Group of order 432 is not simple

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For this blog post, we shall show that a group of order 432=2^4\cdot 3^3 is not simple. We will be using several previous posts as lemmas to prove this nontrivial result.

Suppose to the contrary G is simple. By Sylow’s Third Theorem, n_3\equiv 1\pmod 3, n_3\mid 16. This means that n_3 is 1, 4 or 16.

We recall that if n_3=1, then the Sylow 3-subgroup is normal.

Let Q_1 and Q_2 be two distinct Sylow 3-subgroups of G such that |Q_1\cap Q_2| is maximum.

Using our previous lemma regarding index of intersection of Sylow subgroups, we split our analysis into three cases, the hardest of which is Case 3.

Case 1) If |Q_1\cap Q_2|=1, [Q_1:Q_1\cap Q_2]=3^3. Thus n_3=1\pmod {27}, which allows us to conclude n_3=1.

Case 2) If |Q_1\cap Q_2|=3, [Q_1:Q_1\cap Q_2]=3^2. Thus n_3=1 \pmod 9. Similarly, we can conclude n_3=1 and we are done.

Case 3) If |Q_1\cap Q_2|=9, then [Q_1:Q_1\cap Q_2]=3.

By another previous lemma regarding index of least prime divisor, Q_1\cap Q_2\trianglelefteq Q_1. Thus, Q_1\leq N_G(Q_1\cap Q_2). Thus |N_G(Q_1\cap Q_2)|=3^3\cdot 2, 3^3\cdot 2^2, 3^3\cdot 2^3,\text{or }3^3\cdot 2^4.

If N_G(Q_1\cap Q_2)=G, then Q_1\cap Q_2\trianglelefteq G which is a contradiction. Hence we suppose N_G(Q_1\cap Q_2)\neq G. Let [G:N_G(Q_1\cap Q_2)]=k. The possible values of k are k=2, 2^2, 2^3, 2^4.

Next, we use the fact that if G is a simple group and H is a subgroup of index k, then |G| divides k!.

Thus, 432\mid k!, which forces k=16.

But then Q_1=N_G(Q_1\cap Q_2) and similarly Q_2=N_G(Q_1\cap Q_2). Thus Q_1=Q_2. This is a contradiction to |Q_1\cap Q_2|=9.

Therefore all cases lead to contradiction and thus G is not simple.

Author: mathtuition88

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