Let \(G\) be a group with order 56. We show that \(G\) is not simple.
Note that \(|G| = 2^3\cdot7\).
By Sylow Theorem (3), we see that \(n_7 \mid 8\) and \(n_7 \equiv 1 \mod 7\). Looking at the divisors of \(8,\) we see that \(n_7 = 1\) or \(n_7 = 8.\)
Then, we are done.
Thus, there are \(8\) Sylow-\(7\) subgroups of \(G\). Now, note that each such Sylow-\(7\) subgroup has order \(7\), a prime and thus, the intersection of two distinct Sylow-\(7\) subgroups must be trivial, id est, \((1)\). Thus, the number of elements having order \(7\) equals \(8(7-1) = 48.\)
This leaves us with \(8\) remaining elements which are not part of any Sylow-\(7\) subgroup.
Now, by Sylow Theorem (1), we know that \(n_2 \ge 1\). However, no non-identity element can be part of a Sylow-\(2\) as well as a Sylow-\(7\) subgroup. Thus, the remaining \(8\) elements form the unique Sylow-\(2\) subgroup which gives us that \(n_2 = 1\) and thus, we are done.