Here's an informal argument:
You want to show that $G$ is isomorphic to a subgroup of $S_n$ (for $n=\vert G\vert$). Well, how do you show that something is isomorphic to a subgroup of $S_n$? Actions! Elements of $S_n$ are understood by how they permute the set $[n]=\{1, 2, . . . , n\}$. So in order to show that $G$ is isomorphic to a subgroup of $S_n$, what we need is a machine for doing the following:
The hope is that the map $h: G\rightarrow S_n: g\mapsto \pi_g$ will be a homomorphism from $G$ to $S_n$, and that it will be injective; then $G$ will be isomorphic to $im(h)$.
So this boils down to:
How can I think of an element of $g$ as permuting some $n$-element set?
This is hard because $G$ is an abstract group - we have no sense of what $G$ is "meant to do." So we have to cook up some interpretation of $G$ from scratch!
Luckily, there is a natural $n$-element set lying around: $G$ itself! So, is there a way we can think of an element of $G$ permuting the whole set $G$?
The answer is: yes! Given an element $g$, consider the map from $G$ to $G$ defined as $$\pi_g: a\mapsto g\cdot a.$$ Each $\pi_g$ is a permutation of $G$, we have $\pi_g=\pi_h$ iff $g=h$, and $\pi_g\circ \pi_h=\pi_{gh}$ (do you see why these facts are true?); so the map $h: g\mapsto \pi_g$ is in fact an injective group homomorphism from $G$ to the group of permutations of $G$ . . .
. . . but this latter group is just $S_n$, renamed!
Best Answer
You ask for a simple example, so here is one: Consider the group $\mathbb{Z} / n\mathbb{Z}$. This is a finite abelien group under addition of order $n$.
Let $$ \phi: \mathbb{Z} / n\mathbb{Z} \longrightarrow S_n $$ given by $$ \phi([1]) = (1\; 2\; 3\; \dots \; n ). $$ So $$ \phi([m]) = (1\; 2\; 3\; \dots \; n )^m. $$ Then $\phi$ is a injective. Note that $S_n$ is a multiplicative group.
Note that all you had to do here was to find an element of order $n$ in $S_n$. In fact, the above example is easily extended to embedding any cyclic group in some permutation group (since any cyclic group is isomorphic to a $\mathbb{Z} / n\mathbb{Z}$ for some $n$).