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.
Asked
Active
Viewed 4,999 times
1
Enas
- 351
-
Well, what did you conclude from looking at the $5$-Sylows? – Tobias Kildetoft Nov 27 '13 at 19:12
-
@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? – Enas Nov 27 '13 at 19:34
-
Right. So why did you move on to the $2$-Sylows? – Tobias Kildetoft Nov 27 '13 at 19:34
-
the notes i found in the lecture states that my doctor checks all the sylow supgroups ! i dont know really why?? – Enas Nov 27 '13 at 19:37
-
Was that with this specific example, or another example? – Tobias Kildetoft Nov 27 '13 at 19:40
-
no no! other examples such as 48, 30 , 56 ! but anyway i can stop when i find a normal nontrivial subgroup , that is i dont need to chek all the normal subgroups? – Enas Nov 27 '13 at 19:45
-
Yes. To show that a group is not simple means precisely to show that it has a non-trivial proper normal subgroup. – Tobias Kildetoft Nov 27 '13 at 19:47
4 Answers
8
$|G|=200=2^3\cdot5^2$. 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 $G$.
Tim Ratigan
- 7,247
6
Hints:
$$\begin{align*}\bullet&\;\;200=2^35^2\\{}\\ \bullet&\;\;2\,,\,2^2\,,\,2^3\neq1\pmod 5\;,\;\;\text{so...}\end{align*}$$
DonAntonio
- 211,718
- 17
- 136
- 287
-
4
-
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). – Tobias Kildetoft Nov 27 '13 at 19:36
-
-
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...:) ) – DonAntonio Nov 27 '13 at 19:42
-
@TobiasKildetoft, check this https://etd.ohiolink.edu/ap:0:0:APPLICATION_PROCESS=DOWNLOAD_ETD_SUB_DOC_ACCNUM:::F1501_ID:ysu1337975956,attachment for a graduate thesis about Bender's proof. It really looks messy. – DonAntonio Nov 27 '13 at 19:43
2
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 $200\nmid\frac{8!}{2}$.
Mikasa
- 67,374
1
Alternatively let $P$ be a Sylow $5$-subgroup and show that the normal core of $P$ in $G$ cannot be $1$.
-
Is there actually a simpler argument for this than for the fact that the $5$-Sylow is normal (ie, is its own normal core)? – Tobias Kildetoft Nov 27 '13 at 19:46
-