Tell me more ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.
up vote 1 down vote favorite

Ler G be a group of order 24 with no elements of order 6, Let T be a subgroup of G, is a Sylow 3-subgroup. I have prove that G has no normal subgroup of order 2, so it is also clear that no element of order 4, 8. Now I want to show that the centralizer of T if T itself. I can show that all the elements in Sylow 2-subgroup can not be in the centralizer of T, and also $T\subset C(T)$, if we can show that all the other elements of Sylow 3-subgroup except e, not in T can not be in $C(T)$, then its done. but I can not show this, thanks in advance!

share|improve this question

2 Answers

up vote 2 down vote accepted

Since the prime graph of $G$ is disconnected, and since every group of order $24$ is solvable by Burnside's theorem, $G$ is either Frobenius or $2$-Frobenius. If $G$ were Frobenius, then $G=K\rtimes C$ where $|C|$ and $|K|$ are coprime and $|C|$ divides $|K|-1$, but this is impossible since $|G|=2^3\times 3$. Thus $G$ is $2$-Frobenius. $l_F(G)=3$ so we must then have that $|\text{Fit}(G)|=4$ so that $G/\text{Fit}(G)\cong S_3$. Since $\text{Aut}(\mathbb{Z}_4)\cong \mathbb{Z}_2$, $\text{Fit}(G)$ cannot be cyclic, and thus we have $S_3$ acting faithfully on the Klein $V$ group. Thus $G\cong S_4$, in which of course all the desired properties hold.

share|improve this answer
Thanks very much, actually I want to prove G is $S_{4}$, the reason I want to show that $C(T)=T$ is one step of that, but you have give me the whole proof, I appreciate that, also because of your proof is so short. but one thing is that I am not familiar with the theorem you give above, I will try to figure it out – user53800 Dec 27 '12 at 14:51
I am seriously nuking mosquitos here, my friend. :) – Alexander Gruber♦ Dec 27 '12 at 14:55
can you explain to me that what do you mean by Frobenius and 2-Frobenius and also the reason why $|C|$divides$|K|-1$, we don't learn this in class, so we can not directly use it. – user53800 Dec 27 '12 at 15:02
This was intended as a joke; you shouldn't use this proof for a class. I am using excessively advanced machinery here. But I'd be happy to explain everything in chat if you're interested. – Alexander Gruber♦ Dec 27 '12 at 15:12
1  
Haha, well, since you asked: Frobenius groups are made of two parts, a normal subgroup $K$ and a $C$ that acts on it $K$ fixed point freely (meaning that $c^{-1}kc\not= k\forall k\in K,c\in C$). $\text{Fit}(G)$ is the largest normal nilpotent subgroup of $G$. $2$-Frobenius groups are groups where, if $F_1=\text{Fit}(G)$ and $F_2/F_1=\text{Fit}(G/F)$, $F_2$ and $G/F_1$ are Frobenius groups. Similarly we define $F_i/F_{i-1}=\text{Fit}(G/F_{i-1})$; the smallest $i$ for which $F_i=1$ is $l_F(G)$. All $2$-Frobenius groups have $l_F(G)=3$. Now, the prime graph of a group $G$ is a graph whose vertex – Alexander Gruber♦ Dec 29 '12 at 2:29
show 3 more comments
up vote 5 down vote

Suppose that $H=\{e,a\}$ is a normal subgroup of order $2$. Let $x\in G$ have order $3$. Since $H\lhd G$, therefore $x\{e,a\}=\{e,a\}x$. Hence, $xa=ax$. Thus, $|xa|=\frac{|x||a|}{\gcd(|x|,|a|)}=6$. (contradiction)

share|improve this answer
Thanks very much – user53800 Dec 27 '12 at 14:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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