# Group of order 432 is not simple

Recently, a viewer of my blog found that my recommended books for gifted children was helpful. You may want to check those books out too.

Quote (from comment found on home page): Thanks a lot for the blog. I have a P1 girl whose hobby is to do assessment books but doesn’t like reading story books. She has completed P2 assessment books and doing P3 assessment books. With advises from her school principal, I decided not to let her progress with assessment books and I am really lost as to what to do with her. I shall try out the books that you recommended.

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

http://mathtuition88.com

This site uses Akismet to reduce spam. Learn how your comment data is processed.