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?
Best Answer
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?