0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
7
$\begingroup$

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

$\endgroup$
5
$\begingroup$

Hints:

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

$\endgroup$
  • 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
2
$\begingroup$

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

$\endgroup$
  • $\begingroup$ @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
1
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.