The following is from "Guide to Abstract Algebra, 1st Ed, Whitehead"
Prove that if $G = \langle a \rangle$ is a finite cyclic group of order $n$ then the order of every element of $G$ is a divisor of $n$.
Question: Is the following correct?
I am self-teaching so do miss some key arguments required in apparently simple proofs.
My solution:
The finite cyclic group has order $n$, which means $a^n = e$, the identity element, where $a \in G$.
We construct $a^{jk} = e$, where $jk=n$, for integers $j,k$.
That is, $(a^j)^k = e.$
So the order of $a^j$ is $k$, and $k$ is a divisor of $n$, our objective.
What is left to show is that $a^j$ generates all elements of $G$. We can say it does because $1 \leq j \leq n$ generates all the elements within the first cycle of $G$.
You might need to be clearer about where you start and what follows. You want to startt with an arbitrary element $b\in\langle a\rangle$, which can be represented by $b=a^j$ for some $1\le j\le n$, that at least is correct. But why would $j$ necessarily be a divisor of $n$? In general this is not true. But the statement "find $k$, such that $jk=n$" presumes, that $j$ must be a divisor of $n$. So no, I don't see how this is a full solution. If $j$ does divide $n$, then yes, you found a $k$, that satisfies the condition. But is it the least $k$? That is also required to be shown. I'd say you're not done here.
Instead, let's begin with an arbitrary element $b\in\langle a\rangle$. Then $b$ can be written as a power of $a$ as follows: $b=a^j$ for some fixed $j\le n$. Let's call $k$ the order $b$, i.e. $k$ is the smallest positive integer such that $b^k=e$. We want to know how $k$ relates to $n$. Here we will apply the most useful concept for this proof: Euclidean Division, i.e. division with a remainder. We can always divide $n$ by $k$ with a remainder $r$ that satisfies $0\le r<k$. We know that there is some integer $p$ such that $$n=pk+r.$$ On the other hand we can also divide $jk$ by $n$ with a remainder to find some integer $q$ and some $0\le R<n$ such that $$jk=qn+R.$$ From here we want to use the fact that $n$ is the smallest integer the makes $a^n=e$ true (in particular $R$ can only satisfy $a^R=e$ if $R=0$, since it is smaller than $n$), and also the fact that $k$ is the smallest integer such that $b^k=e$ (and again $b^r=e$ only if $r=0$). Can you see a way to argue that uses both facts and shows that $k$ divides $n$? What would need to be true for $k$ to divide $n$?
Let me know if you needs more clarifications.