Tag Archives: unit

SL(2,IR) is the commutator subgroup of GL(2,IR)

Here is a proof of the above fact.

Let N be the commutator subgroup of the general linear group GL(2,\mathbb R); i.e.,

N=\langle ABA^{-1}B^{-1}:A,B\in GL(2,\mathbb R)\rangle.

First, it is clear that N is contained in the special linear group SL(2,\mathbb R), since \det(ABA^{-1}B^{-1})=1 for any A,B\in GL(2,\mathbb R). Next, we claim that N contains all matrices

\begin{pmatrix} 1 & b\\ 0 & 1\end{pmatrix}.

This follows from noting that

\begin{pmatrix} 1 & b\\ 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & b\\ 0 & b\end{pmatrix}\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}\begin{pmatrix} 1 & b\\ 0 & b\end{pmatrix}^{-1}\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}^{-1}.

By taking transposes, it also follows that N contains all matrices

\begin{pmatrix} 1 & 0\\ c & 1\end{pmatrix}.

Further, N contains all matrices

\begin{pmatrix} a & 0\\ 0 & 1/a\end{pmatrix}

since

\begin{pmatrix} a & 0\\ 0 & 1/a\end{pmatrix}=\begin{pmatrix} a & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix} a & 0\\ 0 & 1\end{pmatrix}^{-1}\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}^{-1}

for any a\neq 0.

Now let

\begin{pmatrix} a & b\\ c & d\end{pmatrix}\in SL(2,\mathbb R).

Then ad-bc=1. Using the above results,

\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} 1 & 0\\ c/a & 1\end{pmatrix}\begin{pmatrix} 1 & ab\\ 0 & 1\end{pmatrix}\begin{pmatrix} a & 0\\ 0 & 1/a\end{pmatrix}\in N

if a\neq 0, and

\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1&-\frac{d}{b}\\ 0&1\end{pmatrix}\begin{pmatrix}1&0\\ ab&1\end{pmatrix}\begin{pmatrix}1/b&0\\ 0&b\end{pmatrix}\in N

if b\neq 0, and the latter since

\begin{aligned}\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}=&\begin{pmatrix}x&y\\0&-x-y\end{pmatrix}\begin{pmatrix}-x-y&0\\ x&y\end{pmatrix}\begin{pmatrix}x&y\\0&-x-y\end{pmatrix}^{-1}\begin{pmatrix}-x-y&0\\ x&y\end{pmatrix}^{-1}\\ \in &N\end{aligned}

for any x,y,x+y\neq 0. Thus SL(2,\mathbb R)\subseteq N, i.e., N=SL(2,\mathbb R).

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Permuting the units mod n

Let n>2 be an integer and a\in(\mathbb Z/n\mathbb Z)^\times. We have a permutation map on (\mathbb Z/n\mathbb Z)^\times given by \pi(a):x\mapsto ax. Then it is easy to see that

\pi:(\mathbb Z/n\mathbb Z)^\times\to S_{\phi(n)}

is an injective homomorphism, where \phi is Euler’s totient function. This is the key idea in the proof of Cayley’s theorem, but we are going to do something a bit different.

Let’s look at \pi(a) in the disjoint cycle notation. One of the cycles is (1\ a\ a^2 \cdots a^{k-1}), where k is the order of a modulo n. What about the others? In fact, every cycle is of the form (b\ ba\ ba^2\cdots ba^{k-1}) for some b\in(\mathbb Z/n\mathbb Z)^\times. So every cycle has length k=\hbox{ord}_n(a). And the number of cycles is \phi(n)/k.

Let’s look at the sign \varepsilon(\pi(a)) of \pi(a). There are \phi(n)/k cycles each of length k, and each cycle of length k can be written as a product of k-1 transpositions, so

\varepsilon(\pi(a))=\displaystyle (-1)^{(k-1)\phi(n)/k}=(-1)^{\phi(n)/k}\quad(*)

where the last equality is obtained by looking at the cases when k is even/odd (and using the fact that \phi(n) is even).

Remark. (*) is saying that the number of transpositions in \pi(a) has the same parity as the number of disjoint cycles in it.

Hence f=\varepsilon\circ\pi, being a composition of homomorphisms, is a homomorphism from (\mathbb Z/n\mathbb Z)^\times to \{\pm 1\}. It might be interesting to know for which n we get all of \{\pm 1\} as \hbox{im}(f), and for which n we just get the trivial group \{1\}.

Fact.  \hbox{im}(f)=\{\pm 1\} if and only if \phi(n)/\hbox{ord}_n(a) is odd for some a\in(\mathbb Z/n\mathbb Z)^\times.

Corollary. If there exists a\in(\mathbb Z/n\mathbb Z)^\times such that v_2(\hbox{ord}_2(a))=v_2(\phi(n)), then there are exactly \phi(n)/2 such a‘s. (v_2(n) is the largest integer k such that 2^k\mid n.)

Proposition. If n is a power of two other than 2^2=4, then \hbox{im}(f)=\{1\}.

Proof. We want to show that \phi(n)/\hbox{ord}_n(a) is even for each a\in(\mathbb Z/n\mathbb Z)^\times. If n=2^k then the integers coprime to n are precisely the odd integers, so \phi(n)=2^{k-1}. So \hbox{ord}_n(a)=2^j for some j\le k-1. If j<k-1 there is nothing to prove. Otherwise, suppose that j=k-1. This means a is a primitive root modulo n. A well-known theorem says that primitive roots exist only for 2,4,p^k,2p^k for p an odd prime. Hence n must be 2 or 4\square

On the other hand:

Proposition. If there is a primitive root modulo n, then \hbox{im}(f)=\{\pm 1\}.

Proof. Let \phi(n)=2^qm, where m is odd. Let a be a primitive root modulo n. Then \hbox{ord}_n(a^m)=2^q, so \phi(n)/\hbox{ord}_n(a^m)=m is odd, and so f(a^m)=-1. \square

So both cases in fact occur infinitely often.

I don’t know whether it might be possible to completely classify the integers based on \hbox{im}(f), nor do I know what any of this actually means, but perhaps it is something worth pondering.

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A generalisation of Pell’s equation

We want to find the solutions to

(*)\qquad x^2+mxy+ny^2=1

over the integers, where m,n\in\mathbb Z.

Let \alpha and \bar\alpha be the roots of x^2+mx+n=0. We can factorise (*) as (x-y\alpha)(x-y\bar\alpha)=1. If \alpha is rational, it must be an integer and then it is easy to find all solutions. So let us consider the more interesting case where \alpha is not rational (i.e. \sqrt{m^2-4n}\not\in\mathbb Z).

Now a solution in the form z=x-y\alpha to (*) is a unit in the ring \mathbb Z[\alpha]. For z=x-y\alpha\in\mathbb Z[\alpha], denote its conjugate by \bar z=x-y\bar\alpha. Then we have the multiplicative norm

N(z)=N(x-y\alpha)=z\bar z=x^2+mxy+ny^2.

It follows that the solutions to (*) form a subgroup S of the multiplicative group of units of \mathbb Z[\alpha].

Case 1: m^2< 4n. Then (*) defines an ellipse, so there are at most finitely many solutions, i.e. S is a finite subgroup of \mathbb C^\times. Hence S is cyclic and every solution is of the form

\displaystyle x-y\alpha=\cos\frac{2\pi t}{k}+i\sin\frac{2\pi t}{k},\quad x-y\bar\alpha=\cos\frac{2\pi t}{k}-i\sin\frac{2\pi t}{k}

\displaystyle \Rightarrow 2x+my=2\cos\frac{2\pi t}{k}\in [-2,2]\cap\mathbb Z\Rightarrow \cos\frac{2\pi t}{k}\in\{0,\pm\frac 12,\pm1\}

\displaystyle\Rightarrow x-y\alpha\in\{\pm 1,\pm i, \frac{1\pm i\sqrt 3}{2}, \frac{-1\pm i\sqrt 3}{2}\}.

Case 2: m^2>4n. Now \alpha is real, so S is a subgroup of \mathbb R^\times.

Claim. S is not dense.

Proof. WLOG S\neq\{\pm 1\}. Then for z=x-y\alpha\in S\backslash\{\pm 1\} we have

\displaystyle \left |z-\frac 1z\right |=|y(\alpha-\bar\alpha)|\ge |\alpha-\bar\alpha|=\sqrt{m^2-4n}=:\varepsilon> 1.

Suppose that 1<z<\varepsilon is a solution. Then

\displaystyle \left |z-\frac 1z\right |=z-\frac 1z<\varepsilon-\frac 1\varepsilon<\varepsilon,

impossible. Thus S\cap (1,\varepsilon)=\emptyset, i.e. S is not dense. \square

It follows that S is either \{\pm 1\} or \langle\delta\rangle\cup -\langle\delta\rangle for some \delta>1 (cf. ordering in groups, corollary 4). Therefore all solutions (x,y) to (*) are either just (\pm 1,0), or are given by

\boxed{x-y\alpha=\pm (x_0-y_0\alpha)^k,\; k\in\mathbb Z}

where x_0-y_0\alpha is the least solution >1.

Corollary. (Dirichlet’s unit theorem for a real quadratic field) If K is a real quadratic field, then \mathcal O_K^*=\{\pm\varepsilon^n: n\in\mathbb Z\} for some \varepsilon\ge 1.

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Filed under Algebra, Number theory