Direct Products and Automorphisms

[STEMucator]

Homework Statement

Doing some suggested exercises out of my textbook today, there were two that I had trouble with.

1. Let G be a group and |G| = n. Suppose k is an integer relatively prime to n. Show that the mapping \phi : G → G \space | \space \phi(g) = g^k is injective. If G is also abelian, show that the map \phi is also an automorphism of G.

2. Suppose that \phi : \mathbb{Z_3} \oplus \mathbb{Z_5} → \mathbb{Z_{15}} \space | \space \phi((2,3)) = 2 \space is an isomorphism. Find an element in \mathbb{Z_3} \oplus \mathbb{Z_5} that maps to 1 in \mathbb{Z_{15}}.

Homework Equations

An automorphism is a bijective homomorphism.
An isomorphism is a bijective map.

The Attempt at a Solution

1. Okay, so G is a group and |G| = n. We also know gcd(n,k) = 1 for some integer k which means that gcd(n,k) = 1 = an + bk for some integers a and b. I'm pretty sure this group is under multiplication, but I could be wrong.

- We want to show the map is one to one. So we assume \phi(r) = \phi(s) and we want to show r = s for any r and s in G.

\phi(r) = \phi(s)
r^k = s^k

Now here is where I get stuck, I'm pretty sure I have to use the fact that n and k are relatively prime here.

Ill save question 2 for after this one is finished. Any help is appreciated.

 

[Dick]

Homework Statement

Doing some suggested exercises out of my textbook today, there were two that I had trouble with.

1. Let G be a group and |G| = n. Suppose k is an integer relatively prime to n. Show that the mapping \phi : G → G \space | \space \phi(g) = g^k is injective. If G is also abelian, show that the map \phi is also an automorphism of G.

2. Suppose that \phi : \mathbb{Z_3} \oplus \mathbb{Z_5} → \mathbb{Z_{15}} \space | \space \phi((2,3)) = 2 \space is an isomorphism. Find an element in \mathbb{Z_3} \oplus \mathbb{Z_5} that maps to 1 in \mathbb{Z_{15}}.

Homework Equations

An automorphism is a bijective homomorphism.
An isomorphism is a bijective map.

The Attempt at a Solution

1. Okay, so G is a group and |G| = n. We also know gcd(n,k) = 1 for some integer k which means that gcd(n,k) = 1 = an + bk for some integers a and b. I'm pretty sure this group is under multiplication, but I could be wrong.

- We want to show the map is one to one. So we assume \phi(r) = \phi(s) and we want to show r = s for any r and s in G.

\phi(r) = \phi(s)
r^k = s^k

Now here is where I get stuck, I'm pretty sure I have to use the fact that n and k are relatively prime here.

Ill save question 2 for after this one is finished. Any help is appreciated.

Use your observation that 1 = an + bk. So r^1=r^(an+bk). Where does that take you?

 

[STEMucator]

Use your observation that 1 = an + bk. So r^1=r^(an+bk). Where does that take you?

Hmmm :

1 = an + bk
r¹ = r^an+bk
r = r^anr^bk
r = (rⁿ)^a(r^k)^b

I'm not sure how this is of assistance? Would I raise both sides to the k now?

A similar case can be observed for s, so ill work on r.

 

[Dick]

Hmmm :

1 = an + bk
r¹ = r^an+bk
r = r^anr^bk
r = (rⁿ)^a(r^k)^b

I'm not sure how this is of assistance? Would I raise both sides to the k now?

A similar case can be observed for s, so ill work on r.

You should know what r^n is. Use that to simplify it.

 

[STEMucator]

You should know what r^n is. Use that to simplify it.

Oh, |G| = n, so rⁿ = e.

So we would get :

r = e^a(r^k)^b
r = (r^k)^b

Similarly : s = (s^k)^b

So we would get :

r^k = s^k
(r^k)^b = (s^k)^b

 

[Dick]

Oh, |G| = n, so rⁿ = e.

So we would get :

r = e^a(r^k)^b
r = (r^k)^b

Similarly : s = (s^k)^b

So we would get :

r^k = s^k
(r^k)^b = (s^k)^b

So I assume you've got the proof?

 

[STEMucator]

So I assume you've got it?

That left a dirty aftertaste for some reason, I thought I may have been wrong. I also made a small error in my last post which I'll try to fix here.

So :

phi(r) = phi(s)
r^k = s^k
(r^bk)^k = (s^bk)^k

Now since 1 = an + bk we have that 1 - an = bk thus :

(r^1-an)^k = (s^1-an)^k

Hmm for some reason I'm not seeing this one. I know every element in G must have the form a^k, but the simplification here isn't very straightforward for me.

 

[Dick]

That left a dirty aftertaste for some reason, I thought I may have been wrong. I also made a small error in my last post which I'll try to fix here.

So :

phi(r) = phi(s)
r^k = s^k
(r^bk)^k = (s^bk)^k

Now since 1 = an + bk we have that 1 - an = bk thus :

(r^1-an)^k = (s^1-an)^k

Hmm for some reason I'm not seeing this one. I know every element in G must have the form a^k, but the simplification here isn't very straightforward for me.

You are overshooting. And there is no error in your last post. You know (r^k)^b=(s^k)^b and you've shown r=(r^k)^b and s=(s^k)^b. There's no simplification needed. Just state the conclusion.

 

[STEMucator]

You are overshooting. You know (r^k)^b=(s^k)^b and you've shown r=(r^k)^b and s=(s^k)^b. There's no simplification needed. Just state the conclusion.

Wow now I feel silly. Thank you, so :phi(r) = phi(s)
r^k = s^k

phi(r) = phi(s)
(r^k)^b = (s^k)^b
r = s

Thus phi is injective.

Now assuming G is abelian, we want to show phi is surjective and a homomorphism which would mean it is an automorphism.

- So we want to show G is onto. That is for s in G we want to find r in G such that phi(r) = s.

Suppose that r = s^b so we get :

phi(r) = phi(s^b) = (s^b)^k = s

Thus phi is onto.

- To show that phi is a homomorphism, we must show phi(rs) = phi(r)phi(s) for any r and s in G.

phi(rs) = (rs)^k = r^ks^k = phi(r)phi(s)

Thus phi is an automorphism of G.

Lookin good?

EDIT : I didn't even have to use the fact G was abelian... still wondering if this is okay.

 

[Dick]

Wow now I feel silly. Thank you, so :phi(r) = phi(s)
r^k = s^k

phi(r) = phi(s)
(r^k)^b = (s^k)^b
r = s

Thus phi is injective.

Now assuming G is abelian, we want to show phi is surjective and a homomorphism which would mean it is an automorphism.

- So we want to show G is onto. That is for s in G we want to find r in G such that phi(r) = s.

Suppose that r = s^b so we get :

phi(r) = phi(s^b) = (s^b)^k = s

Thus phi is onto.

- To show that phi is a homomorphism, we must show phi(rs) = phi(r)phi(s) for any r and s in G.

phi(rs) = (rs)^k = r^ks^k = phi(r)phi(s)

Thus phi is an automorphism of G.

Lookin good?

You still have some issues. Showing it's a homomorphism is a great idea and that part looks good. And if I were grading this I'd like to see the first part written up in a form that looks more like a proof, but you can probably handle that. It's the surjective part that needs work. That doesn't make any sense at all. You can't 'assume' r=s^b, now can you? Look, G is a finite set. Can you have a mapping from G->G that is injective without being surjective? Think about it.

 

[STEMucator]

You still have some issues. Showing it's a homomorphism is a great idea and that part looks good. And if I were grading this I'd like to see the first part written up in a form that looks more like a proof, but you can probably handle that. It's the surjective part that needs work. That doesn't make any sense at all. You can't 'assume' r=s^b, now can you? Look, G is a finite set. Can you have a mapping from G->G that is injective without being surjective? Think about it.

Ahhhh G is a finite group, so any map from a finite group to itself is injective if and only if it is surjective. That would take care of it I believe wouldn't it?

These aren't for grades so I'm being a bit sloppy I'll admit, but it's the concept I'm concerned with.

Actually I can even reference a previous result from another exercise that I've done to save myself the work of proving the if and only if.

Would that be good afterwards though?

 

[Dick]

Ahhhh G is a finite group, so any map from a finite group to itself is injective if and only if it is surjective. That would take care of it I believe wouldn't it?

These aren't for grades so I'm being a bit sloppy I'll admit, but it's the concept I'm concerned with.

Actually I can even reference a previous result from another exercise that I've done to save myself the work of proving the if and only if.

Would that be good afterwards though?

Oh, I think it's generally ok for practice. But with your "EDIT : I didn't even have to use the fact G was abelian... still wondering if this is okay." That would NOT be ok. Yes, you did use that G is abelian. Where?

 

[STEMucator]

Oh, I think it's generally ok for practice. But with your "EDIT : I didn't even have to use the fact G was abelian... still wondering if this is okay." That would NOT be ok. Yes, you did use that G is abelian. Where?

I believe it was this step here :

phi(rs) = (rs)^k = (Right here ) r^ks^k = phi(r)phi(s)

 

[Dick]

I believe it was this step here :

phi(rs) = (rs)^k = (Right here ) r^ks^k = phi(r)phi(s)

That's the one alright.

 

[STEMucator]

That's the one alright.

Sweet.

Now for the second one. First we note :

Z₃ \oplus Z₅ = { (0,0), (0,1), (0,2), (0,3), (0,4), (1,0), (1,1), (1,2), (1,3), (1,4), (2,0), (2,1), (2,2), (2,3), (2,4) }

We also know that phi((2,3)) = 2 where (2,3) is in Z₃ \oplus Z₅ and 2 is in Z₁₅.

So we seek some (a,b) in Z₃ \oplus Z₅ such that phi((a,b)) = 1.

Now, note for the case of (2,3) which get mapped to 2, we have that the order of 2 in Z₁₅ is infinity so it must be the case that the order of (2,3) is also infinity.

Now in Z₁₅, the order of 1 is 15, so we seek (a,b) in Z₃ \oplus Z₅ such that |(a,b)| = 15.

Notice that we want 15 = |(a,b)| = lcm(|a|,|b|). So we want |a| = 3 and |b| = 5 or |a| = 5 and |b| = 3. In Z₃ the order of 1 is 3 and in Z₅ the order of 1 is 5. Thus phi((1,1)) = 1 and we are done.

 

[STEMucator]

Is that good ^?

 

[Dick]

Is that good ^?

No, it's not. How can you even believe it? What do you mean the order of 2 in Z_15 is infinity?? Z_15 is a finite group. How can it have an element of order infinity? This one doesn't even need any deep analysis. phi((2,3))=2. So what's phi((2,3)+(2,3))? phi((2,3)+(2,3)+(2,3))? Keep going.

 

[STEMucator]

No, it's not. How can you even believe it? What do you mean the order of 2 in Z_15 is infinity?? Z_15 is a finite group. How can it have an element of order infinity? This one doesn't even need any deep analysis. phi((2,3))=2. So what's phi((2,3)+(2,3))? phi((2,3)+(2,3)+(2,3))? Keep going.

Ohhh I stopped too early and made an incorrect presumption. The |2| = 15 in Z₁₅, so lcm(|2|,|3|) = lcm(3,5) = 15 for 2 in Z₃ and 3 in Z₅.

Now for what you said :

(2,3)+(2,3) = (1,1)
(2,3)+(2,3)+(2,3) = (0,4)
(2,3)+(2,3)+(2,3)+(2,3) = (2,2)
(2,3)+(2,3)+(2,3)+(2,3)+(2,3) = (1,0)
(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3) = (0,3)

Wouldn't (1,1) work here though regardless?

In Z₃, |2| = 3 and in Z₅, |2| = 5 so (2,2) would also be a candidate.

 

[Dick]

Ohhh I stopped too early and made an incorrect presumption. The |2| = 15 in Z₁₅, so lcm(|2|,|3|) = lcm(3,5) = 15 for 2 in Z₃ and 3 in Z₅.

Now for what you said :

(2,3)+(2,3) = (1,1)
(2,3)+(2,3)+(2,3) = (0,4)
(2,3)+(2,3)+(2,3)+(2,3) = (2,2)
(2,3)+(2,3)+(2,3)+(2,3)+(2,3) = (1,0)
(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3) = (0,3)

Wouldn't (1,1) work here though regardless?

In Z₃, |2| = 3 and in Z₅, |2| = 5 so (2,2) would also be a candidate.

I'm really having a hard time figuring out what is going through your mind. If, as you correctly say, (2,3)+(2,3)=(1,1) then phi((1,1))=phi((2,3)+(2,3))=phi((2,3))+phi((2,3))=2+2=4. 4 isn't equal to 1 mod 15 is it? Why would you think it is? Forget abstract thought for a bit and just do the numbers.

 

[STEMucator]

I'm really having a hard time figuring out what is going through your mind. If, as you correctly say, (2,3)+(2,3)=(1,1) then phi((1,1))=phi((2,3)+(2,3))=phi((2,3))+phi((2,3))=2+2=4. 4 isn't equal to 1 mod 15 is it? Why would you think it is? Forget abstract thought for a bit and just do the numbers.

Wow. Once again I feel silly.

So :

(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3) = (1,4)

phi((1,4)) = phi((2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)) = 2+2+2+2+2+2+2+2 = 16 = 1

Since 16mod15 = 1.

 

[Dick]

Wow. Once again I feel silly.

So :

(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3) = (1,4)

phi((1,4)) = phi((2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)+(2,3)) = 2+2+2+2+2+2+2+2 = 16 = 1

Since 16mod15 = 1.

Right. Now try and go back and figure out what ideas you had that made you think (1,1) would be a correct answer now that you know it's not, and figure out what's wrong with them.

 

[STEMucator]

Right. Now try and go back and figure out what ideas you had that made you think (1,1) would be a correct answer now that you know it's not, and figure out what's wrong with them.

So (1,1) = (2,3) + (2,3), but phi((1,1)) = phi((2,3) + (2,3)) = 2+2 = 4mod15 = 4.

So clearly (1,1) is not what we are looking for.

Once again thanks for your help btw.

 

[Dick]

So (1,1) = (2,3) + (2,3), but phi((1,1)) = phi((2,3) + (2,3)) = 2+2 = 4mod15 = 4.

So clearly (1,1) is not what we are looking for.

Once again thanks for your help btw.

You're welcome. But you should still figure out why you thought (1,1) would be correct. You have some wrong ideas. If you can figure out what they are then maybe you wouldn't have to feel silly again.