which are $\mathbb{Z}_8, \mathbb{Z}_2 \times \mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_2$. And only one of these groups have all non-identity elements of order $2$. This tells you that two groups are isomorphic. Commented Jan 2, 2021 at 21:28
• @Erika Since I thought that you were trying to prove that they are isomorphic, my hint did not include the isomorphism function. Now, I edited my answer accordingly. Sorry about that. Commented Jan 2, 2021 at 21:38

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

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

• You should only get $|GL_3(2)|$ many isomorphisms, as the quotient of two isomorphisms is an automorphism. Commented Jan 3, 2021 at 0:26
• @ahulpke Thanks for the correction. Could you explain a little more?
– user403337
Commented Jan 3, 2021 at 0:49
• If $\phi,\psi\colon G\to H$ are isomorphisms, then $\phi/\psi$ is an automorphism of $G$. The number of isomorphisms between isomorphic groups thus is the order of the automorphism group of $G$. Commented Jan 3, 2021 at 3:26
• @ahulpke thanks alot! I actually had been thinking along those lines, but failed to note that not every permutation corresponds to an automorphism here, as it does for the Klein four group.
– user403337
Commented Jan 3, 2021 at 3:35