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.

  • $\begingroup$ Well, what did you conclude from looking at the $5$-Sylows? $\endgroup$ – Tobias Kildetoft Nov 27 '13 at 19:12
  • $\begingroup$ @TobiasKildetoft there is only one 5-subgroup of order 5 , so it is normal being a unique one ! can i stop here having a nontrivial normal subgroup? $\endgroup$ – Enas Nov 27 '13 at 19:34
  • $\begingroup$ Right. So why did you move on to the $2$-Sylows? $\endgroup$ – Tobias Kildetoft Nov 27 '13 at 19:34
  • $\begingroup$ the notes i found in the lecture states that my doctor checks all the sylow supgroups ! i dont know really why?? $\endgroup$ – Enas Nov 27 '13 at 19:37
  • $\begingroup$ Was that with this specific example, or another example? $\endgroup$ – Tobias Kildetoft Nov 27 '13 at 19:40

$|G|=200=2^3\cdot5^2$. 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 $G$.



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

  • 4
    $\begingroup$ Wow, some rush to downvote...and also the other answer. Oh, well. $\endgroup$ – DonAntonio Nov 27 '13 at 19:35
  • $\begingroup$ BTW, about the Burnside Theorem discussion: There is a purely group theoretic proof due to Bender, but as far as I understand it, that proof is way more complicated that the usual one, even though a few improvements have been made over the original (which was in 1972 I think). $\endgroup$ – Tobias Kildetoft Nov 27 '13 at 19:36
  • $\begingroup$ I upvoted again +1, clearly Enas sincerely tries to solve it. $\endgroup$ – Nicky Hekster Nov 27 '13 at 19:41
  • $\begingroup$ Thanks @NickyHekster, and yes: I think Enas honestly tries. I though cannot understand the downvote ( not that I give much of a rat's book in group theory, of course...:) ) $\endgroup$ – DonAntonio Nov 27 '13 at 19:42
  • $\begingroup$ @TobiasKildetoft, check this etd.ohiolink.edu/… for a graduate thesis about Bender's proof. It really looks messy. $\endgroup$ – DonAntonio Nov 27 '13 at 19:43

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 $200\nmid\frac{8!}{2}$.

  • $\begingroup$ @Serkan: Yes. That $H$ was in my mind. fixed. thanks $\endgroup$ – mrs Nov 28 '13 at 3:23
  • $\begingroup$ Maybe is very late to talk the Lemma, but how you would prove it? $\endgroup$ – MonsieurGalois Jan 23 '17 at 6:05

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

  • $\begingroup$ Is there actually a simpler argument for this than for the fact that the $5$-Sylow is normal (ie, is its own normal core)? $\endgroup$ – Tobias Kildetoft Nov 27 '13 at 19:46
  • $\begingroup$ @Serkan: Have a look at mine, plz. ;-) $\endgroup$ – mrs Nov 28 '13 at 3:13

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