# I Groups of Automporphisms - remarks by Anderson and Feil ...

1. Jun 25, 2017

### 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 $\text{Aut}( \mathbb{R} )$ 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 $1$ ...!!! 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 $\text{Aut}( \mathbb{R} )$ is the trivial group ...?

Peter

NOTE: Anderson and Feil also claim that the groups $\text{Aut}( \mathbb{Z} )$ and $\text{Aut}( \mathbb{Q} )$ are the trivial group ... but how can this be ...?

2. Jun 25, 2017

### Staff: Mentor

Do they mean ring, resp. field automorphisms or automorphisms of the additive groups $(\mathbb{R},+)\, , \,(\mathbb{Q},+)\; , \;(\mathbb{Z},+)\,$?

3. Jun 25, 2017

### Math Amateur

It appears from the text that they mean ring automorphisms ... they define $\text{Aut} ( R )$ as the set of all ring isomorphisms of $R$ onto itself ... and then claim that $\text{Aut} ( R )$ is a group under functional composition ...

Peter

4. Jun 25, 2017

### Staff: Mentor

O.k., then you have basically three defining properties for an automorphism $\sigma \in \operatorname{Aut}(R)$.
• $\sigma (a+b) = \sigma(a) + \sigma(b)$
• $\sigma (a \cdot b) = \sigma(a) \cdot \sigma(b)$
• $\sigma$ is a bijective function.
It also follows from these conditions, that $\sigma(0) = 0$ and $\sigma(1)=1$. You should try to prove this as an exercise.
With that, you can try to figure out, what it means for an element $a \notin \{0,\pm 1\}$ to be mapped on an element $b \neq a$.

5. Jun 25, 2017

### Infrared

As for why $\text{Aut}(\mathbb{R})$ is trivial:

Any automorphism $f:\mathbb{R}\to\mathbb{R}$ must be the identity on $\mathbb{Q}$. To see this, show that $f(n)=n$ for all $n\in\mathbb{Z}$ and $f(1/m)=1/m$ for $m\in\mathbb{Z}\setminus\{0\}$ using the additivitivity of $f$ and the claims in the last part of @fresh_42's post. Conclude that $f(m/n)=m/n$. This should also answer your questions about $\text{Aut}(\mathbb{Z})$ and $\text{Aut}(\mathbb{Q})$.

Next, $f$ must take positive numbers to positive numbers. To show this, suppose $x>0$. Then $x$ has a real nonzero square root, so we can write $x=y^2$ for some real $y\neq 0$. This means $f(x)=f(y)^2$ is positive, since it's a nonzero square.

This implies that $f$ is increasing since if $x<y$, then $y-x>0$ and so $f(y)-f(x)=f(y-x)>0$.

So, we have found that $f$ is an increasing function that restricts to the identity on $\mathbb{Q}$. Can you see why this implies that $f$ is the identity function on $\mathbb{R}$?

6. Jun 26, 2017

### Math Amateur

Thanks fresh_42 and Infrared (not sure where post gone ... )

Still thinking over fresh_42 challenge ... namely:

"... ... try to figure out, what it means for an element $a \notin \{0,\pm 1\}$ to be mapped on an element $b \neq a$ ... ... "

But Infrared asked to show $f(n) = n$ for $n \in \mathbb{Z}$ and $f(1/m) = 1/m$ for $m \in \mathbb{Z}\{0}$ ... for an automorphism f ... ...

... ... to show $f(n) = n$ ... ...

$f(1) = 1$ ... ... property of automorphism

$f(2) = f(1 + 1) = f(1) + f(1) = 1 + 1 = 2$

$f(3) = f( 1 + 2 ) = f(1) + f(2) = 1 + 2 = 3$

... and so on ... so $f(n) = n$

... ... to show $f(1/m) = 1/m$ ... ...

$f(1) = f(m \cdot m^{-1} ) = f(m) \cdot f( m^{-1} ) = 1$

$\Longrightarrow f( m^{-1} ) = 1 / f(m) = 1/m$ ...

... that is ... $f ( 1/m ) = 1/m$ ... ...

Now reflecting on what it means for an element $a \notin \{0,\pm 1\}$ to be mapped on an element $b \neq a$ ... ...

Peter

7. Jun 27, 2017

### Staff: Mentor

I removed it as it was too much help at this early stage of the thread. It's important that you get a "feeling" about what ring homomorphisms are. This question is a good exercise to get more accustomed to them. Infrared's post spoiled this before you had the chance to find it out yourself.
Don't take this too literally. It was just a choice as to why $f = 1$. In the cases $R=\mathbb{Z}\, , \,R=\mathbb{Q}$ it can probably be done constructively (formally by induction) without indirect proof. In the case $R=\mathbb{R}$ one cannot construct all numbers this way. An indirect proof would be needed.
Do you know why $f(0)=0$ and $f(1)=1$ is a property of ring homomorphisms? What's needed in the case above, that it can be deduces from the defining properties? And what about negative $m$ and other fractions $\frac{n}{m}$? Those are needed to show $\left.f\right|_\mathbb{Z}=1$.

For the case $R=\mathbb{R}$ you should use the bijection property, e.g. to consider the element that is mapped on $1$ and the $\operatorname{ker} f$. Butr these are only ideas. Try to figure it out, which properties help the best.

8. Jun 27, 2017

### WWGD

Maybe a good way of seeing it is because the requirements of the image for the unity determine the rest : $f(n)=f(1)+f(1)+..f(1)$( n times) EDIT: Remember that Rationals are dense in the Reals. These two restrict the options you have for automorphisms of the Reals.

9. Jun 28, 2017

### Math Amateur

Hi fresh_42, Infrared, WWGD ... thanks for the help ...

Now, fresh_42 writes:

"... ... Do you know why $f(0)=0$ and $f(1)=1$ is a property of ring homomorphisms? ... ... "

Yes ... did some revision reading on this ... among a number of texts, Anderson and Feil (Theorem 16.1) states and proves both and gives the conditions for $f(1) = 1$ to be true as well ... and certainly automorphisms of rings with unity satisfy $f(1) = 1$ ...

Now ... in a previous post I indicated why $f(n) = n$ and $f( 1/m ) = 1/m$ ... I did not explicitly treat the case of $-n$ and $-1/m$ ... but this case is covered by the fact that if $f$ is an automorphism of $\mathbb{R}$ then $f(-a) = - f(a)$ for all $a \in \mathbb{R}$ ... so we can take it that $f$ is fixed on the rational numbers $\mathbb{Q}$ ...

More later ...

Peter

***EDIT***

in above post ... didn't quite finish argument ...should have included the following at the end ...

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

Peter

Last edited: Jun 29, 2017
10. Jun 29, 2017

### mathwonk

it still doez not seem obvious to me but i think you can prove using squares that the image of a square is a square hence the image of a positive number is positive and hopefully then that the image of two numbers preserves their order, hence you should be able to use the density of the rationals to get thw result. i.e tryb to prove evert automorphism is oirder opreserving, i.e. increasing....,.

11. Jun 30, 2017

### Math Amateur

Hi fresh_42, Infrared, WWGD, mathwonk ... thanks for the help ...

I have shown, somewhat informally, that any automorphism of $\mathbb{R}$ is fixed on the rational numbers $\mathbb{Q}$ ... and mathwonk indicates that the path onward to show that $\text{Aut} ( \mathbb{R} )$ is a group consisting of one element, involves showing that $a \ge b \Longleftrightarrow f(a) \ge f(b)$ for any $a, b \in \mathbb{R}$ and any automorphism $f$ of $\mathbb{R}$ ... (NOTE I am happy with the proof in Anderson and Feil that $\text{Aut} ( \mathbb{R} )$ is a group ... and so I am focused on showing that $\text{Aut} ( \mathbb{R})$ consists of only one element, namely $f = 1$ ... )

We have for $a, b \in \mathbb{R}$ and any automorphism $f$ of $\mathbb{R}$:

$a \ge b$

$\Longleftrightarrow a - b = x^2$ for some $x \in \mathbb{R}$ ... ... note: I think this is true but I'd hate to hate to have to rigorously prove it ...

$\Longleftrightarrow f(a - b) = f(x^2)$

$\Longleftrightarrow f(a - b) = f(x) f(x) \ge 0$

$\Longleftrightarrow f(a) - f(b) \ge 0$

$\Longleftrightarrow f(a) \ge f(b)$

... ...

But where to from here ... ? I know from hints that I have received that I should use what I have just proved, together with the fact that f is fixed on $\mathbb{Q}$ and the fact that between any two real numbers is a rational number ... i.e. that $\mathbb{Q}$ is dense in $\mathbb{R}$ ... but I cannot see how to do this ...

Peter

Last edited: Jun 30, 2017
12. Jun 30, 2017

### WWGD

I don't mean to derail your argument, but here is another idea: you have shown f must fix the integers, the fractions of the form 1/m, from which it follows that all Rationals are fixed, since f(b/m)=f(b)f(1/m)=b/m . By density of the Rationals, I think you can argue that Irrationals are then fixed too, and you are done. Just a different idea to consider. Let me see if I can make it more precise.

13. Jun 30, 2017

### Staff: Mentor

I've put post #5 back in place, as it contains an essential hint at the end, which could be used together with the fact that $f(q)=q$ for all $q \in \mathbb{Q}$ and the Archimedean order.

14. Jun 30, 2017

### mathwonk

if x is irrational, and r,s are rationals with r < x < s, then what?

this is in fact an attempt to use WWGD's idea on density. I.e. density does you no good unless you can prove the automorphism is continuous, but it seems a bijection from the reals to the reals, is continuous if and only if it is either order preserving or order reversing.

15. Jun 30, 2017

### WWGD

Yes, I don't see a clear solution from there either; I think Infrared's post #5 is a nice answer, and to the point.

16. Jul 1, 2017

### Math Amateur

Thanks everyone ...

Have received some advice that I should attempt to follow Infrared's advice at the stage where I have shown that any automorphism $f$ on the reals $\mathbb{R}$ fixes $\mathbb{Q}$ and is order preserving ...

"... ... So, we have found that $f$ is an increasing function that restricts to the identity on $\mathbb{Q}$. Can you see why this implies that $f$ is the identity function on $\mathbb{R}$? ... ... "

BUT ... while Infrared's hint refers to something plausible, I cannot see how to formally and rigorously prove that $\text{Aut} ( \mathbb{R} )$ consists only of the identity function ...

I think I see the relevance of the fact that an automorphism $f$ has to be order preserving ... indeed if we switch any two functional values of $\mathbb{R}$ ... say $f( \sqrt{2} ) = \sqrt{2}$ to $f( \sqrt{2} ) = \sqrt{3}$ and change $f( \sqrt{3} ) = \sqrt{3}$ to $f( \sqrt{3} ) = \sqrt{2}$ ... we lose the order preserving property and the new function is not an automorphism ... in fact, thinking about it further ... even just changing $f( \sqrt{2} ) = \sqrt{2}$ to $f( \sqrt{2} ) = \sqrt{3}$ would destroy the order preserving property ...

I must say in passing that I do not see how to bring in the fact that an automorphism of $\mathbb{R}$ is fixed on $\mathbb{Q}$ ... even given the density of $\mathbb{Q}$ in $\mathbb{R}$ ... ...

But how to formulate a proof that $\text{Aut} ( \mathbb{R} )$ consists only of the identity function eludes me ...

Hope someone can help ...

Peter

Last edited: Jul 1, 2017
17. Jul 1, 2017

### WWGD

My bad, I thought the solution would be immediate. There is a result that a monotone Real-valued function can only have countably-many jumps, so we cannot yet guarantee this function is continuous, unless we can use the fact that it fixes Rationals and it is increasing forces it to be continuous. Let me think.

18. Jul 1, 2017

### Infrared

I guess I'll give the answer I intended.

Fix $\varepsilon>0$ and let $x\in\mathbb{R}$ be arbitrary. Choose $a,b\in\mathbb{Q}$ such that $x-\varepsilon<a<x<b<x+\varepsilon$ (using density of $\mathbb{Q}$). Using monotonicity of $f$, we have $f(a)=a<f(x)<b=f(b)$ and hence $x-\varepsilon<f(x)<x+\varepsilon$. As $\varepsilon>0$ was arbitrary, this means $f(x)=x$.

19. Jul 1, 2017

### WWGD

Ah, yes, squeeze any Irrational using Rationals at both ends. Fixing the Rationals will prevent the jumps.. Any jump at an irrational will, informally, increase the value at the right-end, contradicting that the Rationals are fixed. EDIT: If , wolg, for y irrational, f(y)=y+a; a>0 and x is a Rational with x>y and x-y<a/2 , then f(x)<f(y) while x>y . EDIT2: Sorry, Infrared, did not see your argument; mine is essentially the same as yours.

Last edited: Jul 1, 2017
20. Jul 2, 2017

### Math Amateur

Thanks Infrared ... really impressive proof ...

... ... great! ... ...

Peter

21. Jul 2, 2017

### Math Amateur

Thanks for the help WWGD ...

Peter

22. Jul 4, 2017

### mathwonk

oops. i didn't even notice that infrared had already given the proof i suggested 5 posts earlier.

23. Jul 4, 2017

### Infrared

Oh, that's not your fault- fresh42 initially deleted it since he saw it as too much of a hint and reinstated it later (after you posted).

24. Oct 9, 2017

### lpetrich

One can prove this result by starting from the nonnegative integers N. For mapping S from N onto some subset of some superset of N, with S(a+b) = S(a)+S(b) and S(a*b) = S(a)*S(b), it is easy to prove that S(0) = 0 and S(1) = 1.

Let us now do Peano's unary construction of all the elements of N. They are all 1 + 1 + 1 + ... + 1. This can be done with mathematical induction. For a in N,
S(a+1) = S(a) + S(1) = S(a) + 1
If S(a) = a, then S(a+1) = a + 1, thus S(a) = a for all a in N.

Thus, S is the identity mapping for N.

For integers Z, we need additive inverses: S(a + (-a)) = S(0) = 0 = S(a) + S(-a) = a + S(-a)
Thus, S(-a) = - a, and S is the identity mapping for Z.

For rational numbers Q, we need multiplicative inverses: S(a*(1/a)) = S(1) = 1 = S(a)*S(1/a).
Thus, S(1/a) = 1/a, and S(a/b) = a/b, and S is the identity mapping for Q.

For real numbers R, one can use Cauchy sequences of rational numbers: x ~ {x1, x2, x3, ...}
Thus, S(x) ~ {S(x1), S(x2), S(x3), ...} = {x1, x2, x3, ...}
Meaning that S(x) = x and that S is the identity mapping for R.