Nonexistence of a simple group of order 420 From Dummit & Foote, Abstract Algebra, $\S6.2$, Exercise 17(a).

Prove that there is no simple group of order 420.

Suppose not; label such group $G$. the number of Sylow 7-subgroups of $G$ is 15. Let $G$ act on the set of Sylow 7-subgroups (denoted hereon by "letters") by conjugation. There is only one orbit of size 15 (by Sylow 2nd), thus each stabilizer on one letter has size 420/15 = 28.
The action induces an injective homomorphism of $G$ into $A_{15}$. Each stabilizer on a single letter should have an element of order 7 (by Cauchy), permuting the remaining 14 letters. Naturally, this element then generates the unique Sylow 7 within the stabilizer.

I now assume that the element of order 7 is a product of 2 disjoint 7-cycles. Is this valid, and why? In particular, am I able to eliminate the possibility of the element being a single 7-cycle?

If the above assumption is valid, then I am now able to eliminate the possibility of order 14 and 28 elements, since order 14 implies single 14-cycle (odd) or product of single 7-cycle and some 2-cycles (2nd power is single 7-cycle), likewise for order 28.
Now use the fact that the Sylow 7 is normal within the stabilizer: the permutations that sends a 7-cycle to its power by conjugation is either a product of 3 2-cycles (sends to inverse), 2 3-cycles (sends to 2nd/4th power), or a 6-cycle (sends to 3rd/5th power). By similar calculations, permutations that switch between the two 7-cycles are either 7 2-cycles, a 2-cycle and 3 4-cycles, a 2-cycle and 2 6-cycles, a 2-cycle and a 12-cycle, or a 14-cycle. Since the remaining elements are either order 2 and 4, our choices are either 6 2-cycles, 7 2-cycles (odd), or a 2-cycle and 3 4-cycles (third power breaks the pairing of the two 7-cycles).

Are my calculations and reasoning correct here?

The pair of fixed letters in the 2 7-cycles then determines the whole of the permutation, but noting that the two different fixed letter 2-cycles (per 1 7-cycle) result in a product of a 7-cycle, we conclude that the pairings of fixed letters must be disjoint for two different elements. Then, the number of possible remaining elements is 7 out of a required 21; contradiction.

In general, is there a cleaner way to go about this exercise, or proving nonexistence of groups of some highly composite order? I only know the method of embedding the group in an alternating group and trying to derive a contradiction from there (outside the usual repertoire).

 A: As an answer for you first questions: let me call the homomorphism $\phi:G\rightarrow A_{15}$, and the order 7 element $g$, generating the Sylow 7-subgroup $P$, which has normaliser (the stabilisers of your action are called normalisers) $N_G(P)$ with order 28. First, you state that $P$ is the unique order 7 subgroup of $N_G(P)$, which is true, because an order 28 group has only 1 Sylow 7-subrgroup. Note that $\phi(g)^7=\phi(g^7)=\phi(1)=\text{id}$, so $\phi(g)$ (which permutes 14 letters as you say) has order dividing 7, i.e. either 1 or 7. Therefore it is either the identity permutation, a single 7-cycle, or 2 disjoint 7-cycles. But in any case apart from the last case, $\phi(g)$ fixes at least 7 other letters, in other words it normalises some Sylow 7-subgroups that are not $P$; call one of these $Q$. But now $N(Q)$ contains 2 distinct order 7 subgroups, namely $P$ and $Q$, which contradicts your point that an order 28 group can have only 1 order 7 subgroup. Therefore we get the result that you want; $\phi(g)$ is 2 disjoint 7-cycles.
EDIT: As an answer to the main problem: I don't think I would go down your route directly (mainly because I don't like composing permutations, which I find very fiddly), so here is a way to continue with minimal fiddly permutation computations.
Let's look at the structure of $N_G(P)$; we know it has at least one order 4 (Sylow 2) subgroup, so call this $H$. now we have $P\unlhd N_G(P)$ and $H\le N_G(P)$, so $PH\le N_G(P)$. But $P,H\le PH$ so the size of $PH$ is divisible by 4 and 7, so it has size (at least) 28, and hence $PH=N_G(P)$; we can represent all elements of $N_G(P)$ as the product of an element of $P$ with one of $H$, and this representation is unique because there are only $7\cdot4=28$ possible representations. $P\cong C_7$, and either $H\cong C_4$ or $H\cong V_4$. Then we have (by $P$ normal in its normaliser) that $N_G(P)$ is isomorphic to a semidirect product, $C_7\rtimes C_4$ or $C_7\rtimes V_4$, determined by a homomorphism $\psi:H\rightarrow\text{Aut}(P)\cong\text{Aut}(C_7)\cong C_6$. $\ker(\psi)$ has size 1, 2 or 4. But ${H\over\ker(\psi)}\cong\text{im}(\psi)\le C_6$, so it cannot have size 1. If it has size 4, then the homomorphism is trivial, the semidirect product is direct, and $N_G(P)$ is abelian, and hence either $C_{14}\times C_2$ or $C_{28}$. But as you say, it cannot have an order 14 or order 28 element, contradiction. So $\vert\ker(\psi)\vert=2$, and so the image is the order 2 subgroup of $\text{Aut}(P)$, namely the trivial automorphism and the inversion automorphism (call this automorphism $\varphi$).
First, suppose $H\cong V_4$, and call the 3 non-identity elements $x$, $y$ and $xy$. If $\psi(x)=\psi(y)=\varphi$, then $\psi(xy)=\varphi^2=\text{id}$, so in any case we have non-identity element $h\in H$ such that $\psi(h)=\text{id}$, i.e. $h$ commutes with $g$. But then (easily verified) $gh$ has order 14, contradiction again.
So we must have $H\cong C_4$, so there exists $h\in H$ with order 4. If $\psi(h)=\text{id}$, then $\varphi$ is not in the image of $\psi$ ($h$ generates $H$), contradiction. So $\psi(h)=\varphi$.
EDIT 2: as pointed out in the comment, at this point I was already done, because $h^2$ now commutes with $g$, so as before we have an order 14 element $gh$ which is a contradiction. If you want to read a far more convoluted solution, here it is:
By semidirect product, $hgh^{-1}=g^{-1}$. Then finally we do have to look at permutations again; let $\phi(h)=\sigma$, which should be an order 4 permutation by the injection $\phi$. We have $\sigma\phi(g)\sigma^{-1}=\phi(g)^{-1}$, where w.l.o.g. $\phi(g)$, two disjoint 7-cycles, sends $i$ to $i+1$ (mod 7) and $i'$ to $i'+1'$ (mod 7') as a permutation of the letters 0 through 6 and 0' through 6'. Working modulo 7 and 7':
If $\sigma(i)=j$ then $\sigma(i+1)=[\sigma\phi(g)\sigma^{-1}](j)=[\phi(g)^{-1}](j)=j-1$, and by induction $\sigma(i+k)=j-k$ for all $k$ modulo 7. Let $k=j-i$, and we get $\sigma(j)=i$. Similarly, if $\sigma(i')=j'$, then $\sigma(j')=i'$. 
By the same inductive step we get that if $\sigma(i)=j'$ then $\sigma(i+k)=j'+k'$ for each $k$. But now $\sigma(j')=i+n$ for some particular $n$, which by the same induction again gives $\sigma(j'+k')=i+n+k$ for each $k$. Now $\sigma(i+n)=j'+n'$, and then by having order 4, $i=\sigma^{4}(i)=\sigma^{3}(j')=\sigma^{2}(i+n)=\sigma(j'+n')=i+n+n=i+2n$ so in fact we must have $n=0$, by inverse of 2 existing modulo 7. Hence $\sigma(i)=j'$ gives $\sigma(j')=i$. Again similarly, we also get that if $\sigma(i')=j$ then $\sigma(j')=i$.
This covers all cases for where $\sigma$ sends the letters 0 through 6 and 0' through 6', so $\sigma$ is composed only of disjoint transpositions, and hence has order 2, the final contradiction we needed, therefore $G$ is not simple.
A: Firstly, we have the following factorization $420=2^2.3.5.7$. I want to work on Sylow $5-$subgroups. I will use the $P_q$ for Sylow $q-$subgroups of G, and $n_q$ for the number of Sylow $q-$subgroups.
From the Sylow Theorems we have $n_q \equiv 1\pmod q $ and $n_q ||G| $. Thus, we have following possibilities for $n_5$:
$$n_5 =
\begin{cases}
6   \\[2ex]
21
\end{cases}$$
In each case we have $3|n_5$, and this means that $N_{G}(P_5)$ is not multiple by $3$. Thus, we cannot have an element of order $15$, as if we have such an element then we should have an element of order $3$ in the $N_{G}(P_5)$. This consequence can be improved.
Lemma 1: Simple group $G$ of order 420 does not have any subgroup of order $15$.
This is because every group of order $15$ is cyclic as it is with the form of $pq$ and $q  $ is not $ 1\pmod p $.
Now we want to use Sylow $3-$subgroups.
$$n_3 =
\begin{cases}
4   \\[2ex]
7   \\[2ex]
28
\end{cases}$$
We know that $n_3$ could not be $4$, since we don't have any subgroups with index less than $7$. In both other cases we will have a subgroup of order $15$. We know that $P_3$ is a normal subgroup of $N_G(P_3)$, and in both cases $5|N_G(P_3)$. Therefore, we have a subgroup of order $5$ in $N_G(P_3)$, and actually its is the Sylow $5-$subgroup of $G$. But, $P_3$ is a normal subgroup in $N_G(P_3)$; as a result, $P_3P_5$ is a subgroup of $N_G(P_3)$. This means $G$ has a subgroup of order $15$. Contradiction
Edited Part: I missed $n_3=10$. In such condition, $P_7$ is a normal subgroup itself. As if $n_3=10$ this means that $|N_G(P_3)|=2.3.7$. In this group $P_7$ is normal in $N_G(P_3)$ and $P_3P_7$ is a cyclic subgroup. Thus, $3.7| |N_G(P_3)|$. Moreover, $n_7|2^2.5$ and $n_7$ is $1 \pmod 7$. Therefore, $n_7 = 1$.
