Relationships between Symmetric Groups Here is a question that I can't find the answer to in my notes or my textbook.
Is there any relation between $\text{Sym}_n$ and $\text{Sym}_k$ where $k < n$? Is $\text{Sym}_k \subset \text{Sym}_n$? Since $k! |  n!\text{ } \forall  k < n $ they satisfy Lagrange's theorem, but of course that doesnt guarantee the existence of a subgroup
I'm trying to use this result to show that that there is a transitive action of a finite group $G$ on a smaller finite set $S$ by showing that $G$ or a subgroup of $G$ is isomorphic to $\Sigma (S)$ (The permutation group of $S$) which has a natural transitive action. 

$S_n$ is not contained in $S_m$ when $n < m$ if you define $S_k$ to be the set of all bijections $\{1,\dots,k\}\to\{1,\dots,k\}$, simply because an element of $S_n$ has domain $\{1,\dots,n\}$ and elements of $S_m$ have domain $\{1,\dots,m\}$, and the two sets are different.
Now this is a silly problem, so surely there is some way out...
Suppose again that $n < m$. There is a map $\phi:S_n\to S_m$ such that whenever $\pi\in S_n$, so that $\pi:\{1,\dots,n\}\to\{1,\dots,n\}$ is a bijection, then $\phi(\pi):\{1,\dots,m\}\to\{1,\dots,m\}$ is the map given by $$\phi(\pi)(i)=\begin{cases}\pi(i),&\text{if $i\leq n$;}\\ i,&\text{if $i > n$.}\end{cases}$$ You can easily check that $\phi$ is well-defined (that is, that $\phi(\pi)$ is a bijection for all $\pi\in S_n$) and that moreover $\phi$ is a group homomorphism which is injective.
It follows from this that the image of $\phi$ is a subgroup of $S_m$ which is isomorphic to $S_n$. It is usual to identify $S_n$ with its image $\phi(S_n)\subseteq S_m$, and then we can say that «$S_n$ is a subgroup of $S_m$», but this is nothing but a façon de parler. 
It should be noted, though, that there are many injective homomorphisms $S_n\to S_m$, so there are many ways to identify a subgroup of $S_m$ with $S_n$. The one I described above is nice because it looks very natural.
