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Direct Products and Automorphisms

  1. Nov 23, 2012 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    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 [itex]\phi : G → G \space | \space \phi(g) = g^k[/itex] is injective. If G is also abelian, show that the map [itex]\phi[/itex] is also an automorphism of G.

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

    2. Relevant equations

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

    3. 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 [itex]\phi(r) = \phi(s)[/itex] and we want to show r = s for any r and s in G.

    [itex]\phi(r) = \phi(s)[/itex]
    [itex]r^k = s^k[/itex]

    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.
     
  2. jcsd
  3. Nov 23, 2012 #2

    Dick

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    Use your observation that 1 = an + bk. So r^1=r^(an+bk). Where does that take you?
     
  4. Nov 23, 2012 #3

    Zondrina

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    Hmmm :

    1 = an + bk
    r1 = ran+bk
    r = ranrbk
    r = (rn)a(rk)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.
     
  5. Nov 23, 2012 #4

    Dick

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    You should know what r^n is. Use that to simplify it.
     
  6. Nov 23, 2012 #5

    Zondrina

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    Oh, |G| = n, so rn = e.

    So we would get :

    r = ea(rk)b
    r = (rk)b

    Similarly : s = (sk)b

    So we would get :

    rk = sk
    (rk)b = (sk)b
     
  7. Nov 23, 2012 #6

    Dick

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    So I assume you've got the proof?
     
  8. Nov 23, 2012 #7

    Zondrina

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    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)
    rk = sk
    (rbk)k = (sbk)k

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

    (r1-an)k = (s1-an)k

    Hmm for some reason I'm not seeing this one. I know every element in G must have the form ak, but the simplification here isn't very straightforward for me.
     
  9. Nov 23, 2012 #8

    Dick

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    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.
     
    Last edited: Nov 23, 2012
  10. Nov 23, 2012 #9

    Zondrina

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    Wow now I feel silly. Thank you, so :


    phi(r) = phi(s)
    rk = sk

    phi(r) = phi(s)
    (rk)b = (sk)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 = sb so we get :

    phi(r) = phi(sb) = (sb)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 = rksk = 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.
     
  11. Nov 23, 2012 #10

    Dick

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    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.
     
  12. Nov 23, 2012 #11

    Zondrina

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    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?
     
  13. Nov 23, 2012 #12

    Dick

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    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?
     
  14. Nov 23, 2012 #13

    Zondrina

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    I believe it was this step here :

    phi(rs) = (rs)k = (Right here ) rksk = phi(r)phi(s)
     
  15. Nov 23, 2012 #14

    Dick

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    That's the one alright.
     
  16. Nov 23, 2012 #15

    Zondrina

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    Sweet.

    Now for the second one. First we note :

    Z3 [itex]\oplus[/itex] 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 [itex]\oplus[/itex] Z5 and 2 is in Z15.

    So we seek some (a,b) in Z3 [itex]\oplus[/itex] 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 [itex]\oplus[/itex] 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.
     
    Last edited: Nov 23, 2012
  17. Nov 23, 2012 #16

    Zondrina

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    Is that good ^?
     
  18. Nov 23, 2012 #17

    Dick

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    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.
     
    Last edited: Nov 23, 2012
  19. Nov 23, 2012 #18

    Zondrina

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    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.
     
  20. Nov 23, 2012 #19

    Dick

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    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.
     
  21. Nov 23, 2012 #20

    Zondrina

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    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.
     
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