My book states that any finite cyclic group $=\langle a\rangle$ of order $n$ is isomorphic to $\mathbb{Z_n}$ under the mapping $a^k$ → $k$ mod $n$. This correspondence is one-to-one. I was wondering exactly why $a^k$ → $k$ mod $n$ is one-to-one. For example, if I work in $\mathbb{Z_5}$, then
$1^5$ → $5$ mod $5$,
and $1^{10}$ → $10$ mod $5$.
So, $5$ mod $5$ =$10$ mod $5$
would mean that $5=10$ which is a contradiction. Hope someone can help.
Think this way: since $\langle a \rangle$ has order $n$, the elements of $\langle a \rangle$ are $$ e, a, a^2, \ldots, a^{n-1}. $$ Define the asserted isomorphism using only those exponents, so you map $a^k$ to $k$ where $k$ is assumed to be on the list $0, 1, 2, \ldots, n-1$. This makes it really obvious that it's a bijection. Now, to prove it's an iso, you have to prove that the exponents act mod $n$ when you operate $a^ka^j=a^{k+j}$, but they do because of the Division Algorithm.
Edit for example: suppose $a$ has order 12. Then two such elements, as I've advertised them, are $a^{8}$ and $a^{10}$. What is $a^8a^{10}$? Well, it's $a^{18}$, but this is not one of the magic powers I am allowing you to use. That's okay, because I can long divide $18$ by $12$ and get $18 = 12+6$ so $$ a^8a^{10} = a^{18} = a^{12+6} = a^{12}a^6 = ea^6 = a^6. $$ So, all along $a^{18}$ was one of the magic powers in disguise, namely $a^6$. Notice that this is exactly $18 \equiv 6 \bmod 12$ which explains the isomorphism.