Let \(G\) be a group with order 144. We show that \(G\) is not simple.
Note that \(|G| = 2^4\cdot3^2\).
By Sylow Theorem (3), we get that \(n_3 \in \{1, 4, 16\}.\)
In this case, we are done.
In this case, let \(G\) act on Syl\(_3(G)\) by conjugation.
Consider the corresponding natural homomorphism \(m:G\to S_4\).
By Sylow Theorem (2), ker \(m \neq G\) as the action is transitive.
Assume ker \(m = (1)\). Then, \(m\) is an injective map and thus, \(m(G)\) has \(144\) elements. However \(m(G) \le S_4\) and \(|S_4| = 24\), a contradiction.
Thus, ker \(m\) is a proper nontrivial subgroup of \(G\).
As kernels of homomorphisms are always normal, we are done.
Thus, the number of elements in the union of all the Sylow-\(3\) subgroups is \(16(9-1) = 128,\) after removing the identity.
Now, note that any Sylow-\(2\) subgroup and any Sylow-\(3\) subgroup can intersect only trivially. Thus, the remaining \(144 - 128 = 16\) elements must form the unique Sylow-\(2\) subgroup.
This gives us that \(n_2 = 1\) and thus, we are done.
Let \(P_1\) and \(P_2\) be the two subgroups in question.
Note that \(|P_1| = |P_2| = 9\) and thus, \(P = P_1 \cap P_2\) must contain exactly \(3\) elements.
Now, let \(N = N_G(P),\) the normaliser of \(P\) in \(G.\)
As groups of order square of a prime are abelian (proof), we get that \(P_1 \cup P_2 \subset N.\)
Now, observe that if \(N\) contains \(P_1\) and \(P_2,\) then \(N\) must contain the set \(P_1P_2 = \{p_1p_2 : p_1 \in P_1,\; p_2 \in P_2\}.\) (This set need not be a subgroup.)
| This gives us that $$ | N | \ge | P_1P_2 | = \dfrac{ | P_1 | P_2 | }{ | P_1 \cap P_2 | } = 27.\(But\)27\(does not divide\)144$$. We are done! |