Let $G$ be a finite group. Show that $G$ is isomorphic to a subgroup of $S_n$ (symmetric group). I am solving a geometry exercise in which I have to proof the following: 
Let $G$ be a finite group with $n=\vert G \vert$ elements. Show that $G$ is isomorph to a subset of $S_n$ (symmertric group).
Problem: Since we only had 4 geometry lectures we have not covered a lot. We get an exercise sheet every 2nd week and have lectures every week. We have not covered cayley's theorem yet. Is this theorem necessary to solve the exercise? I looked up cayley's theorem and its proof in M.Artin - Algebra and Fraleigh's First Course in Abstract Algebra. I tried it on my own but I would be glad if you could correct or enhance my 'proof'.
My Attempt: 
Let $|G|=n$ . We want $|G|$ to me isomorph to $S_n$. So we can number the elements of $G$:
$ \begin{array}{ccc} 1 & 2 & \ldots & n\\ a_1 & a_2 & \ldots & a_n \end{array}  $
we can now choose $x\in G$ and do a left multiplication, which leads to the following  
$\Rightarrow$
$ \begin{array}{ccc} 1 & 2 & \ldots & n\\ x \cdot a_1 & x \cdot a_2 & \ldots &  x \cdot a_n \end{array}  $
so we can say therefore the elements will be newly arranged in $G$ after the multiplication and because we numbered the elements of $G$ we can 'look up' where they have been before. That actually means:
$x \cdot a_i = a_j$ so therefore x sends element $a$ from place $i$ to place $j$ (permutation). So we define the following definition:
$\lambda_x=\begin{cases} \{1,2,\ldots,n\} \rightarrow \{1,2,\ldots,n\} \\ \lambda_x(i) \quad \qquad \mapsto \ j \end{cases}$
so therefore we can write:
$ \begin{array}{ccc} 1 & 2 & \ldots & n\\ a_{\lambda_x(1)} & a_{\lambda_x(2)} & \ldots & a_{\lambda_x(n)} \end{array}  $
now we can state a new function:
$\phi=\begin{cases} G \quad \rightarrow \quad S_n \\ \phi(\lambda(x)) \mapsto \lambda_x \end{cases}$
which basically means: we link the permutation that the left multiplication by x does to its x. Now we have to proof the isomorphism between $G$ and $S_n$ therefore it needs to be bijective and homomorphic. Well $\phi$ needs to be a group homomorphism and bijective which means:
homomorphism:
Let $\lambda(i),\lambda(j) \in G$ , $(G, \star)$ and $(S_n, *)$ be the two groups then: 
$\phi(\lambda(i) \star \lambda(j))= \phi(\lambda(i)) * \phi(\lambda(j)) = \lambda_i * \lambda_j$
bijective:
injective: $\forall \lambda(i),\lambda(j) \in G\ , \ \phi(\lambda(i)) = \phi(\lambda(j)) \Rightarrow \lambda_i=\lambda_j$
surjective: $ \forall \lambda_x \in S_n \ , \ \exists \lambda(x) \in G : \phi(\lambda(x)) = \lambda_x \qquad  \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$
 A: Following mark Brandenburg's comment, I suggest to first note the following:

If $A,B$ are two sets and $f\colon A\to B$ is a bijection (so $f^{-1}$ exists), then we obtain a group isomorphism $\phi_f\colon \operatorname{Sym}(A)\to \operatorname{Sym}(B)$ by letting $\phi_f(\sigma)=f\circ \sigma\circ f^{-1}$ for each $\sigma\in \operatorname{Sym}(A)$. 

To see that $\phi_f$ is a homomorphism note that $$\phi_f(\sigma)\phi_f(\tau)=f\circ \sigma\circ f^{-1}\circ f\circ \tau\circ f^{-1}=f\circ \sigma\circ \tau\circ f^{-1}=\phi_f(\sigma\tau);$$ to see that $\phi_f$ is an isomorphism note that we can explicitly name the inverse homomoprhism $\operatorname{Sym}(B)\to\operatorname{Sym}(A)$, namely we have $(\phi_f)^{-1}=\phi_{f^{-1}}$: $$\phi_{f^{-1}}(\phi_f(\sigma))= f^{-1}\circ \phi_f(\sigma)\circ f = f^{-1}\circ f\circ \sigma\circ f^{-1}\circ f=\sigma,$$ so $\phi_{f^{-1}}\circ\phi_f=\operatorname{Id}_{\operatorname{Sym}(A)}$ and similarly $\phi_{f}\circ\phi_{f^{-1}}=\operatorname{Id}_{\operatorname{Sym}(B)}$. 
Now if $|G|=n$, this just states that there exists a bijection $G\to \{1,\ldots,n\}$ of the underlying set of $G$ with the set of the first $n$ naturals. By waht we just saw, $\operatorname{Sym}(G)$ is isomorphic to $\operatorname{Sym}(\{1,\ldots,n\})=S_n$. Therefore, to show that $G$ is isomorphic to a subgroup of $S_n$, it is sufficient to show that $G$ is isomorphic to a subgroup of $\operatorname{Sym}(G)$: Any injective group homomorphism $G\hookrightarrow\operatorname{Sym}(G)$ produces an injective group homomorphism $G\hookrightarrow S_n$ via the isomorphism.
That being said, you essentially did the right thing: you considered left multiplication, i.e. for $a\in G$ the left multiplication by $a$, i.e. the map
$$\begin{align}\lambda_a\colon G&\to G\\
x&\mapsto ax\end{align} $$  We immediately verify that $\lambda_a(\lambda_b(x))=abx=\lambda_{ab}(x)$ (the law of associativity hides in here) so that $$\tag1\lambda_a\circ\lambda_b=\lambda_{ab}.$$ Moreover $\lambda_1$ is the identity map: $\lambda_1(x)=1x=x$). We conclude that $\lambda_a\circ\lambda_{a^{-1}}=\lambda_{a^{-1}}\circ \lambda_a=\lambda_1=\operatorname{Id}_G$, i.e. $\lambda_a$ is bijective with inverse $\lambda_{a^{-1}}$. Thus for each $a\in G$ we have an element $\lambda_a\in\operatorname{Sym}(G)$. Moreover, the map 
$$\begin{align}\lambda\colon G&\to \operatorname{Sym}G\\
a&\mapsto \lambda_a\end{align} $$ is a homomorphism because of $(1)$. This homomorphism has trivial kernel because $\lambda_a(1)=a\ne 1$ whenever $a\ne 1$. Thus we have our desired injective gorup homomorphism $\lambda\colon G\hookrightarrow \operatorname{Sym}(G)$.
