No group of order 200 is simple

Question

Show that any group of order 200 is not a simple group.

I have started with the sylow 5-subgroup, but the sylow 2 subgroups i found that they are 25 subgroup but i couldnt proceed in the proof any more! Please help.

Answer 1

$|G|=200=2^3\cdot5^2$. If you want the easy way out, Burnside's Theorem trivially proves that this is not a simple group.

By Sylow's third theorem, the amount of $5$-subgroups $n_5$ must divide $2^3=8$. However, $n_5\equiv 1\pmod 5$ so $n_5=1$.

Since conjugation preserves the order of elements and all the elements in the $5$-subgroup are the only elements which can have order that divides $5^2$, the subgroup must be normal in $G$.

Answer 2

Hints:

$$\begin{align*}\bullet&\;\;200=2^35^2\\{}\\ \bullet&\;\;2\,,\,2^2\,,\,2^3\neq1\pmod 5\;,\;\;\text{so...}\end{align*}$$

Answer 3

Alternatively let $P$ be a Sylow $5$-subgroup and show that the normal core of $P$ in $G$ cannot be $1$.

Answer 4

There is a nice way for seeing that the group $G$ is simple:

Lemma: Let $G$ be a simple group and $H< G$ such that $[G:H]=n$. Then $G\hookrightarrow A_n$.

Indeed, about the $5-$sylow subgroup of $G$ we see that $200\nmid\frac{8!}{2}$.