Prove that any two cyclic groups of the same order are isomorphic? 
Prove that any two cyclic groups of the same order are isomorphic.

Let the groups be $G,H$ with order $k$. Let $G=<a>$ and $H=<b>$. Thus we have $|a|=|b|=k$ and by definition, $G=\{a^0,a^1,...,a^{k-1}\}$ and $H=\{b^0,b^1,...,b^{k-1}\}$. Next I set up the bijection $\theta : G \rightarrow H: \theta(a^n)=b^n$. I think I should proceed by showing $\theta$ is an isomorphism but I'm not sure how to do that. What's a good way to go about showing $\theta$ is an isomorphism?
 A: Not to detract from the other answers, but here is a more abstract proof that I prefer:
If $G$ is a cyclic group with generator $g$, then there is a surjective homomorphism $(\mathbb{Z},+)\to G$ sending $1$ to $g$.
By the First Isomorphism Theorem, $G\cong\mathbb{Z}/H$, where $H$ is a subgroup of $\mathbb{Z}$. But we can classify the subgroups of $\mathbb{Z}$ as $0\mathbb{Z},1\mathbb{Z},2\mathbb{Z},3\mathbb{Z},\ldots$.
Since the quotient $\mathbb{Z}/n\mathbb{Z}$ has order $n$ if $n\neq 0$, and order $\infty$ if $n=0$, the order of $G$ uniquely determines its isomorphism class.
In other words, if $|G|=n$ is finite, then $G\cong \mathbb{Z}/n\mathbb{Z}$.  If $|G|$ is infinite, then $G\cong\mathbb{Z}$.
A: HINT: Firstly, you prove that $\varphi (a^n) = b^n$ for all $n \in \mathbb{Z}$.
After that, you can prove that $\varphi(xy) = \varphi(x)\varphi(y)$, for all $x$, $y \in G$. 
A: Recall that you need to demonstrate three things: injectivity, surjectivity, and the homomorphism property.  Injectivity follows from the fact that $\theta(a^i) = \theta(a^j)$ that means that $b^i = b^j$ and we can reduce these down to $\tilde{i} \equiv i \mod k$ and $\tilde{j} \equiv j \mod k$, so $b^{\tilde{i}} = b^{\tilde{j}}$, but this only occurs if $\tilde{i} = \tilde{j}$ which means that $a^{\tilde{i}} = a^i = a^j = a^{\tilde{j}}$.  Surjectivity should be easier to prove.  
The homomorphism property should be easier since any element of $G$ can be written $a^{i+j}$ we have
$$
\theta(a^{i+j}) \;\; =\;\; b^{i+j} = b^i b^j \;\; =\;\; \theta(a^i) \theta(a^j).
$$
