Automorphism on integers

Is multiplying by a constant m (integer) on group of set of all integers on addition an automorphism?

If so why does the 2nd example in says that the unique non trivial automorphism is negation?

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Automorphism is a permutation of a set which respects some structure on the set. What structure? It varies. Automorphism is a general term and does not apply simply to groups, or rings.

In the context of $(\mathbb Z,+)$ as an additive group, we say that $f\colon\mathbb {Z\to Z}$ is an automorphism if:

  1. $f$ is a bijection,
  2. $f(m)+f(n)=f(m+n)$,
  3. $f(0)=0$.

Now suppose that $f$ is an automorphism like that. Well, $f(0)=0$. If $f(1)=1$ then $f$ has to be the identity, because for $n>0$ we have $$f(n)=f(\underbrace{1+\ldots+1}_{n \text{ times}})=\underbrace{f(1)+\ldots+(1)}_{n \text{ times}}=\underbrace{1+\ldots+1}_{n \text{ times}}=n$$ and if $n<0$ then $$0=n+(-n)=f(n)+f(-n)=f(n)+(-n)$$ and therefore $f(n)=n$ as well.

Similarly if $f(1)=-1$ then $f(n)=-n$ for all $n$. So these are two automorphisms.

If $f(1)=n$ for some $n>1$, since $f$ is a bijection we have some $k\in\mathbb Z$ such that $f(k)=1$. We can write, if so, $f(1)=f(k)+\ldots+f(k)=f(k+\ldots+k)$. However $f$ is injective and therefore $k+\ldots+k=1$, which can only happen if $k=1$ and $n=1$. In a similar way we show that if $f(1)$ is negative then it has to be $-1$.

All that is well, but we can have more to say here. Suppose now we take $\mathbb Z$ as an ordered group. This means that now we add the ordering $\leq$ and require it plays nice with the addition, as it does with the standard addition and ordering.

In this case it is easy to show that $f(1)=1$ is the only automorphism possible. Why? Because $0<1$ implies that $f(0)<f(1)$, since now $f$ has to preserve the order too, so the first part shows that you cannot have a group automorphism of $\mathbb Z$ except $n\mapsto\pm n$, and if you want order preservation too you can only have the identity.

If on the other hand, you only want to consider $(\mathbb Z,\leq)$ as an ordered set, then you only wish to preserve $n<m\iff f(n)<f(m)$. In this case, it is not hard to show that for every $k\in\mathbb Z$ the function $f(n)=n+k$ is an automorphism.

To sum up, this is really about the structure you wish to preserve. The standard sets you know, $\mathbb {N,Z,R,Q,C}$ are all sets which have several different structures arising naturally from their properties as we know them. Each of these structures has a different property to preserve, and some would allow us rich automorphism groups, while other structures will limit us to have only a few automorphisms, if not one.

You might have heard of the term “homomorphism” of groups. This is a map between the groups that behaves nicely with respect to the group operations.

i.e. $f(gh) = f(g)f(h)$ for all $g\in G$, $h\in H$.

Now a bijective homomorphism is an “isomorphism”. This is essentially capturing the ability to relabel the elements of one group to get the other group with the products matching. For example here are two groups:

1) $\{0,1\}$ under addition mod $2$.

2) $\{a,b\}$ with $a^2=a, ab=b, ba=b, b^2=a$.

These are essentially the same group if we replace $0$ with $a$ and $1$ with $b$. They are isomorphic

Now we might ask if there are any ways to relabel elements of a group as elements of the same group. In other words are there any isomorphisms from a group to itself? These are called the “automorphisms” of a group. (in fact they form a group themselves under composition).

So for example with $\mathbb{Z}$ we have the identity automorphism $n\mapsto n$. We could also relabel the integers as going from right to left instead of the usual left to right. In other words we can consider $n\mapsto -n$. This is another automorphism corresponding to multiplication by $-1$.

Now these are in fact the only two automorphisms of $\mathbb{Z}$ (so that the automorphism group is isomorphic to the cyclic group of order $2$).

Why is this? Well $\mathbb{Z}$ is a cyclic group under addition with generator $1$, so any automorphism is completely determined by its value at $1$. But if you send $1$ to say the integer $5$ then the outputs of this homomorphism will all be multiples of $5$. So we will NOT have relabelled properly, some integers will be missing as outputs.

It is clear that $1$ can only be sent to $\pm 1$ in order for this to work.

Let $f:\mathbb Z\to \mathbb Z$ be a homomorphism. By definition, for each $m\in \mathbb Z$, $$f(m)=f(1+\cdots+1)=f(1)+\cdots+f(1)=m f(1)=f(1)m$$ Hence any homomorphism $f:\mathbb Z\to \mathbb Z$ is just multiplication by $f(1)$ that is the map $m\mapsto f(1) m$. Now it is easy to see that every such non zero homomorphism is injecitve. Indeed the equation $f(1)m=f(1)n$ implies that $m=n$ since $f(1)\not =0$ by hypothesis and also because $\mathbb Z$ is an integral domain. Now the only cases where these homomorphisms are surjective correspond to when $f(1)=1$ or $f(1)=-1$ hence if we call the case $f(1)=1$ the trivial case then the only remaining case is when $f(1)=-1$ and the corresponding automorphism $m\mapsto -m$ is called the negation.