# 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

https://mathtuition88.com/

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