Let $H$ be a subgroup of a group $G$ of (finite) index $n$. Show that there is a homomorphism $f: G \rightarrow > \mathscr{S}_{n}$ with $\operatorname{kernel}(f) \subseteq H$.
Am I right that I basically have to show the homomorphism condition here?
For all $g_{1}, g_{2} \in G$, $f\left(g_{1} \circ g_{2}\right)=f\left(g_{1}\right) \circ$ $f\left(g_{2}\right)$ $f\left(g_{1} \circ g_{2}\right)$ is the permutation $\sigma_{g_{1} \circ g_{2}}$, which maps the subclass $i H$ to the subclass $\left(g_{1} \circ g_{2}\right) i H$ for $i=1,2, \ldots, n$.
$f\left(g_{1}\right) \circ f\left(g_{2}\right)$ are the permutations $\sigma_{g_{1}}$ and $ \sigma_{g_{2}}$, which map the minor classes $i H$ to the minor classes $g_{1} i H$ and $g_{2} i H$ for $i=1,2, \ldots, n$. If $i \in\{1,2, \ldots, n\}$, then $\sigma_{g_{1} \circ g_{2}}(i)=j$ if and only if $\left(g_{1} \circ g_{2}\right) i H=j H$. On the other hand, $\sigma_{g_{1}}\left(\sigma_{g_{2}}(i)\right)=\sigma_{g_{1}}(j)$, where $j=\sigma_{g_{2}}(i)$ and thus $j$ is the element to which $g_{2} i H$ is mapped under $g_{1}$.
With the compatibility of the group operations, $\left(g_{1} \circ g_{2}\right) i H=g_{1}\left(g_{2} i H\right)$ applies, since $j$ is the element to which $g_{2} i H$ is mapped under $g_{1}$, i.e. $j=$ $g_{1} g_{2} i H$. This means that $\sigma_{g_{1} \circ g_{2}}(i)=j=\sigma_{g_{1}}\left(\sigma_{g_{2}}(i)\right)$ for all $i \in\{1,2, \ldots, n\}$ and therefore $f\left(g_{1} \circ g_{2}\right)=f\left(g_{1}\right) \circ f\left(g_{2}\right)$.
The kernel $\operatorname{ker}(f)$ of $f$ now consists of the elements $g \in G$, for which $\sigma_{g}$ is the identity permutation. This means that $g$ maps to every element of the secondary classes $i H$, which is why $g$ is in $H$. Therefore, $\operatorname{ker}(f) \subseteq H$.
Can I argue like this (could I do this without subclasses)?
When you see a group homomorphism from $G$ to $\mathscr{S}_n$, it would be natural to consider a group action of $G$ on some finite set of size $n$. (We know that $\mathscr{S}_X\cong\mathscr{S}_n$ naturally if $|X|=n$.)
Do we have such a finite set now? The answer is yes. Note that $[G:H]=n$, so the set $G/H$ of left cosets of $H$ in $G$ is finite with cardinality $n$.
Do we have an action on $G/H$ by $G$? Yes again. We let $G$ act on $G/H$ by left multiplication: $$g\cdot aH:=(ga)H.$$ It would be a good exercise for you to check that such map is well-defined and truly a group action on $G/H$ by $G$.
Then such group action induces a group homomorphism $f:G\to\mathscr{S}_{G/H}$ whose kernel is contained in $H$. By composing it with the isomorphism $\mathscr{S}_{G/H}\to\mathscr{S}_n$, the desired homomorphism follows. The construction here is somewhat equivalent to your map above.