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Homomorphisms with unknown groups

  1. Feb 19, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    1)Let p,q be primes. Show that the only group homomorphism $$\phi: C_p \mapsto C_q$$ is the trivial one (i.e ## \phi (g) = e = e_H\,\forall\,g##)

    2)Consider the function $$det: GL(n,k) \mapsto k^*.$$ Show that it is a group homomorphism and identify the kernel and image.


    3. The attempt at a solution

    I cannot make much progress with either of these questions essentially because I do not know what the groups are. I don't know what ##C_p## or ##C_q## are. Any ideas?

    For the second one, I can at least make a start:
    Take A,B in GL(n,k). GL(n,k) is a group under multiplication of matrices, so det(g1 * g2) = det(AB) = (detA)(detB) from linear algebra = det(g1)* det(g2) since det(gi) ##\neq## 0 for all gi. Again, what is ##k^*##? From the homomorphism I just showed, it is probably a group under multiplication of numbers and GL(n,k) = set of invertible nxn matrices with real number entries, but I want to make sure.
    So,
    Ker (det) = {##g \in GL(n,k) \mid det(g) = e_H = 1##} = SL(n,k)
    Im (det) = {##h \in k^* \mid \exists g \in GL(n,k) , det(g) = h##} = ##k^* ##\## {0}##. I don't know what ##k^*## is so whether I need to exclude 0 or not, I don't know.

    Many thanks.
     
  2. jcsd
  3. Feb 19, 2013 #2

    jbunniii

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    You should check to make sure, but I assume these are cyclic groups of order ##p## and ##q##. Suppose you did have a homomorphism ##\phi : C_p \rightarrow C_q##. Think about what the kernel and image of ##\phi## can be.
     
  4. Feb 19, 2013 #3

    Zondrina

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    Hmm for the first one, I THINK they might be modulo groups. Contradiction is a good approach to it. That is, assume that phi is not the only group homomorphism and then proceed to show that phi is indeed the only one.
     
  5. Feb 19, 2013 #4

    jbunniii

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    ##k^*## means the multiplicative group of nonzero elements of the field ##k##. GL(n,k) refers to the multiplicative group of INVERTIBLE nxn matrices over the field k, i.e. the matrices with nonzero determinant. It would not be a group if you included matrices with determinant zero, because these don't have inverses.

    Your kernel looks fine, and so does your image. (Note that ##k^* \setminus \{0\} = k^*##.) You may want to give an specific example to show that for any ##h \in k^*##, you can find a matrix ##A## with ##det(A) = h##.
     
  6. Feb 19, 2013 #5

    CAF123

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    So assuming they are cyclic groups,I have ##Ker \phi = \left\{g^{nq}\right\}, n \in \mathbb{N}, q## the order of ##C_q## and ## Im \phi = \left\{e_H\right\}## because the element in Ker phi will give the identity. I don't know if I can say for sure if there exists more elements to Im phi. Does this help me?
     
  7. Feb 19, 2013 #6

    jbunniii

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    I'm not sure how you got this.

    Instead of trying to construct the kernel explicitly, consider what its order can be. (Use Lagrange's theorem.) Do the same thing for the image. What are the possibilities?
     
  8. Feb 19, 2013 #7

    CAF123

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    I thought I could perhaps construct an element in ##C_p## such that when it is raised to some multiple of the order of q, I would get the identity in ##C_q##.

    Ker ##\phi \leq C_p\,\Rightarrow\, |C_p| = p = a |Ker \phi|\, \Rightarrow\, |Ker \phi| = p/a## so a = 1 or p meaning it's order must be 1 or p, since p is prime. Similarly, for ##Im \phi##, we have its order 1 or q.
     
  9. Feb 19, 2013 #8

    jbunniii

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    OK, so far so good.

    If ##|ker \phi| = p##, then what is ##ker \phi##, and what is ##\phi##?
     
  10. Feb 19, 2013 #9

    CAF123

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    Does this mean Ker ##\phi## = ##C_p## so phi sends all elements of ##C_p## to the identity in ##C_q##? So I am done if I can show that it's order can't be 1. I know that it is at least 1 since the identity in ##C_p## always maps to the identity in ##C_q##. Can I argue that another element mapping to ##e_H## is a element of the form ##g^{nq}##? I am not sure, because this element may not exist.
     
    Last edited: Feb 19, 2013
  11. Feb 19, 2013 #10

    jbunniii

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    Right, so ##\phi## is the trivial map, as indicated in the problem statement.

    So now all we have to do is rule out the other possibility, that ##|ker \phi| = 1##. If this were true, what does it imply about the map ##\phi##? Injective, surjective, something like that?
     
  12. Feb 19, 2013 #11

    CAF123

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    |Ker ##\phi##| = 1 would mean that only the identity in ##C_p## maps to the identity in ##C_q##. I have proven this in another exercise that we have injectivity. I am not sure about surjectivity since all other elements map elsewhere, but we can't say for sure if they cover all of H.

    But if we have injectivity, then order of g in G = order of ##\phi(g)## in H. By assumption, p and q were different and so this is a contradiction. Hence, |Ker \phi| = 1 is ruled out.
    EDIT: Question should say 'p,q are different primes'
     
    Last edited: Feb 19, 2013
  13. Feb 19, 2013 #12

    jbunniii

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    Right, ##\phi## must be injective. This in turn implies that ##|im(\phi)|## must equal ##|C_p| = p##. This gives you the contradiction you need, assuming, as you pointed out, ##p \neq q##.
    Yes, good catch. Otherwise the identity map is a counterexample.
     
  14. Feb 19, 2013 #13

    micromass

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    It would be an interesting exercise to see how many homomorphisms [itex]\varphi:C_p\rightarrow C_p[/itex] there are.

    Since they are all bijections except for the trivial map, we can look at the set of all bijective homomorphisms. This forms a group. It would be a good exercise to find out which group it is (answer: it will be the cyclic group on p-1 elements).
     
  15. Feb 19, 2013 #14

    CAF123

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    Is the way I gave okay as well? (about the order of g in H). Also, would I be correct in saying that we can't deduce whether phi is surjective?
     
  16. Feb 19, 2013 #15

    jbunniii

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    I think it's OK. The key is that if ##\langle g \rangle## denotes the subgroup generated by ##g##, then ##\phi(\langle g \rangle)## is a subgroup of ##C_q##, so its order must divide ##q##. But ##\phi## is an injection, so if ##g \neq e## we must have ##|\phi(\langle g \rangle)| = p##, and ##p## does not divide ##q##.
    Well, we can deduce it after the fact: the only homomorphism is the trivial map, and this is not surjective.

    If we wanted to prove it directly, we could suppose that ##\phi## is a surjective homomorphism. Then ##im(\phi) = q##. But by the first isomorphism theorem, we have ##C_p / ker(\phi) \cong im(\phi)##, so ##|im(\phi)|## must divide ##p##. This is a contradiction.
     
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