Groups of Automorphisms - remarks by Anderson and Feil ....

[Math Amateur]

I am reading Anderson and Feil - A First Course in Abstract Algebra.

I am currently focused on Ch. 24: Abstract Groups ... ...

I need some help in understanding some claims in Chapter 24 by Anderson and Feil ... ...Anderson and Feil claim that \(\displaystyle \text{Aut}( \mathbb{R} )\) is the trivial group ...

This surprises me a great deal as it seems to be saying that the number of permutations of the real number is \(\displaystyle 1\) ...! It seems to me that this should be an infinite number ... but obviously my intuition is way out ... ! Can someone please explain to me why \(\displaystyle \text{Aut}( \mathbb{R} )\) is the trivial group ...?Peter
NOTE: Anderson and Feil also claim that the groups \(\displaystyle \text{Aut}( \mathbb{R} )\) and \(\displaystyle \text{Aut}( \mathbb{R} )\) are the trivial group ... but how can this be ...

 

[Euge]

Hi Peter,

You're confusing permutation with field automorphism. Permutations on a set $S$ are bijections from $S$ onto itself, whereas automorphisms of a field $F$ are bijections that preseve addition and multiplication in $F$. The dilation map $\Bbb R\to \Bbb R$, $x\mapsto 2x$ is a permuation on $\Bbb R$, but not a field automorphism since it does not map $1$ to $1$.

Let $\sigma \in \operatorname{Aut}(\Bbb R)$. Use induction arguments to show that $\sigma(q) = q$ for every rational number $q$. Then prove that $\sigma$ is a continuous. By density of the rationals in the reals, it will follow that $\sigma$ is the identity.

 

[Math Amateur]

Hi Peter,

You're confusing permutation with field automorphism. Permutations on a set $S$ are bijections from $S$ onto itself, whereas automorphisms of a field $F$ are bijections that preseve addition and multiplication in $F$. The dilation map $\Bbb R\to \Bbb R$, $x\mapsto 2x$ is a permuation on $\Bbb R$, but not a field automorphism since it does not map $1$ to $1$.

Let $\sigma \in \operatorname{Aut}(\Bbb R)$. Use induction arguments to show that $\sigma(q) = q$ for every rational number $q$. Then prove that $\sigma$ is a continuous. By density of the rationals in the reals, it will follow that $\sigma$ is the identity.

Thanks for the help Euge ... I think i can establish that an automorphism \(\displaystyle f \) on \(\displaystyle \mathbb{R}\) is fixed on the rational numbers \(\displaystyle \mathbb{Q}\) ... ...
Suppose \(\displaystyle f\) is an automorphism on \(\displaystyle \mathbb{R}\) ...

Then \(\displaystyle f(n) = n\) and \(\displaystyle f(1/m) = 1/m\) for all \(\displaystyle m, n \in \mathbb{Z}\) ...... ... to show \(\displaystyle f(n) = n\) ... ...

\(\displaystyle f(1) = 1\) ... ... property of automorphism\(\displaystyle f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2\)\(\displaystyle f(3) = f( 1 + 2 ) = f(1) + f(2) = 1 + 2 = 3\)... and so on ... so \(\displaystyle f(n) = n\)
... ... to show \(\displaystyle f(1/m) = 1/m\) ... ...\(\displaystyle f(1) = f(m \cdot m^{-1} ) = f(m) \cdot f( m^{-1} ) = 1\)\(\displaystyle \Longrightarrow f( m^{-1} ) = 1 / f(m) = 1/m\) ...... that is ... \(\displaystyle f ( 1/m ) = 1/m\) ... ...
The case of \(\displaystyle -n\) and/or \(\displaystyle -m\) is covered by the fact that \(\displaystyle f( -a) = - f(a) \) for all \(\displaystyle a \in \mathbb{R}\) ... ... So ... \(\displaystyle f\) is fixed on the rational numbers \(\displaystyle \mathbb{Q}\) ...Is that correct?Not sure how to proceed ... can you help further ...

Peter

 

[Euge]

You're almost there. Rational numbers are generally in the form $n/m$ where $n$ and $m$ are integers with $m \neq 0$ (they're not just in the form $n/1$ or $1/m$). So the missing piece of your argument is establishing $f(n/m) = n/m$ for all such $n$ and $m$.

To proceed after that point, show first that $f(x) > 0$ for all $x > 0$, then prove $\sigma$ is strictly increasing. This will help in constructing a (short) $\epsilon-\delta$ argument to prove $\sigma$ is continuous.

 

[Math Amateur]

You're almost there. Rational numbers are generally in the form $n/m$ where $n$ and $m$ are integers with $m /neq 0$ (they're not just in the form $n/1$ or $1/m$). So the missing piece of your argument is establishing $f(n/m) = n/m$ for all such $n$ and $m$.

To proceed after that point, show first that $\sigma(x) > 0$ for all $x > 0$, then prove $\sigma$ is strictly increasing. This will help in constructing a (short) $\epsilon-\delta$ argument to prove $\sigma$ is continuous.

Yes, didn't quite finish ...

We have \(\displaystyle f(n/m) = f(n \cdot 1/m ) = f(n) \cdot f(1/m) = n \cdot 1/m = n/m\) ... using multiplicative property of automorphism ...

Peter

 

[Euge]

Yes, didn't quite finish ...

We have \(\displaystyle f(n/m) = f(n \cdot 1/m ) = f(n) \cdot f(1/m) = n \cdot 1/m = n/m\) ... using multiplicative property of automorphism ...

Peter

Did you also prove that $f$ is continuous?