If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple.

Question

If $G$ is a group of order $250,000 = 2^4 5^6$, show that $G$ is not simple.

By the Sylow theorem, we have that the number of $2$-sylow subgroups of $G$ $n_2$ satisfy that $$ n_2 \equiv1\mod2\mbox{ and } n_2\mid5^6 $$ Similarly for $n_5$ we have, $$ n_5 \equiv1\mod5\mbox{ and } n_5\mid2^4 $$ Hence, $$ n_2 \in \{1,5,5^2,5^3,5^4,5^5,5^6\}, \mbox{ and } n_5 \in \{1,16\} $$

But assuming that none of the $n_p'$s are one and using sylow theorems, I can't surpass the order of $G$ as the professor show us in class with one example. Now I am pretty sure I will need another approach but nothing comes to my mind. Any help would be appreciated.

Answer 1

If $n_5=1$, you are done.

If $n_5=16$, take one such 5-subgroup $H$. Its index is $2^4=16$. The index of its normal core $Core (H) $, per well-known Poincare's theorem, divides $16! $, thus it divides $\gcd (2^45^6,16!)=2^45^3$. In other words, $Core (H) $ is of index at most $2^45^3$ so it cannot be trivial - this gives you a nontrivial normal subgroup of $G$.

Answer 2

As you state, $n_5$ is either $1$ or $16$. If it's the former, we are done. If it is $16$, then there is a homomorphism from $G\to S_{16}$ given by the fact that $G$ acts on the $5$-sylow subgroups by conjugation. But notice that the prime factorization of $|S_{16}|=16!$ contains exactly three copies of $5$ (coming from $5$, $10$ and $15$), but $|G|=2^45^6$ contains six copies of $5$. So this homomorphism cannot be injective so that it's kernel is a normal subgroup of $G$. So $G$ is not simple.

This expounds the well known aphorism that "Groups, as men, will be known by their actions".