Show that all groups of order $3$ are isomorphic with each other.

Question

I wanted to find an isomorphism between two arbitrary groups of order $3$ and then conclude but , I don’t know how I can concretely define this application.

Answer 1

Let the elements be $\{e,a,b\}$ with $e$ the identity. Show there is only one way to fill in the addition table that satisfies the group axioms.

Answer 2

Let $G$ and $H$ be two groups of order $3$.
Let $g\in G$ such that $g\neq 1$.
Method 1: By using Lagrange's Theorem, $|\langle g\rangle|$ divides $|G|=3$. Since $\langle g\rangle$ is nontrivial, it must have size $3$ and therefore $G=\langle g\rangle$.
Method 2: Let $G=\{1,g,f\}$. Since the Cayley table of a group is a Latin square, by filling the Cayley table of $G$, we can obtain that $f=g^2$ and $g^3=1$.

So by either method, we get that $G=\{1,g,g^2\}$. Similarly, we get $H=\{1,h,h^2\}$.
Thus the obvious guess for the isomorphism from $G$ to $H$ will be $\phi:G\rightarrow H$ where $$1\mapsto 1,\ g\mapsto h,\ g^2\mapsto h^2.$$ The map is clearly a bijection. To complete the proof, we need to show that it is a homomorphism. If you know modular arithmetic, you can show by the method below: Let $g^i,g^j\in G$, $$\phi(g^ig^j)=\phi(g^{(i+j)\bmod n})=h^{(i+j)\bmod n}=h^ih^j.$$ Otherwise, you can prove by considering all the possibilities. Since $G$ is abelian, it is sufficient to prove for the following cases: $$\phi(1\cdot 1),\ \phi(1\cdot g),\ \phi(1\cdot g^2),\ \phi(g\cdot g),\ \phi(g\cdot g^2),\ \phi(g^2\cdot g^2).$$

Answer 3

Let $G$ be a group of order $3$. Take $g\ne e$. Then $G=\langle g \rangle $, by Lagrange. Define $\varphi: G\to\Bbb Z/3\Bbb Z$ as the unique homomorphism determined by $\varphi(g)=1$. It is easy to check that it is bijective. Thus we get an isomorphism.

Essentially the same proof works for any prime $p$.