# Elements of Aut(Z/4Z)

## Homework Statement

Find, with justification, all the elements of $\operatorname{Aut} (\mathbb{Z}/4\mathbb{Z})$. Which ones are in $\operatorname{Inn} (\mathbb{Z} / 4\mathbb{Z})$?

## The Attempt at a Solution

First we note that $\operatorname{id}_{\mathbb{Z} / 4\mathbb{Z}} \in \operatorname{Aut} (\mathbb{Z}/4\mathbb{Z})$ and that it is the only element in $\operatorname{Inn} (\mathbb{Z} / 4\mathbb{Z})$. This is because any inner automorphism will have the form $c_g (x) = g+ x - g$. But since $\mathbb{Z} / 4\mathbb{Z}$ is abelian the g's cancel out and we're left with the identity.

Now, we claim that there is only one other automorphism. This comes from the the fact that $|0| = 1,~|1|=4,~|2| = 2,~|3|=4$. Since order must be preserved by a homomorphism, $3$ must either mapped to $1$ or itself. Likewise, $1$ must be mapped to itself or $3$. $0$ must be mapped to itself (as it is a property of homomorphisms). $2$ must also be mapped to itself because it is the only element of order 2. This leaves us with the identity map and the map where $f(0)=0,~f(1)=3,~f(2)=2,~f(3)=1$ as possibilities for automorphisms. This map is certainly bijective, so we must only check that it is a homomorphism. Since $\mathbb{Z} / 4\mathbb{Z}$ is abelian we only have to check three sums: $f(1+2) = f(3) = 1~|~1=3+2=f(1)+f(2)$
$f(1+3) = f(0) = 0~|~0=4=3+1=f(1)+f(3)$
$f(2+3) = f(5)= f(1) = 3~|~3=2+1=f(2)+f(3)$. Hence, the map $f$ is an automorphism.

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fresh_42
Mentor

## Homework Statement

Find, with justification, all the elements of $\operatorname{Aut} (\mathbb{Z}/4\mathbb{Z})$. Which ones are in $\operatorname{Inn} (\mathbb{Z} / 4\mathbb{Z})$?

## The Attempt at a Solution

First we note that $\operatorname{id}_{\mathbb{Z} / 4\mathbb{Z}} \in \operatorname{Aut} (\mathbb{Z}/4\mathbb{Z})$ and that it is the only element in $\operatorname{Inn} (\mathbb{Z} / 4\mathbb{Z})$. This is because any inner automorphism will have the form $c_g (x) = g+ x - g$. But since $\mathbb{Z} / 4\mathbb{Z}$ is abelian the g's cancel out and we're left with the identity.

Now, we claim that there is only one other automorphism. This comes from the the fact that $|0| = 1,~|1|=4,~|2| = 2,~|3|=4$. Since order must be preserved by a homomorphism, $3$ must either mapped to $1$ or itself. Likewise, $1$ must be mapped to itself or $3$. $0$ must be mapped to itself (as it is a property of homomorphisms). $2$ must also be mapped to itself because it is the only element of order 2. This leaves us with the identity map and the map where $f(0)=0,~f(1)=3,~f(2)=2,~f(3)=1$ as possibilities for automorphisms. This map is certainly bijective, so we must only check that it is a homomorphism. Since $\mathbb{Z} / 4\mathbb{Z}$ is abelian we only have to check three sums: $f(1+2) = f(3) = 1~|~1=3+2=f(1)+f(2)$
$f(1+3) = f(0) = 0~|~0=4=3+1=f(1)+f(3)$
$f(2+3) = f(5)= f(1) = 3~|~3=2+1=f(2)+f(3)$. Hence, the map $f$ is an automorphism.
Correct. Here are my thoughts:

I find it easier to use the multiplicative notation $\mathbb{Z}_4=\{\,1,g,g^2,g^3\,\}$ but that doesn't matter. Now you have only to check what $f(g)$ could be. With $f\neq 1$ we get $f(g)\in \{\,g^2,g^3\,\}$ and as you said, $f(g)=g^2$ can be ruled out. Either by your argument or simply by $f(g^2)=g^4=1$ and a non trivial kernel. So $f(g)=g^3$ is left. To see it's a homomrphism $f(g^n\cdot g^m)=g^{3n+3m}=f(g)^n\cdot f(g)^m$ is easy.
Injectivity is sufficient and $g^{3n}=1$ implies $4\,|\,3n$ and so $4\,|\,n$, i.e. $g^n=1$.

Correct. Here are my thoughts:

I find it easier to use the multiplicative notation $\mathbb{Z}_4=\{\,1,g,g^2,g^3\,\}$ but that doesn't matter. Now you have only to check what $f(g)$ could be. With $f\neq 1$ we get $f(g)\in \{\,g^2,g^3\,\}$ and as you said, $f(g)=g^2$ can be ruled out. Either by your argument or simply by $f(g^2)=g^4=1$ and a non trivial kernel. So $f(g)=g^3$ is left. To see it's a homomrphism $f(g^n\cdot g^m)=g^{3n+3m}=f(g)^n\cdot f(g)^m$ is easy.
Injectivity is sufficient and $g^{3n}=1$ implies $4\,|\,3n$ and so $4\,|\,n$, i.e. $g^n=1$.
A few questions. Why do you only have to check what $f(g)$ could be? Also, what are you trying to show in the last line of your response on injectivity?

How would I approach this problem in general? For example, how could I characterize the automorphisms of $\mathbb{Z}/8 \mathbb{Z}$? Or $\mathbb{Z}/n \mathbb{Z}$? Is the answer very much related to generators and mapping to other generators?

fresh_42
Mentor
A few questions. Why do you only have to check what $f(g)$ could be?
As $g$ generates the cyclic group all other values are given by $f(g^n)=f(g)^n$.
Also, what are you trying to show in the last line of your response on injectivity?
$g^n$ is an arbitrary element. If we assume $g^n \in \operatorname{ker}(f)$ we get $f(g^n)=g^{3n}=1$ and from that $g^n=1$. Since the kernel of a homomorphism is trivial, the homomorphism is injective. Since it is a map between sets of equal finite order (4), it has to be surjective as well.
How would I approach this problem in general?
What you did was absolute o.k. I just wanted to add a different notation. This was an easy group. To determine an Automorphism group in general could be rather difficult and there is probably no golden way. At least as long as we will not use heavy machinery. It also depends on how the group is given. E.g. a presentation $G= \langle r,s\,|\,r^2=s^n\, , \,srsr \rangle$ can be difficult. (IIRC then even the word problem in an arbitrary group is NP complete.)
For example, how could I characterize the automorphisms of $\mathbb{Z}/8 \mathbb{Z}$? Or $\mathbb{Z}/n \mathbb{Z}$?
Similar to what you did. Abelian groups have no inner automorphisms, so all are outer. Then cyclic groups are generated by a single element, which means $G=\langle a\,|\,a^n\rangle$. If you look closer into this situation, you will find out, which powers of $a$ are allowed to be $f(a)=a^k$. Hint: It has to do with whether $k$ and $n$ are coprime or not. With this result, your question above is easy: $f(g)=g^k$ and $\operatorname{gcd}(k,4)=1$ can only be, if $k=1$ (identity) or $k=3$ (your automorphism $f$). This would have been really short. I suggest to prove this as an exercise. You already did it for $n=4$ and in general the arguments are the same (no induction!).
Is the answer very much related to generators and mapping to other generators?
In this case (cyclic), yes. I explained above why it is sufficient to define $f(a)$, and you can think about, why the image of $a$ has to generate the group as well, and why all coprime powers are generators.

As $g$ generates the cyclic group all other values are given by $f(g^n)=f(g)^n$.$g^n$ is an arbitrary element. If we assume $g^n \in \operatorname{ker}(f)$ we get $f(g^n)=g^{3n}=1$ and from that $g^n=1$. Since the kernel of a homomorphism is trivial, the homomorphism is injective. Since it is a map between sets of equal finite order (4), it has to be surjective as well.
What you did was absolute o.k. I just wanted to add a different notation. This was an easy group. To determine an Automorphism group in general could be rather difficult and there is probably no golden way. At least as long as we will not use heavy machinery. It also depends on how the group is given. E.g. a presentation $G= \langle r,s\,|\,r^2=s^n\, , \,srsr \rangle$ can be difficult. (IIRC then even the word problem in an arbitrary group is NP complete.)

Similar to what you did. Abelian groups have no inner automorphisms, so all are outer. Then cyclic groups are generated by a single element, which means $G=\langle a\,|\,a^n\rangle$. If you look closer into this situation, you will find out, which powers of $a$ are allowed to be $f(a)=a^k$. Hint: It has to do with whether $k$ and $n$ are coprime or not. With this result, your question above is easy: $f(g)=g^k$ and $\operatorname{gcd}(k,4)=1$ can only be, if $k=1$ (identity) or $k=3$ (your automorphism $f$). This would have been really short. I suggest to prove this as an exercise. You already did it for $n=4$ and in general the arguments are the same (no induction!).

In this case (cyclic), yes. I explained above why it is sufficient to define $f(a)$, and you can think about, why the image of $a$ has to generate the group as well, and why all coprime powers are generators.
So here is my argument for characterizing $\operatorname{Aut} (\mathbb{Z}_n)$.

Since $\mathbb{Z}_n = \langle~ g|~g^n=1 \rangle$, each map from $\mathbb{Z}_n$ to $\mathbb{Z}_n$ will be uniquely determined by where it takes $g$. Since it is necessary that automorphisms preserve order, we suspect that all of the automorphisms will be of the form $f(g)=g^k$ where $\operatorname{gcd} (k,n)=1$. First, we show that this is a homomorphism: $f(g^pg^q) = f(g^{p+q}) = (g^{p+q})^{k} = (g^p)^k(g^q)^k = f(g^p)f(g^q)$. Next, we show that $f$ is a bijection. The kernel of $f$ is trivial because if $f(g^p)=1$ then $g^{kp} =1$, and so $n~|~kp$, but $\operatorname{gcd} (k,n)=1$, so $n~|~p$. But $p \in \{0,1, \dots, n-1\}$, so it must be the case that $p=0$, and so the kernel is trivial. Hence $f$ is injective, and since this is a map between two sets of equal finite order, it is also surjective. So $f$ is an automorphism and characterizes every automorphism.

fresh_42
Mentor
So here is my argument for characterizing $\operatorname{Aut} (\mathbb{Z}_n)$.

Since $\mathbb{Z}_n = \langle~ g|~g^n=1 \rangle$, each map from $\mathbb{Z}_n$ to $\mathbb{Z}_n$ will be uniquely determined by where it takes $g$. Since it is necessary that automorphisms preserve order, we suspect that all of the automorphisms will be of the form $f(g)=g^k$ where $\operatorname{gcd} (k,n)=1$. First, we show that this is a homomorphism: $f(g^pg^q) = f(g^{p+q}) = (g^{p+q})^{k} = (g^p)^k(g^q)^k = f(g^p)f(g^q)$. Next, we show that $f$ is a bijection. The kernel of $f$ is trivial because if $f(g^p)=1$ then $g^{kp} =1$, and so $n~|~kp$, but $\operatorname{gcd} (k,n)=1$, so $n~|~p$. But $p \in \{0,1, \dots, n-1\}$, so it must be the case that $p=0$, and so the kernel is trivial. Hence $f$ is injective, and since this is a map between two sets of equal finite order, it is also surjective. So $f$ is an automorphism and characterizes every automorphism.
Yes, correct. Maybe the step for $n\,|\,p$ could be more detailed, but that's a matter of taste. We basically use that $\operatorname{gcd}(k,n)=1=a\cdot k + b\cdot n$, multiply by $p$ and take the module by $n$.

What you have shown now is, that $f(g)=g^k$ is an automorphism if $\operatorname{gcd}(k,n)=1$. What's left to show is, why this is also a necessary condition. Why is $f$ no automorphism if $\operatorname{gcd}(k,n) > 1$.

Yes, correct. Maybe the step for $n\,|\,p$ could be more detailed, but that's a matter of taste. We basically use that $\operatorname{gcd}(k,n)=1=a\cdot k + b\cdot n$, multiply by $p$ and take the module by $n$.

What you have shown now is, that $f(g)=g^k$ is an automorphism if $\operatorname{gcd}(k,n)=1$. What's left to show is, why this is also a necessary condition. Why is $f$ no automorphism if $\operatorname{gcd}(k,n) > 1$.
Didn't I show the necessary condition (maybe somewhat implicitly) by using the fact that any isomorphism must preserve order? Doesn't this necessitate that $\operatorname{gcd}(k,n)=1$?

fresh_42
Mentor
Didn't I show the necessary condition (maybe somewhat implicitly) by using the fact that any isomorphism must preserve order? Doesn't this necessitate that $\operatorname{gcd}(k,n)=1$?
In a way, yes, but this is very implicit in my opinion. Why do elements for which $k$ and $n$ share a common factor, say $k=r\cdot s$ and $n=r\cdot t$, do not generate the group? However, you are right, this is a matter of how detailed a proof should be.

In a way, yes, but this is very implicit in my opinion. Why do elements for which $k$ and $n$ share a common factor, say $k=r\cdot s$ and $n=r\cdot t$, do not generate the group? However, you are right, this is a matter of how detailed a proof should be.
Don't they not generate the group because they have a different order than the generator $g$, since $|g^k| = \frac{n}{\operatorname{gcd} (k,n)}$? Order is the same iff $\operatorname{gcd} (k,n) = 1$

fresh_42
Mentor
Yes, in case you have this formula, which I think is not so obvious that it doesn't need proof or at least mention. However, I repeat myself, it's a matter of taste, resp. audience.