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Math Help - automorphism and dihedral group

  1. #1
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    automorphism and dihedral group

    1) Descibe explicitly all homomorphisms
    h: C_6 ---> Aut(C_12)

    C_6 ={1,x,....,x^5} x^6 = 1

    We know that Aut(C_12) = {q_1,q_5,q_7,q_11} =~ C_2*C_2

    where * = "cross" is the direct product

    Then, h: C_6 ---> C_2*C_2

    So, I found that the possibilities for h(x) are:

    h_1(x) = 1
    h_2(x) = q_5
    h_3(x) = q_7
    h_4(x) = q_11

    Is that the right answer ? If wrong, please correct it

    2)Describe explicitly all homomorphisms
    h: C_5 ---> Aut(C_11)

    Since 11 is prime number, so Aut(C_11) =~ C_10

    I worked out that there are 5 homomorphism for C_5 ---> Aut(C_11). Is that true?

    3)This question I dont know how to do it.

    Show that Aut(C_2*C_2) =~ D_6 ( from cyclic group to dihedral group)

    Thank you in advanced for your time
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  2. #2
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    Quote Originally Posted by knguyen2005 View Post

    1) Descibe explicitly all homomorphisms
    h: C_6 ---> Aut(C_12)

    C_6 ={1,x,....,x^5} x^6 = 1

    We know that Aut(C_12) = {q_1,q_5,q_7,q_11} =~ C_2*C_2

    where * = "cross" is the direct product

    Then, h: C_6 ---> C_2*C_2

    So, I found that the possibilities for h(x) are:

    h_1(x) = 1
    h_2(x) = q_5
    h_3(x) = q_7
    h_4(x) = q_11

    Is that the right answer ? If wrong, please correct it
    i don't know what you mean by q_5, q_7, q_{11}, but yes, there are exactly 4 homomorphisms, because a homomorphism can send x, the generator of C_6, to any of 4 elements of C_2 \times C_2.


    2)Describe explicitly all homomorphisms
    h: C_5 ---> Aut(C_11)

    Since 11 is prime number, so Aut(C_11) =~ C_10

    I worked out that there are 5 homomorphism for C_5 ---> Aut(C_11). Is that true?
    correct!



    3)This question I dont know how to do it.

    Show that Aut(C_2*C_2) =~ D_6 ( from cyclic group to dihedral group)

    Thank you in advanced for your time
    let G=C_2 \times C_2=\{1,x,y,xy\}, where x^2=y^2=1, \ xy=yx. let f \in \text{Aut}(G). then f(1)=1, \ f(xy)=f(x)f(y). now f(x) can be defined to be any element of the set G-\{1\}.

    since f is one-to-one, f(y) can be defined to be any element of the set G-\{1,f(x)\}. so there are 3 possibilities for f(x) and 2 possibilities for f(y). therefore |\text{Aut}(G)|=6.

    define the automorphisms f,g by: f(1)=1, \ f(x)=y, \ f(y)=x, \ f(xy)=xy, and g(1)=1, \ g(x)=xy, \ g(y)=x, \ g(xy)=y. then f \neq fg and o(f)=o(fg)=2. hence \text{Aut}(G)

    is not cyclic. now use this fact that if p is a prime number, then any group of order 2p is either cyclic or it's isomorphisc to D_{2p}.

    alternatively, see that o(g)=3 and D_6 \simeq H=\{\text{id}_G, f, g, g^2, fg, fg^2 \} \subseteq \text{Aut}(G). [you need to check that all elements of H are distinct.] but we have |\text{Aut}(G)|=6. so: \text{Aut}(G)=H.
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  3. #3
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    Question 1, cuz I was in rush, so I just used the short term

    thus, Aut(C_12) = {q_1,q_5,q_7,q_11} =~ C_2*C_2
    = { 1, a, b, ab} where a,b are generators of C_12
    and a^2 = b^2 = 1
    ab = ba
    So, the homomorphism are:
    h_1(x) = 1
    h_2(x) = a
    h_3(x) = b
    h_4(x) = ab

    Does that make sense now?

    Question 3. Firstly, I'd like to thank NonCommAlg , the work is very good but there are some points I dont understand and I hope to get the explanation.

    a) let where
    I was taught that Aut (C_n) is Abelian but in general Aut(G) is non-abelian , e.g. Aut(C2*C2) =~D_6 . So, if you let G = C2*C2 then it is abelian group because xy=yx (commutative) . I'm confused

    b) can be defined to be any element of the set so there are 3 possibilities for and 2 possibilities for therefore
    When you said How do you work out the order of this group is 6 ( Id is also included, right).

    c) then and hence is not cyclic
    How do you know Aut(G) is not cyclic group and why do you have to use 2 automorphisms f and g. Is that because you want to find the generators of D_6.

    d) why is ? and How do you come up with this?

    I'm sorry because I got too many questions, I just want to make it clear.

    Thank you so much

    KN
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  4. #4
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    ok, i'll answer some of your questions. you need to find the answer to the rest yourself by reading my solution more carefully!


    Quote Originally Posted by knguyen2005 View Post
    Question 1, cuz I was in rush, so I just used the short term

    thus, Aut(C_12) = {q_1,q_5,q_7,q_11} =~ C_2*C_2
    = { 1, a, b, ab} where a,b are generators of C_12
    and a^2 = b^2 = 1
    ab = ba
    So, the homomorphism are:
    h_1(x) = 1
    h_2(x) = a
    h_3(x) = b
    h_4(x) = ab

    Does that make sense now?
    yes, it's correct! just note that a,b are the generators of C_2 \times C_2 not C_{12}.


    a) let where
    I was taught that Aut (C_n) is Abelian but in general Aut(G) is non-abelian , e.g. Aut(C2*C2) =~D_6 . So, if you let G = C2*C2 then it is abelian group because xy=yx (commutative) . I'm confused
    the automorphism group of an abelian group might be abelian or non-abelian. the purpose of this exercise is to show you that there are abelian groups G such that Aut(G) is non-abelian.


    b) can be defined to be any element of the set so there are 3 possibilities for and 2 possibilities for therefore
    When you said How do you work out the order of this group is 6 ( Id is also included, right).
    an automorphism f is completely determined if we know f(x) and f(y), because f(1) has to be 1 and if we know f(x),f(y), then we'll also know f(xy)=f(x)f(y). now f(x) has 3 possible values

    and f(y) has two possible values. in order to define an automorphism, we have to choose f(x) from one of those 3 possibilities and f(y) from those two possibile values. obviously that can happen

    in 6 ways.


    c) then and hence is not cyclic

    How do you know Aut(G) is not cyclic group and why do you have to use 2 automorphisms f and g. Is that because you want to find the generators of D_6.
    because a cyclic group of order 6 has exactly one element of order 2.
    Last edited by NonCommAlg; February 1st 2009 at 02:33 PM.
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