Results 1 to 12 of 12

Math Help - Any group of 3 elements is isomorphic to Z3

  1. #1
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318

    Any group of 3 elements is isomorphic to Z3

    Prove that any group with three elements is isomorphic to Z3




    Take a group of three elements {e,a,b}. Since the order of every element must be three, we have that b=aČ. Thus the group is {e,a,aČ}.

    Define the map G --> Z3 by
    e ---> 0
    a ---> 1
    b ---> 2


    We need to show 1-1, onto, and the homomorphism property
    I'm having problems with all of the properties
    Assume c(a)=c(b)
    This implies 1=2, so c(a) is not equal to c(b).
    Does this show 1-1?

    Onto:
    Need to show y=c(x) and solve for x
    y=c(a)=1
    y=1
    I'm confused on this

    Homomorphism:
    c(a)c(b)
    1*2
    2=c(ab)
    I know this should equal c(ab), but I didn't see why 2=c(ab)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kathrynmath View Post
    Prove that any group with three elements is isomorphic to Z3




    Take a group of three elements {e,a,b}. Since the order of every element must be three, we have that b=aČ. Thus the group is {e,a,aČ}.

    Define the map G --> Z3 by
    e ---> 0
    a ---> 1
    b ---> 2


    We need to show 1-1, onto, and the homomorphism property
    I'm having problems with all of the properties
    Assume c(a)=c(b)
    This implies 1=2, so c(a) is not equal to c(b).
    Does this show 1-1?

    Onto:
    Need to show y=c(x) and solve for x
    y=c(a)=1
    y=1
    I'm confused on this

    Homomorphism:
    c(a)c(b)
    1*2
    2=c(ab)
    I know this should equal c(ab), but I didn't see why 2=c(ab)
    How much group theory do you know? Because this isn't necessarily the easiest route.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    This is my first semester of abstract algebra. Basically, I know groups and subgroups, and a little about cyclic groups.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kathrynmath View Post
    This is my first semester of abstract algebra. Basically, I know groups and subgroups, and a little about cyclic groups.
    Do you know Lagrange's theorem? If so, note that if G=\{e,a,b\} then \langle a\rangle is a subgroup of G. But, by Lagrange's theorem |\langle a\rangle|\mid 3 and since |\langle a\rangle|\geqslant (since e,a\in\langle a\rangle) it follows that |\langle a\rangle|=3 and thus \langle a\rangle=G. If you haven't done Lagrange's I'm not sure there's an easier method than to just write out the possible Cayley tables.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    I have done Lagranges Theorem, but I'm not sure how that would show an isomorphism. Doesn't
    an isomorphism exist if it is 1-1, onto, and a homomorphism? I don't see how those properties follow.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kathrynmath View Post
    I have done Lagranges Theorem, but I'm not sure how that would show an isomorphism. Doesn't
    an isomorphism exist if it is 1-1, onto, and a homomorphism? I don't see how those properties follow.
    Ok, so clearly if G=\langle g\rangle is a cyclic group of order 3 then G\cong \mathbb{Z}_3 via the isomorphism f:\mathbb{Z}_3\to\langle g\rangle:k\mapsto g^k as you can clearly check.

    But, what I showed above proves that every group of order 3 (and in fact any prime) is cyclic...sooo
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    then it's isomorphic to Z3?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by kathrynmath View Post
    then it's isomorphic to Z3?
    Yes, via the mapping \theta:\mathbb{Z}_3\to\langle g\rangle given by (maybe this will help)

    \begin{array}{c|c}k & \theta(k)\\ \hline 1 & g\\ 2 & g^2\\ 3 & g^3\end{array}
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,536
    Thanks
    1391
    Quote Originally Posted by kathrynmath View Post
    Prove that any group with three elements is isomorphic to Z3




    Take a group of three elements {e,a,b}. Since the order of every element must be three, we have that b=aČ. Thus the group is {e,a,aČ}.

    Define the map G --> Z3 by
    e ---> 0
    a ---> 1
    b ---> 2


    We need to show 1-1, onto, and the homomorphism property
    I'm having problems with all of the properties
    Assume c(a)=c(b)
    This implies 1=2, so c(a) is not equal to c(b).
    Does this show 1-1?

    Onto:
    Need to show y=c(x) and solve for x
    y=c(a)=1
    y=1
    I'm confused on this
    You already have that c(e)= 0, c(a)= 1, and c(b)= 2. That proves both "one-to-one" and "onto" immediately- two different members of {3, a, b} are not mapped to the same thing and there is something mapped to each of {0, 1, 2}

    Homomorphism:
    c(a)c(b)
    1*2
    2=c(ab)
    No, the operation in Z_3 is addition (modulo 3) NOT multiplication. 1+ 2= 0 in Z_3. You need to prove that c(ab)= 0 and that is true because, since this is a group
    1) ab cannot be equal to a since if a= ab then e= b which is not true.
    2) ab cannot be equal to b since if b= ab then e= a which is not true.
    That leaves only ab= e and c(e)= 0.
    I know this should equal c(ab), but I didn't see why 2=c(ab)
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    Can I prove the homomorphism property with c(ab)=c(a)c(b)
    c(a)=1
    c(b)=2
    1+2
    3
    e
    c(ab)


    So the 1-1, onto is just based on how I defined the group. i don't have to do any of the assuming c(a)=c(b) stuff?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,536
    Thanks
    1391
    Quote Originally Posted by kathrynmath View Post
    Can I prove the homomorphism property with c(ab)=c(a)c(b)
    c(a)=1
    c(b)=2
    1+2
    3
    There is NO "3" in Z_3! The elements of Z_3 are 0, 1, 2 and 1+ 2= 0.

    e
    c(ab)


    So the 1-1, onto is just based on how I defined the group. i don't have to do any of the assuming c(a)=c(b) stuff?
    Not how you defined the group but how you defined the isomorphism: you said c(e)= 0, c(a)= 1, c(b)= 2. Clearly c(a) is NOT equal to c(b)!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Senior Member
    Joined
    Apr 2008
    From
    Vermont
    Posts
    318
    c(a)c(b)
    =1+2
    =0
    =c(e)
    =c(ab)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 7
    Last Post: November 16th 2011, 11:39 PM
  2. Determine if group / isomorphic
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: September 26th 2011, 12:45 PM
  3. isomorphic group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 5th 2011, 10:51 PM
  4. Prove a group is isomorphic to another
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: February 8th 2010, 04:19 PM
  5. Prove a group H is isomorphic to Z.
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: September 23rd 2009, 11:08 PM

Search Tags


/mathhelpforum @mathhelpforum