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!
Group of order $24$ containing no elements of order $6$
2 Answers
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.
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0proof, try to work through it and feel free to message me if you get stuck. – 2012-12-29
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)