Let \(G\) be a group with order 80. We show that \(G\) is not simple.
Note that \(|G| = 2^4\cdot5\).
By Sylow Theorem (3), we see that \(n_5 \mid 16\) and \(n_5 \equiv 1 \mod 5\). Looking at the divisors of \(16,\) we see that \(n_5 = 1\) or \(n_5 = 16.\)
Then, we are done.
Thus, there are \(16\) Sylow-\(7\) subgroups of \(G\). Now, note that each such Sylow-\(5\) subgroup has order \(5\), a prime and thus, the intersection of two distinct Sylow-\(5\) subgroups must be trivial, id est, \((1)\). Thus, the number of elements having order \(5\) equals \(16(5-1) = 64.\)
This leaves us with \(16\) remaining elements which are not part of any Sylow-\(5\) 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-\(5\) subgroup. Thus, the remaining \(16\) elements form the unique Sylow-\(2\) subgroup which gives us that \(n_2 = 1\) and thus, we are done.