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I was trying to work out the Sylow p-subgroups for general linear groups over arbitrary fields, and was running into some trouble with non-algebraically closed fields. The real numbers, R, were particularly troubling:

Does the group GL(2,R) of 2×2 invertible matrices with real entries have a single conjugacy class of maximal 2-subgroups?

The specific thing that bothers me is that every 2nth root of unity has a quadratic minimal polynomial over R, and so we get cyclic 2-subgroups of arbitrary (finite) size, but I'm not sure I understand how to write down the elements of the union of this subgroup, and so I am not sure what its normalizer looks like.

Is there a description of this subgroup that allows one to determine if any particular matrix is contained in it?

In particular, I can write the subgroup down as a direct limit of reasonably explicit matrix groups, but the maps are not the standards inclusions, and so I cannot easily describe the limit as an actual union.

Using trig identities we can write down a recursive formula for the minimal polynomial of the 2nth root of unity over R, and then the rational canonical form gives a generator for a cyclic group of order 2n. A generator for 2n is mapped to the 2mth power of the 2n+m generator, which for m = 1 (the only map a sane person cares about) is easy to write down explicitly. However, as a whole the direct limit seems quite inexplicit, since no particular matrix is ever actually in the direct limit, only a conjugate of it by an infinite sequence of change of bases.

I would like to give fairly constructive descriptions of the Sylow subgroups, and in particular, would like an algorithm that, for an arbitrary p-subgroup, finds a conjugator into a "standard" copy of the Sylow p-subgroup. However, right now I cannot even find the standard copy.

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1 Answer 1

Jack, I can see why you are confused:

The union of the cyclic 2-subgroups has order 3.

Define $G_n$ to be the cyclic group generated by the companion matrix of $\zeta_{2^n}$,

$$G_n = \left\langle \begin{bmatrix} 0 & 1 \\ -1 & x_n \end{bmatrix} \right\rangle$$

where $x_n$ is defined recursively by $x_2 = 0$, $x_{n+1}^2 = x_n+2$, so that $$x_3 = -\sqrt{2}, x_4 = -\sqrt{2-\sqrt{2}}, x_5 =-\sqrt{2-\sqrt{2-\sqrt{2}}}, etc.$$

$$\displaystyle\varinjlim G_n = \varinjlim \left\langle \begin{bmatrix} 0 & 1 \\ -1 & x_n \end{bmatrix} \right\rangle = \left\langle \begin{bmatrix} 0 & 1 \\ -1 & \lim_{n\to\infty} x_n \end{bmatrix} \right\rangle = \left\langle \begin{bmatrix} 0 & 1 \\ -1 & -1 \end{bmatrix} \right\rangle \cong C_3$$

where we calculate the limit of $x_n$ as in this question, by recognizing that the members of the sequence are all negative and the limit is therefore the negative root of $x^2 = x+2$, so that $0=x^2-x+2=(x-2)(x+1)$.

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This is just a silly answer for finals week Friday. –  Jack Schmidt Dec 16 '11 at 19:40
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A comment to make sure that you're sure you're sane and not only talking to yourself in this thread :) –  t.b. Dec 16 '11 at 21:22

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