Let \(G\) be a group with order 180. We show that \(G\) is not simple.
Note that \(|G| = 2^2\cdot3^2\cdot5\).
Let us assume that \(G\) is simple. We shall arrive at a contradiction.
By our assumption, we get that \(n_3 > 1\) and \(n_5 > 1.\)
By Sylow Theorem (3), the choices for \(n_3\) are \(4\) and \(10\) and the choices for \(n_5\) are \(6\) and \(36.\)
Claim 1. \(n_3 \neq 4\)
Proof. Suppose not.
Let \(G\) act on Syl\(_3(G) = \{P_1, \ldots, P_4\}\) 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 \(180\) elements. However \(m(G) \le S_4\) and \(|S_4| = 4!\), a contradiction as \(180\not\mid4!.\) This shows us that ker \(m\) cannot be \((1).\)
Thus, ker \(m\) is a proper nontrivial subgroup of \(G.\)
As kernels of homomorphisms are always normal, we arrive at a contradiction as \(G\) was supposed to be simple.
Claim 2. \(n_5 \neq 6\)
Proof. Suppose not.
Consider the similar action as in the previous claim’s proof.
By our assumption of simplicity, we get that the kernel of the corresponding homomorphism \(m':G\to S_6\) is trivial.
Thus, \(G\) is isomorphic to \(m(G) \le S_6.\)
Let \(H = m(G).\) As \(H\) is simple and \(|H| > 2,\) it follows that \(H\) is a subgroup of \(A_6\) as well. (See the proof of claim here.)
However, observe that \(|H| = 180 = |A_6|/2.\)
This means that \([A_6:H]=2.\)
As subgroups of index 2 are normal, we get that \(H\) is normal in \(A_6.\)
However, this is a contradiction as \(A_6\) is simple.
(\(A_n\) is simple for \(n \ge 5.\))
Thus, we now arrive at the conclusion that \(n_3 = 10\) and \(n_5 = 36.\)
Note the following:
The above facts follow by considering the fact that the intersection would be a subgroup of the two bigger subgroups and would have to divide their orders.
Thus, the union of all Sylow-\(3\) and Sylow-\(5\) subgroups contains \(36(5-1) + 10(9-1) + 1 > 180\) elements, a contradiction.
Let \(P_1\) and \(P_2\) be the Sylow-\(3\) 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 \(|N| \ge |P_1P_2| = \dfrac{|P_1||P_2|}{|P_1 \cap P_2|} = 27.\)
Thus, the set of cosets \(G/N\) has cardinality at most \(180/27 < 7.\) (\(G/N\) need not a group.)
Let \(n = |G/N|.\) We have shown that \(n \le 6.\)
Now, let \(G\) act on \(G/N\) by left multiplication.
As this action is transitive, the kernel of the corresponding homomorphism \(\varphi:G\to S_n\) cannot be the whole group \(G.\)
However, as \(G\) is simple, this only leaves us with the choice that ker \(\varphi = (1).\)
Thus, \(\varphi\) is an injective map and \(|G| = 180 \mid n!.\)
We already knew \(n \le 6\) but the above now forces \(n = 6.\)
But the above has now left us in the same situation as the proof of Claim 2, where we showed that \(G\) cannot be isomorphic to a subgroup of \(S_6.\)
Thus, we are done!