Every Finite Cyclic Group Of Order N Is Isomorphic To Zn

The Cayley graphs of both $\mathbb{Z}_7$ and $H_7$ are simple cycles of length 7. We could even draw them inline using $$1\to 2\to 3\to 4\to 5\to 6\to 7\to 1\quad \text{and} \quad \phi_1\to \phi_2\to \phi_3\to \phi_4\to \phi_5\to \phi_6\to \phi_7\to \phi_1,$$ however, we'd need an arrow from 7 back to 1, rather than writing 1 twice. We drew these graphs together in class.

We now consider the function $f:H_n\to \mathbb{Z}_n$ defined by $f(\phi_k)=k$. Note first that we know $H_n=\{\phi_0, \phi_1,\phi_2, \ldots, \phi_{n-1}\}$. This means that given $x\in H_n$, we know $x=\phi_k$ for some integer $k$ with $0\leq k< n$. This also means that $f(x)$ is well defined, as the output definitely lies in $\mathbb{Z}_n$. To show that $f$ is an isomorphism, we must show that $f$ is a bijection and that $f$ preserves the group structure. We first show that $f$ is a bijection. Let $g:\mathbb{Z}_n\to H_n$ be defined by $g(k)=\phi_k$ for each $k\in\mathbb{Z}_n$. We then compute $f(g(k))=f(\phi_k)=k$ for each $k\in\mathbb{Z}_n$. Given $x\in H_n$, we know $x=\phi_k$ for a unique $k\in \mathbb{Z}_n$. We then compute $g(f(x))=g(f(\phi_k))=g(k)=\phi_k=x$ for each $x\in H_k$. This shows that $g$ is the inverse of $f$ and so $f$ is a bijection. We now must show that $f$ preserves the group structure. Given $x,y\in H$, we find integers $j$ and $k$ so that $x=\phi_j$ and $y=\phi_k$. Note that we know $\phi_j\circ\phi_k=\phi_{j+k}$, and we also know that $\phi_m=\phi_{m\pmod n}$ for any integer $m$. We can then compute $$f(x\circ y)=f(\phi_j\circ\phi_k)=f(\phi_{j+k})=f(\phi_{(j+k)\pmod n}) = (j+k)\pmod n = f(x)+f(y).$$

We now generalize the above result to work for any cyclic group $G$. Let $G$ be a cyclic group of order $n$ with generator $a$. Consider the function $f:\mathbb{Z}_n\to G$ defined by $f(k)=a^k$. We must show that $f$ is an isomorphism. Note that we have already shown that $G=\langle a \rangle=\{a^0, a^1, \ldots, a^{n-1}\}$ and that the $n$ elements listed to the left are distinct. This means that given any $x\in G$, there is a unique nonnegative integer $k$ that is less than $n$ such that $x=a^k$. The inverse of $f$ is the function $g:G\to \mathbb{Z}_n$ defined by $g(x)=k$ where $k$ is the unique integer with $0\leq k<n$ such that $x=a^k$. You can check that $f(g(x))=x$ and $g(f(k))=k$ for each $x\in G$ and $k\in\mathbb{Z}$ which means that $f$ is bijective as $g$ is its inverse.

We now need to show that $f$ preserves the group structure. Let $j,k\in\mathbb{Z}$. If $j+k<n$, then we know $(j+k)\pmod n=j+k$. Otherwise, we know that $(j+k)\pmod n=j+k-n$. In either case, we know that there exists an integer $q$ so that $(j+k)\pmod n=j+k+qn$. We can now compute $$ f((j+k)\pmod n) =a^{(j+k)\pmod n} =a^{j+k+qn} =a^ja^ja^{qn} =a^ja^je =a^ja^j =f(j)f(k) .$$ This shows that $f$ preserves the group structure, and complete the proof that $f$ is an isomorphism.