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Is every maximal $p$-subgroup of $\operatorname{PGL}(2,K)$ conjugate, where $p$ is an odd prime not equal to the characteristic of $K$?

Here $\operatorname{PGL}(2,K)$ is the quotient group of the group of $2×2$ invertible matrices with entries from the field $K$ modulo the group of nonzero $K$-scalar of multiplies the identity matrix.

The only $p$-subgroups are isomorphic to the locally cyclic groups of $p^n$th roots of unity in $F$, where $[F :K]≤2$, that is, $F$ is a quadratic field extension of $K$ or $F = K$. Assuming the $p$-subgroup generates $F$ over $K$ and is not the identity when $F = K$, the normalizers are (generalized) dihedral groups, $\operatorname{Dih}(F^×)$, which proves the maximality of the full groups of roots of $p^n$th roots of unity in $F$.

In the finite field case, things are easier. $F^×/K^×$ has order $q+1$ and $K^×$ has order $q−1$, so either $p$ divides the first, or $p$ divides the second, and thus all $p$-subgroups of the same order are conjugate just by bringing their torus into line. This is not surprising since the maximal $p$-subgroups are conjugate by Sylow's theorem, and cyclic, so have a single subgroup of each order.

In the infinite field case, I am not quite as certain. For one thing, the quadratic $F$ is no longer uniquely defined by $K$, so I'm not sure to what extent the non-split tori are conjugate. Even worse, I'm not even sure that one cannot have a maximal $p$-subgroup of $K$-type and a maximal $p$-subgroup of $F$-type.

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up vote 5 down vote accepted

I think the answer is yes. I am not certain I understand it perfectly, but let me at least make an attempt at answering it. The main idea is to try and show that your extension field $F$ is uniquely determined by $K$. Apologies if I have got this wrong!

Let $q=p^n$ be a nontrivial power of the odd prime $p$. Suppose that $K$ contains a primitive $q$-th root of 1 (and hence it contains all $q$-th roots of 1). Suppose that $g \in {\rm PGL}(2,K)$ has order $q$, and let $h \in {\rm GL}(2,K)$ be an inverse image of $g$. So $h$ satisfies the equation $x^q-t=0$ for some $t \in K$. If $h$ acts irreducibly, then this equation factorizes into linear factors over the extension field $F$ centralizing $h$. But then over $K$ it factorizes into linear and quadratic factors, and since $q$ is odd, at least one of the factors must be linear. But then all roots of the equation lie in $K$, contradiction. So $h$ must act reducibly, and $g$ fixes exactly 2 points in its projective action.

Suppose first that $K$ contains primitive contains $p^n$-th roots of 1 for all $n$. Then, by the above argument, all $p$-elements in ${\rm PGL}(2,K)$ fix precisely 2 points, and it follows easily that any maximal $p$-subgroup fixes two points. It then follows from the 2-transitivity of ${\rm PGL}(2,K)$ that all maximal $p$-subgroups are conjugate.

So we assume from now on that there is a maximal $m \ge 0$ such that $K$ contains primitive $p^m$-th roots of 1.

The result follows similarly if the inverse images $h$ of all nontrivial $p$-elements $g$ act reducibly.

The final case is when some $h$ acts irreducibly. Again $h$ satisfies a polynomial equation $x^q-t=0$ for some $t \in K$. But the minimal polynomial of $h$ over $K$ is quadratic, so $F$ contains two roots $h_1$ and $h_2$ of this equation, and $h_1h_2^{-1} = w$ is an $r$-th root of 1, for some divisor $r=p^k$ of $q$. If $w \in K$ then, since $h_1h_2 \in K$, we have $h_1^2 \in K$, but then $h_2 = -h_1$, contradicting $p$ odd.

So $F$ contains primitive $r$-th roots of 1, but $K$ does not. So we must have $F = K[w]$ where $w$ is a primitive $p^{m+1}$-th root of 1. So in fact $F$ is uniquely determined by $K$. Since the maximal $p$-subgroups of ${\rm PGL}(2,K)$ all arise in the same way from this specific field $F$, I think it now follows that they are all conjugate.

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I think this does it, but infinite groups still worry me unduly. After I write up your argument carefully (which should be easy, thanks for being so clear!), I'll accept the answer. –  Jack Schmidt Jun 7 '11 at 18:38
Ok, I have now written up this much: (1) There are at most two types of p-subgroup: those contained in a 2-point stabilizer ("K-type"), and those acting irreducibly with a unique up to K isomorphism centralizer field F ("F-type"). (2) All K-type p-subgroups of a fixed order are conjugate, and all F-type p-subgroups of a fixed order are conjugate. –  Jack Schmidt Jun 8 '11 at 14:59
However, I have not shown that if an F-type p-subgroup exists, then every K-type subgroup is contained in (a conjugate of?) an F-type subgroup. In other words, right now I have possibly two conjugacy classes. –  Jack Schmidt Jun 8 '11 at 14:59
I should also have shown that if a $K$-type subgroup of order $p^n$ exists, then $K$ contains a primitive $p^n$-th root $w$ of 1. That follows from the fact that a matrix mapping onto an element of order $p^n$ in a $K$-type subgroup must be representable as a diagonal matrix with entries $t,tw$ for some $t \in K$. So, if $m$ is above, then maximal $K$-type subgroups have order $p^m$. But now, it follows from the above that the subgroup of order $p^m$ in a maximal $F$-type subgroup is a maximal $K$-type subgroup. –  Derek Holt Jun 8 '11 at 16:05
Thanks! I had all those pieces in my writeup, but didn't put them together. Do we have any examples where m is positive, but an F-type subgroup exists? I thought I proved the intersection of a K-type and an F-type was the identity in PGL2. If I'm right, then m=0, but I might be wrong. –  Jack Schmidt Jun 8 '11 at 16:10
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