Show that $|C_G(u)| = 12$ by “counting involutions”

Let $G$ be a transitive permutation group acting on $\Omega$ such that every non-trivial element fixing some point has exactly three fixed points. Suppose $G_{\alpha} \cong A_5$ for some point stabilizer, and that $G_{\alpha}$ has non-regular orbits of length $30$, $20$ and $12$ on $\Omega \setminus \{\alpha\}$.

Then $|C_G(u)| = 12$ for any involution $u \in G_{\alpha}$.

Why does this hold? The argument is by "counting involutions", but I do not know what this should mean?

• Is action of $G$ transitive ? – mesel Jan 15 '16 at 15:48
• From the information about the orbits of $G_\alpha$, an involution fixes exactly three points ($\alpha$ and two in the orbit of length $30$). Its centralizer in $A_5$ has order $4$, and $C_G(u)$ acts transitively on its fixed point set, so by the Orbit-Stabilizer Theorem $|C_G(u)|=12$. I am not sure what "counting involutions" means here. – Derek Holt Jan 15 '16 at 16:58
• The orbit of length $30$ is the only one having stabilizer with order divisible by $2$. The fact that $C_G(u)$ is transitive on its fixed points follows from the fact that all involutions in $A_5$ are conjugate. – Derek Holt Jan 15 '16 at 21:49
• This is a very standard argument in permutation group theory, and I am surprised that you have not seen it before. Suppose that $\alpha,\beta \in {\rm Fix}(u)$ and let $g \in G$ with $\beta^g=\alpha$. Then $u^g$ fixes $\alpha$, so $u^g$ is conjugate to $u$ in $G_\alpha = A_5$. Can you finish it now? – Derek Holt Jan 16 '16 at 9:40
• Looking at what you wrote again, you seem to have almost proved it yourself. Let $\alpha,\beta,\gamma$ be the fixed points of $u$. Then you know that there is an element in $C_G(u)$ fixing $\alpha$ and interchanging $\beta,\gamma$. Similarly there is an element fixing $\beta$ and interchanging $\alpha,\gamma$, and you are done. – Derek Holt Jan 16 '16 at 12:51

I collect and give more details to the hints given by Derek in his comments.

First a note on $A_5$. The elements of $A_5$ are the identity, the $3$-cycles, the $5$-cycles and the double transpositions, hence double transpositions are the only involutions. By trial-and-error we see that all the $\frac{1}{2} \binom{5}{2}\binom{3}{2} = 15$ double transposition are conjugate, hence $15 = |A_5 : C_{A_5}(u)|$ for some double transposition $u$, which gives $|C_{A_5}(u)| = 4$.

In the given situation, $C_G(u)$ acts transitive on the three fixed points of $u \in G_{\alpha}$. Let $\beta$ be a fixed point different from $\alpha$, then $\beta^g = \alpha$ for some $g \in G$ and $u^g$ fixes $\alpha$, i.e. $u^g \in G_{\alpha}$. Now choose some $h \in G_{\alpha}$ such that $u^{gh} = u$. Then $\beta^{gh} = \alpha^h = \alpha$ and $gh \in C_G(u)$.

With orbit-stabilizer we have $$3 = |C_G(u) : C_G(u) \cap G_{\alpha}| = |C_G(u)| / 4$$ hence $|C_G(u)| = 12$.

Also for $u \in G_{\alpha}$ an involution, the two other fixed points lie in the orbit of size $30$, as for example suppose it would fix a point $\beta$ in the orbit of size $12$, then $u \in G_{\alpha} \cap G_{\beta}$, but $12 = |G_{\alpha} : G_{\alpha}\cap G_{\beta}|$, and hence with $|G_{\alpha}| = 60$ the two-point stabilizer is not divisible by $2$, giving $u \notin G_{\alpha}\cap G_{\beta}$.

Just an addition: With the last paragraph, all the $15$ involutions in $G_{\alpha}$ have at most $1 + 30$ fixed points, conversely, if $u$ fixes $\beta$ and $\gamma$, then $u^g$ fixes $\beta^g, \gamma^g$, and so if $g$ runs through all $g \in G_{\alpha}$, if for example $\beta$ lies in the orbit of size $30$, it goes through all its points, and is a fixed point of $u^g$, hence the involutions have at least $30 + 1$ fixed points. Taken together, this gives every involution corresponds uniquely to two distinct points which it fixes (and the two fixed points corresponding to distinct involutions are all different), and each two points fixed on the orbit of size $30$ correspond to a unique involution from $G_{\alpha}$. With this observation the transitivity of $C_G(u)$ is also implied, as if $\alpha, \beta \in \mbox{Fix}(u)$ and $\alpha^g = \beta$ with $g \in G_{\alpha}$ (which could be choosen as they all lie in one orbit), then $u^g$ fixes also $\beta$ and lies in $G_{\alpha}$ as $g \in G_{\alpha}$, and this gives $u^g = u$ immediately as if $u^g \ne u$ they would have four different fixed points with the above remarks.

An argument similar as the above leads also to my conclusion in one comment, and as noted by Derek with this conclusion it also follows by noting that for $\beta \ne \alpha$ we also have $G_{\beta} \cong A_5$ as the action is transitive, and the situation is the same. Hence in the centralizer is an element fixing $\beta$ and interchanging the other two points, see the comments.