Let \(G\) be a group with order 90. We show that \(G\) is not simple.
Note that \(|G| = 2\cdot3^2\cdot5\).
Suppose that \(G\) is not simple. Then, we have that \(n_3 > 1\) and \(n_5 > 1.\)
By the constraints forced by Sylow Theorem (3), we get that \(n_3 = 10\) and \(n_5 = 6.\)
Let Syl\(_3(G) = \{P_1, \ldots, P_{10}\}\) and Syl\(_5(G) = \{Q_1, \ldots, Q_6\}.\)
It is clear that \(Q_i \cap Q_j = (1)\) for all \(i, j \in \{1, \ldots, 6\}\) and \(i \neq j.\) (\(Q_i \cap Q_j\) would have to be a proper subgroup of \(Q_i\) which has prime order.)
Also, \(Q_i \cap P_j = (1)\) for all \(i \in \{1, \ldots, 6\}\) and for all \(j \in \{1, \ldots, 10\}.\) (\(Q_i \cap P_j\) would have to be a subgroup of both \(Q_i\) and \(P_j.\) In particular, the order of their intersection would have to divide \(5\) and \(9.\))
Thus, \(|Q_1 \cup \cdots \cup Q_6 \cup P_1 \cup \cdots \cup P_{10}| = 6(5-1) + 10(9-1) + 1 = 24 + 80 + 1 > 90,\) a contradiction.
Let \(P = P_1 \cap P_2.\)
Then, \(P\) is a proper subgroup of \(P_1\) and not equal to \((1).\) As \(|P|\) divides \(|P_1|,\) we get that \(|P| = 3.\)
Let \(N = N_G(P),\) the normaliser of \(P\) in \(G.\)
Note that \(P_1\) and \(P_2\) were groups of order \(9 = 3^2.\) Thus, they were abelian. (See proof here.)
Thus, \(P_1 \cup P_2 \subset N.\)
\(\implies |N| \ge 9+9-3 = 15.\)
Also, \(P_1 \subset N \implies 9 \mid |N|\) and thus, \(|N| \ge 18.\)
In particular, \(90/|N| \le 5.\) (This will help us later.)
Now consider the set of all cosets \(G/N.\) (Note that this need not be a group but that is okay with us.)
Consider the action of \(G\) on \(G/N\) by left multiplication. That is, let \(\varphi:G\times G/N\to G/N\) be defined as \(\varphi(g, C) = gC = \{gc \in G : c \in C\}.\) It is simple to check that this is a valid group action.
Now, consider the corresponding natural homomorphism \(m:G\to S_n\) where \(n = [G:N],\) the index of \(N\) in \(G.\)
As the action is transitive, ker \(m \neq G.\) Also, \(n = 90/|N| \le 5\) and hence, \(90 \not\mid n!.\)
This gives us that \(m\) cannot be injective, for otherwise \(m(G)\) would be a subgroup of \(S_n\) with order \(90\) and would have to divide \(|S_n| = n!.\)
Thus, ker \(m\) is a proper nontrivial subgroup of \(G.\)
As kernels of homomorphisms are normal, we are done!