How to prove that a finite group of order $280$ is not simple? 
Let $G$ be a finite group of order $280$. How to prove that $G$ is not simple?

A way to do it is to prove that there exists a p-Sylow subgroup of G that is normal, ie that there is a unique p-Sylow subgroup in G.
Here is what I have:
We have $|G| = 280 = 2^3 \cdot\ 5 \cdot\ 7 $
For each prime $p_i$ in the decomposition of 280, we have that the number of p_i-Sylow subgroups $n_{p_{i}}$ divides the product of the other primes and is congruent to 1 modulo $p_i$.
So that $n_5 \equiv 1 (\mod 5)$ and $n_5 | 56$ so $n_5$ is 1 or 56.
Let us say it is 56.
Then, with the same reasoning, $n_3$ is 1 or 40.
Let us say it is 40.
We have 56(5-1)+8(7-1)=272 elements of G who are elements of order 5 or 7. So those that remain are of order 2. There are 8 of them and we need to prove that it makes $n_8=1$ but how to proceed?
 A: 
Goal: Show that $G$ must have exactly $1$ Sylow $p$-subgroup for some prime $p$ dividing $280$.

Consider the number, $n_5$, of Sylow $5$-subgroups of $G$.  The third theorem tells us that $n_5$ must divide $|G|/5 = 56$ and that $n_5 \equiv 1 \pmod{5}$.
The divisors of $56$ are $\{1, 2, 4, 7, 8, 14, 28, 56 \}$.  Of these, only $56$ and $1$ are equivalent to $1 \!\pmod{5}$.  As you've pointed out, if $n_5 = 1$, we're done.  So assume that $n_5 = 56$.  Note that all the $5$-subgroups are necessarily disjoint save for the identity (why?).  So if there are $56$ of them, then they account for $56 \times 4 = 224$ of the elements of $G$.  
If we now consider the number of Sylow $7$-subgroups, we find that there's either $1$ or $8$ of them.  If there's only $1$, then we're done, so assume that there's $8$.  For the same reason as in the previous paragraph, the various Sylow $7$- subgroups share only the identity element in common.  Moreover, the same is true for $p$-subgroups and $q$-subgroups whenever $p$ and $q$ are distinct primes (again, why?).  So how many distinct elements of $G$ have now been accounted for?
Finally, if there are $56$ Sylow $2$-subgroups and $8$ Sylow $7$-subgroups, how many Sylow $2$-subgroups can there be?  
A: What are the possible numbers of subgroups of order $5$? If not one, how many elements of order $5$ are there. Similar question for $7$. How many elements left over?
A: If you know just a little about permutation groups, there's a more elegant solution.  Since $n_7>1$, $n_7\equiv 1 \mod 7$, and $n_7$ divides $40$, we have $n_7=8$.  The $8$ $7$-Sylow subgroups are permuted by $G$ and the action is faithful, since $G$ is simple, so $G$ embeds in $S_8$.  Note that the normalizer $N(P_7)$ of a $7$-Sylow has order $35$, and (since the $7$-Sylow subgroups of $S_8$ also have order $7$) its image, also of order $35$, in $S_8$ must be a subgroup of the normalizer of a $7$-Sylow in $S_8$, but that has order $42$, so this is not possible.
Note that we can use this strategy to show more, namely that any group of order $280$ has a normal $5$-Sylow.  For, again, if it doesn't have a normal $7$-Sylow, it must have $8$ $7$-Sylows.  But the argument in the preceding paragraph, one (and thus every) $5$-Sylow must be in the kernel of that action (so that the image of $N(P_7)$ has order dividing $42$). Also, since this action is transitive and since the elements of a $7$-Sylow are not in the kernel, both $8$ and $7$ divide the order of the image.  So the kernel must have order $5$, so the $5$-Sylow is normal, thus unique.
Now suppose the $7$-Sylow is normal.  The quotient has order $40$ and one easily sees by Sylow's counting theorem that the quotient has a normal $5$-Sylow.  Its inverse image in $G$ is thus a normal subgroup $H$ of order $35$.  Sylow again shows that $H$ has a normal $5$-Sylow, which is characteristic in $H$ and thus normal in $G$.  
Thus any group of order $280$ has a normal $5$-Sylow subgroup, which is clearly the unique $5$-Sylow.
A: Another way is to note that the normalizer $N(P_7)$ of a 7-Sylow has order 35, but the only group of order 35 up to isomorphism is the cyclic one. That implies the existence of elements of order 35 in $G$, and $A_8$, which $G$ injects into, cannot have elements of order 35, for that would imply that there are at least 12 letters that are permuted by $A_8$, for example, (1 2 3 4 5 6 7)(8 9 10 11 12).
