Proving that if $|G| =n$ and $A_d=\{g\in G : |g|=d\}$ then $A_d = \emptyset$ if $d$ is not a divisor of $n$. Reading a solution/proof to an exercise in Fraleigh's Algebra, I've found myself a bit stuck (probably it's trivial, so excuse me as I'm at an entry level on Abstract Algebra) on the following stage of the proof to it :

Let $G$ be a finite group of order $n$. Then it's obviously $|G| = n$. If we consider the group :
  $$A_d=\{g\in G : |g|=d\}$$
  then $A_d = \emptyset$ if $d$ is not a divisor of $n$.

How would one proof the statement made with simple abstract algebra stuff (considering it's my first course) ?
 A: To show: $d \not \mid  n \implies A_d = \emptyset.$ It suffices to show that $A_d \neq\emptyset \implies d\mid n.$ So, let $A_d \neq\emptyset.$ Then there exists $g \in A_d.$ This means that $|g|=d.$ 
Since $g \in G,$ therefore, $\langle g\rangle$ is a subgroup of $G.$ By Lagrange's theorem, $$d=|g|=|\langle g\rangle| \text{ divides } n$$
A: Here are two proofs, one with less machinery and one with more.

Proof 1:

I think the trick you will want to use here is division for ordinary integers: the rule is if you have integers $m \leq n$, you can divide $n$ by $m$, writing $n$ as a quotient-plus-remainder: 
  $$n = qm + r$$
where the remainder satisfies $0 \leq r < m$. (This is a generally useful tool for group theory.)

In this case, suppose you have a group $G$ with $n$ elements. Pick $g\in G$ and suppose $g$ has order $k$. This means that $g^k = e$, and that $k$ is minimal (!): $k$ is the smallest positive integer for which $g^k=e$. (There are other larger integers, for example $g^{2k}, g^{3k},$ etc.)

Now use the division rule to divide $n$ by $k$: we can write $n$ as a quotient:
$$n = qk + r$$
where $0 \leq r < k$. 
Our goal is to show that the order $k$ evenly divides the order of the group, $n$. To do so, we must prove that the remainder $r=0$, i.e. that $k$ divides $n$ without remainder.
To do so, remember that $g^n = e$ (because the group has order $n$). We can compute $g^n$ another way using our division rule: $g^n = g^{qk+r} = g^{qk}g^r = (g^k)^q g^r = e^q g^r = g^r.$
But these two computations imply that $g^r = e$. The remainder $r$ is less than $k$ (the order of $g$), because $r$ is the remainder when dividing $n$ by $k$. Yet $k$ was assumed to be the smallest positive integer for which $g^k = e$. Hence it must be that $r=0$ after all.
This completes the proof, because we have shown that there exists an integer $q$ such that $n = kq$; hence the order $k$ of element $g$ necessarily evenly divides the order $n$ of the group.

Proof 2:


*

*Let $G$ be a group of finite order $n$, and let $g$ be an element of $G$.

*Consider the powers of $g$: $g^0, g^1, g^2, g^3, \ldots$. Because the group $G$ itself is finite, these powers aren't all different; there are at most $n$ different elements represented here.

*Now we define an equivalence relation on $G$: say that two elements $x,y \in G$ are related if $x = g^k y$ for some $k$ (positive, negative, zero).

*You can prove that this is an equivalence relation, i.e. reflexive $(x = g^0 x)$, symmetric $(x = g^k y \iff y = g^{-k}x)$, and transitive $(x = g^kg^\ell z)$.

*This means that it partitions $G$ into subsets of "equivalent" elements.

*Suppose there are $q$ such subsets.

*It turns out that every subset has the same number of elements. To prove this, consider any two such subsets: $X = \{xg^k : k\in \mathbb{Z}\}$ and $Y=\{yg^k : k\in \mathbb{Z}\}$. The function $f(h) = yx^{-1}h$ sends $X$ to $Y$, and it's bijective with an inverse $f^{-1}(h) = xy^{-1}h$. Hence $X$ and $Y$ have the same number of elements.

*Call the number of elements in each subset $k$.

*Then the number of elements in $G$ is just the number of subsets $(q)$ times the number of elements in each subset $(k)$, because the partitions are exclusive and exhaustive.

*This shows that $k$ evenly divides $n$, because $n=qk$.

*But the order of $g$ is the same as the number of elements in $\{g^{k} : k\in \mathbb{Z}\}$, which is the number of elements in the partition $\{eg^k : k\in \mathbb{Z}\}$, which is $k$, which is a divisor of $n$.
(The reason the order of $g$ is the same as the number of elements is this: by definition, the order of $g$ is the smallest positive integer $k$ such that $g^k = e$. But this implies that every lower power $g^0,\ldots, g^{k-1}$, is a distinct member of $G$. And if $g^k = e$, then $g^{k+\ell} = g^\ell$. Hence $\{g^m : m\in \mathbb{Z}\}$ is a set of $k$ distinct elements, where $k$ is the order of $g$.)

*This shows that the order of any element of $G$ divides $n$, which implies that there are no elements whose order does not divide $n$.

