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Let $G$ be a finite, transitive, nonregular permutation group on $\Omega$ such that every element of $g \in G^{\#} := G \setminus \{ 1_G \}$ has at most two fixed points. Suppose further that $|\Omega| \ge 4$.

Lemma 1: Let $\alpha \in \Omega$ and let $1 \ne X \le G_{\alpha}$. Then $G_{\alpha}$ contains a subgroup of $N_G(X)$ of index at most $2$.

Proof: As $N_G(X)$ acts on the set of fixed points of $X$ on $\Omega$, it is contained in $G_{\alpha}$ or it induces the symmetric group on two points on the two fixed points of $X$. Therefore $N_{G_{\alpha}}(X)$ has index at most $2$ in $N_G(X)$.

Lemma 2: $|Z(G)| \le 2$

Proof: Let $x \in Z(G)^{\#}$ and assume that $x$ has a fixed point $\omega \in \Omega$. Then $x \in G_{\omega}$ and the transitivity of $G$ forces $x$ to fix every point in $\Omega$. This is a contradiction because $|\Omega| \ge 4$. Therefore $x$ has no fixed points on $\Omega$.

Let $\alpha \in \Omega$ and $1 \ne x \in G_{\alpha}$. Then $Z(G) \le C_G(x)$, but $Z(G) \cap G_{\alpha} = 1$ by the previous paragraph. Therefore $|Z(G)| = |Z(G) : Z(G) \cap G_{\alpha}| \le 2$ by Lemma 1.

In the above two proof I do not understand the lines

"Therefore $N_{G_{\alpha}}(X)$ has index at most $2$ in $N_G(X)$"

and

"Therefore $|Z(G)| = |Z(G) : Z(G) \cap G_{\alpha}| \le 2$ by Lemma 1".

As I understand it, what Lemma 1 says is that if $1\ne X \le G_{\alpha}$, then there exists a subgroup $U$ of $G_{\alpha}$ and of $N_G(X)$, i.e. $U \le G_{\alpha} \cap N_G(X)$, such that $|N_G(X) : U| \le 2$. But regarding the first line, that the index of $N_{G_{\alpha}}$ is bounded, says not that much about the index of subgroups of $N_{G_{\alpha}}(X) = G_{\alpha}\cap N_G(X)$ in $N_G(X)$, as they are smaller their index could be larger?

Also for the second line. If $X = \langle x \rangle$, then by $Z(G) \le C_G(x)$ we have $Z(G) \le N_G(X)$. Now if $U$ is the subgroup proposed by Lemma 1, then I know almost nothing about the relation between $Z(G)$ and $U$, they could be subgroups of each other or not, so how could I derive such a statement about the index? (if I do not know how the index of $U$ relates to the index of $Z(G)$)

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For the first if $N_{G}(X) \leq G_{\alpha},$ then there is nothing to prove. Otherwise, $X$ has just two fixed points, one of which is $\alpha$. If the other one is $\beta$, then $N_{G}(X)$ must contain an element which sends $\alpha$ to $\beta$ (it has to send $\alpha$ to some fixed point of $X$). Hence $N_{G}(X)$ acts as a transitive permutation group on $\{\alpha,\beta \}$ in which a point stabilizer has index $2$. But $N_{G_{\alpha}}(X)$ is a point stabilizer.

For the second, note that $Z(G) \leq N_{G}(X)$. The same argument as above shows that either $Z(G) \leq G_{\alpha}$ or else $[Z(G):Z(G) \cap G_{\alpha}] = 2$, because in the latter case $Z(G)$ acts again as a transitive permutation group on the two fixed points of $X$, and has point stabilizer of index $2$.

For (probably the intended approach to) the second, note also that if $H$ and $K$ are subgroups of a finite group $L$, then we always have $[H: H \cap K] \leq [L:K],$ because the subset (though not necessarily subgroup) $HK$ of $L$ has cardinality $|H||K|/|H \cap K|$ and this has to be at most $|L|.$ We can apply this with $Z(G)$ in the role of $H$, $N_{G}(X)$ in the role of $L$, and $N_{G_{\alpha}}(X)$ in the role of $K$. Then we have $[Z(G): Z(G) \cap G_{\alpha}] = [H: H \cap K] \leq [L:K] \leq 2.$

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  • $\begingroup$ In your first paragraph, the next to last sentence: "[...] in which a point stabilizer has index $2$ [...]". Why that? $\endgroup$ – StefanH Mar 24 '15 at 16:30
  • $\begingroup$ If $G$ is a transitive permutation group on a finite set $\Omega$, and $G_{\alpha}$ is a point stabilizer, then we have $|\Omega| = [G:G_{\alpha}].$ $\endgroup$ – Geoff Robinson Mar 24 '15 at 16:41

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