Group of order $24$ containing no elements of order $6$

Let $G$ 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!

-

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.

-
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
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

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)

-
Thanks very much – user53800 Dec 27 '12 at 14:47