Suppose that G is the simple group of order 168.

Question

Show that G has 8 sylow subgroups. Find the normaliser of one of these. Show that the number of Sylow 3 subgroups is divisible by 7.

So $168=7\cdot 2^3\cdot3$ and is simple.

Then by Sylow's theorem we can say that

$p=7 \Rightarrow n_7=\{\text{factors of}\ 2^3\cdot3=24\}=\{1,2,3,4,6,8,12,24\}\equiv1\pmod7$ so $n_7=1$ or $8$. So we can conclude that $n_7=8$ and there are exactly $8$ sylow $7$ subgroups in $G$.

Then we do the same for $n_3$ and $n_2$, so we find that there are; $4,7,28$ sylow $3$ subgroups and $3,7,21$ sylow $2$-subgroups.

But I don't really know what to do next. Thanks

If $n_p = 1$, then the p-Sylow subgroup is unique, and hence normal. So in all cases, you can discount $n_p=1$. — Prahlad Vaidyanathan, Dec 02 '13 at 14:50

score 1 · Answer 1 · edited Apr 13 '17 at 12:20

You already showed that $n_7(G)=8$ unless we would face to a contradiction. So for your second question we have $$8=n_7(G)=[G:N_G(P)]$$ wherein $P$ is $\text{Syl}_7(G)$. This means that $|N_G(P)|=21$. Now if $N_G(P)$ is abelian so as you can see here or here, it must be $\mathbb Z_{21}$. And if it is not, it is not hard seeing that $$N_G(P)=\mathbb Z_3\ltimes\mathbb Z_7$$ with the following presentation:

$$\langle a,b\mid a^3=1, a^{-1}ba=b^2\rangle$$

An Khuong Doan · Answer 2 · 2013-12-02T15:17:56.753

0

Hint: The normalizer in $G$ of a $Sylow$ $3$- group is isomorphic to $S_3$

By Sylow's Theorem, $n_3= 7$ or $28$ then of course it is divisible by $7$

$n_3$ can't be $4$ as $G$ has no proper subgroup of index less than $7$.

Since if it had then the action of $G$ on the cosets of the subgroup would give a injective homomorphism (Because $G$ is simple) from $G$ into some $S_n$ with $n\leq 6$

edited Dec 02 '13 at 15:17

answered Dec 02 '13 at 14:55

An Khuong Doan

2,107

I've just found written in my notes;"this group must act on 7 things" is this about the G having no proper subgroup of index less than 7? I don't understand where the significance of the 7 comes from? – ZZS14 Dec 02 '13 at 15:17
Sr "7 things" I'm not sure about these words! – An Khuong Doan Dec 02 '13 at 15:22
it's just whatmy lecturer said. 7 cosets? – ZZS14 Dec 02 '13 at 15:22
Oh, Yes, it is right – An Khuong Doan Dec 02 '13 at 15:23
why must the group act on 7 cosets? – ZZS14 Dec 02 '13 at 15:26
You can see more explaination in detail in the textbook ABSTRACT ALGEBRA author David S.Dummit and Richard M.Foote, page $207$ (download free ) – An Khuong Doan Dec 02 '13 at 15:29
thanks! do you have any idea about the finding a normalizer of the 8 sylow 7 subgroups? – ZZS14 Dec 02 '13 at 15:32
In the text book I have just introduced to you, they gave and proved everything about the simple group of order $168$ ,page $207$ – An Khuong Doan Dec 02 '13 at 15:37
ah thats brilliant thanks, just trying to find an online copy as it is currently checked out of my library – ZZS14 Dec 02 '13 at 15:38
Find and enjoy it ! :D – An Khuong Doan Dec 02 '13 at 15:39
http://zeth.ciencias.uchile.cl/~cortiz/Apuntes/ebooks/Abstract%20algebra%20-%20Dummit%20and%20Foote.pdf
This is the link for anyone interested
– ZZS14 Dec 02 '13 at 15:41
Sorry, just another query. Can I say that the sylow 2 subgroups of G are only 7 and 21 not 3 because 3 is less than 7? – ZZS14 Dec 02 '13 at 16:01

Suppose that G is the simple group of order 168.

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