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?
Best Answer
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 $G\cong\mathbb{Z}$.