# 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., $N=SL(2,\mathbb R)$.

Filed under Linear algebra

## 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 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.

Filed under Algebra, Number theory

## 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 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 $\varepsilon\ge 1$.