# Elements of Aut(Z/4Z) and their Inclusion in Inn(Z/4Z)

• Mr Davis 97
In summary, we have found that the only elements in ##\operatorname{Aut} (\mathbb{Z}/4\mathbb{Z})## are the identity and the map where ##f(0)=0,~f(1)=3,~f(2)=2,~f(3)=1##. 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. We can also extend this approach to finding the automorphisms of other cyclic groups, such as ##
Mr Davis 97

## 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.

Mr Davis 97 said:

## 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##.

Mr Davis 97
fresh_42 said:
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?

Mr Davis 97 said:
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.

Mr Davis 97
fresh_42 said:
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.

Mr Davis 97 said:
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##.

fresh_42 said:
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##?

Mr Davis 97 said:
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.

fresh_42 said:
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##

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.

Mr Davis 97

## 1. What is Aut(Z/4Z)?

Aut(Z/4Z) refers to the automorphism group of the integers modulo 4. This group is composed of all the bijective homomorphisms from Z/4Z to itself, which preserve the group structure.

## 2. How many elements are in Aut(Z/4Z)?

There are 2 elements in Aut(Z/4Z). These are the identity function and the map that swaps 1 and 3, and 2 and 0.

## 3. What is an automorphism group?

An automorphism group is a mathematical concept that describes a set of symmetries or transformations that preserve the structure of a mathematical object, such as a group or a ring.

## 4. How is Aut(Z/4Z) related to other automorphism groups?

Aut(Z/4Z) is a subgroup of Aut(Z), the automorphism group of the integers. This means that every automorphism of Z/4Z is also an automorphism of Z. Additionally, Aut(Z/4Z) is isomorphic to Aut(Z/2Z) as both groups have 2 elements.

## 5. What are some applications of Aut(Z/4Z)?

Aut(Z/4Z) has various applications in number theory, abstract algebra, and cryptography. It can be used to study the structure of finite groups, solve algebraic equations, and design secure encryption algorithms.

• Calculus and Beyond Homework Help
Replies
5
Views
1K
• Calculus and Beyond Homework Help
Replies
2
Views
428
• Calculus and Beyond Homework Help
Replies
16
Views
1K
• Calculus and Beyond Homework Help
Replies
3
Views
546
• Calculus and Beyond Homework Help
Replies
1
Views
525
• Calculus and Beyond Homework Help
Replies
3
Views
633
• Calculus and Beyond Homework Help
Replies
1
Views
636
• Calculus and Beyond Homework Help
Replies
1
Views
636
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
1
Views
593