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A finite group G satisfies property $P_n$ if for every prime integer $p$, $G$ has at most $(n−1)$ non-central conjugacy classes the order of the representative element of which is a multiple of $p$. Let $G$ be a non-solvable group satisfying property $P_5$. Also suppose that $N$ be an abelian $p$-group of order $p^r$ with $p\notin \pi(G/N)$ and $N=C_G(N)$. Let $K=G/N\cong A_5$, $Y< K$ be a Frobenius group of order 12‎, ‎$x\in Y$ of order 3 and $P$ a Sylow 2-subgroup of $Y$‎. Then we conclude that $p, 2p, 3p \in \pi_e(G)$ and since $G$ satisfies in property $P_5$, it follows that $G$ has at most two class of elements of order 3p. Clearly all elements of order $3$ lie in one class. Suppose that $y$ and $z$ are elements of order $3p$. The question is whether $P$ acts fixed point freely on $N$ and how the values of $|C_N(x)|, |C_G(x)|$, $|C_G(y)|$ and $|C_G(z)|$ be computed?

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I don't believe that there are any groups $G$ satisfying these conditions. Let $g \in G$ with $|g|=3p$. Then the cyclic group $H = \langle g \rangle$ contains $2(p-1)$ elements of order $3p$. If any of these are conjugate in $G$, then they are conjugate in $N_G(H)$. But, by considering its image in $A_5$, we see that $|N_G(H)/C_G(H)| \le 2$, so there must be at least $p-1$ conjugacy classes of element sof order $3p$ in $G$.

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  • $\begingroup$ Would you please explain more that what did you mean from "by considering its image in $A_5$" and how did you conclude that $|N_G(H)/C_G(H)|\leq 2$? $\endgroup$ – Z. Fo Jul 12 '14 at 9:37
  • $\begingroup$ @B.S. Please explain it. What is the role of $A_5$ in the answer? what can we say if $K=G/N\cong A_5$, $K=G/N\cong L_2(q)$ for $q=7,8,9,11,13,17,19$ or $K=G/N\cong L_3(4)$? $\endgroup$ – Z. Fo Jul 13 '14 at 8:17
  • $\begingroup$ @Derek Holt: Thank you. But I didn't understand What is the role of $A_5$ in the answer? What can we say if $K=G/N\cong A_5$, $K=G/N\cong L_2(q)$ for $q=7,8,9,11,13,17,19$ or $K=G/N\cong L_3(4)$? $\endgroup$ – Z. Fo Jul 14 '14 at 14:24

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