I just wanted to confirm that I completed this proof correctly, as I have not yet encountered any exercise requiring me to "put down on paper" the following argument.
In Pinter's A Book of Abstract Algebra, Chapter 15 Exercise E2 asks the reader to prove the following:
Let $G$ be a group and $H\triangleleft G$. If $(G:H)=m$, then $\forall a \in G$, $\exists n \in \mathbb Z \backepsilon \operatorname{ord}(Ha)*n=m \in \mathbb Z$
or as stated explicitly in the text:
Consider the cyclic group generated by an arbitrary coset $Ha$ belonging to $G/H$...call it $ \langle Ha \rangle$.
Then, consider any other arbitrary coset belonging to $G/H$...call it $Hx$.
Because $\langle Ha \rangle$ is a subgroup, and $Hx$ is an element of the group $G/H$, the following object is a "coset" of $G/H$:
"$\langle Ha \rangle Hx$"...this is sort of like a "coset of the group that has cosets as its elements"
Therefore, with this object established, we can now take advantage of Lagrange's Theorem.
Using this theorem, we know that $(G/H\ :\ \langle Ha \rangle ) = \frac{|G/H|}{|\langle Ha \rangle|}$.
From what is given by assumption, we know that $(G:H)=m$, however, $(G:H)$ is precisely describing the number of distinct cosets that are formed by $H$. Therefore, $(G:H)$ is equivalent to the cardinality of $G/H$...i.e. $|G/H|$.
As such, $|G/H| = m$
So we have $(G/H \ : \ \langle Ha \rangle ) = \frac{m}{|\langle Ha \rangle |}$
Noting that $|\langle Ha \rangle |= \operatorname{ord}(Ha)$ , and then rearranging, we have:
$\operatorname{ord}(Ha) * (G/H \ :\ \langle Ha \rangle) = m$
Letting $(G/H \ :\ \langle Ha \rangle) = n$, we arrive at the conclusion that $\operatorname{ord}(Ha) * n = m$
Since $Ha$ was arbitrarily chosen, we have proven this to be true for all elements belonging to $G/H$.
Is this a correct argument? Cheers~