The Cayley's theorem says that every group $G$ is a subgroup of some symmetric group. More precisely, if $G$ is a group of order $n$, then $G$ is a subgroup of $S_n$.
In the course on group theory, this theorem is taught without applications. I came across one interesting application:
If $|G|=2n$ where $n$ is odd, then $G$ contains a normal subgroup of order $n$.
Q. What are the other elementary applications of Cayley's theorem in group theory, which can be explained to the undergraduates?
Cayley's theorem has an important status in group theory even in the absence of explicit applications: it's a sanity check on the definition of a group.
Before anyone had the idea of writing down the axioms for an abstract group, people studied what one might call concrete groups (I think the actual term was "systems of substitutions" or something like that, though), namely collections of bijections of some set $X$ closed under composition and inverses, or equivalently subgroups of $\text{Sym}(X)$. Cayley's theorem tells you that every abstract group is a concrete group, so the abstract axioms of group theory capture the concrete phenomenon, namely concrete groups, that they were invented to capture.
Similarly, before anyone had the idea of writing down the axioms for an abstract manifold, people studied what one might call concrete manifolds, namely submanifolds of $\mathbb{R}^n$. Theorems like the Whitney embedding theorem tell you that every abstract manifold is a concrete manifold, so again we find that the abstract axioms of manifold theory capture the concrete phenomenon that they were invented to capture.
See also, for example, this question where I asked what "Cayley's theorem for Lie algebras" should be.
Cayley's theorem realizes $G$ as a subgroup of $S_n$ when $|G|=n$. There is a more general fact to this.
Suppose that $G$ is a group, and $H$ is a subgroup of $G$ with finite index $n=|G:H|$. Then there is a morphism $\eta:G\to S_n$, and $\ker\eta\leqslant H$ is a normal subgroup of $G$. In particular, $\eta$ is injective if $H$ (or $G$) is simple$^1$, in particular if $H$ is trivial $\eta$ is the usual Cayley representation of $G$ in $S_n$.
$1.$ A group is simple if it admits no nontrivial normal subgroup, that is, the only normal subgroups of $G$ are $1$ and $G$ itself. Examples of simple groups include $A_n$ for $n>4$ and ${\rm PSL}(2,K)$ for every infinite field $K$.
P Let $\{g_1H,\ldots,g_nH\}$ be the distinct left cosets of $H$ in $G$, where the $g_i$ are representatives of each of the cosets. For $g\in G$, $g(g_iH)=(gg_i)H$ is one of the cosets in the tuple, so there is $\sigma(i)$ for which $g(g_iH)=g_{\sigma (i)}H$. Hence each $g\in G$ defines a permutation of these cosets and in turn of $\{1,\ldots,n\}$, that is, we send $g$ to the permutation $\eta(g)$ afforded by $g$. We need to check that $\eta(hg)=\eta(h)\eta(g)$. I will leave this to you. At any rate, we have built a morphism $\eta:G\to S_n$. Now suppose that $g$ is sent to the identity permutation. Then in partulcar, if we choose the coset $H$, we see that $gH=H$, this means that $g\in H$. Hence $\ker \eta\leqslant H$.
This lemma or whatever you want to call it is actually very useful throughout the basic group theory. For example, the above shows that if $G$ is a simple group, then $|G|\leqslant [G:H]!^2$ for any $H\leqslant G$.
$2.$ This is not an literary exclamation mark!$^3$ We're claiming the order of $G$ is at most the order of $[G:H]$ factorial.
$3$. This one is a literary exclamation mark.
In a similar vein, consider the following solution
Every (nonabelian) group of order $216$ is not simple.
P Assume $G$ is simple. By Sylow's theorem $n_3$, the number of Sylow $3$-subgroups, divides $8$ and is congruent to $1$ modulo $3$. This gives that $n_3$ is either $4$ or $1$. Since $G$ is simple, there must be $4$ Sylow subgroups. Then $G$ acts on the set of four Sylow subgroups (by conjugation) and thus gives us a morphism $\eta: G\to S_4$, which is not trivial. Since $\ker \eta$ is a normal subgroup of $G$, it must be the trivial subgroup, hence $G$ injects into $S_4$. But $G$ has order $216$, and $S_4$ has order $4!=24$. This is absurd, so $G$ must have been non-simple, after all.
The general idea is that if we can make $G$ act on a set of $n$ elements, we can obtain a morphism of $G$ into $S_n$. This represents $G$ as a group of permutations, but not necessarily faithfully (i.e. two different elements of $G$ may act identically on our set) since this representation (the morphism) might have a kernel. Cayley's theorem gives a faithful representation, since we're taking the cosets of the trivial subgroup $\{1\}$.
