Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.