Results 1 to 5 of 5

Math Help - Show a group is isomorphic. Table included!

  1. #1
    Junior Member
    Joined
    Oct 2011
    Posts
    28

    Talking Show a group is isomorphic. Table included!

    Let T be the group with four elements e,x,y,z with multiplication table

    lllll | e | x | y | z |
    l e | e | x | y | z |
    l x | x | e | z | y |
    l y | y | z | x | e |

    l z | z | y | e | x |

    Show that this group is isomorphic to Z4.
    (Z being all integers)
    (e being the identity)

    I know I have to define a bijection between these groups which is a homomorphism but then I'm lost.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,328
    Thanks
    702

    Re: Show a group is isomorphic. Table included!

    Z4 is cyclic of order 4. in Z4, 1 is an element of order 4 which generates the group (3 is also a generator).

    if T and Z4 are to be isomorphic, T has to have an element of order 4.

    well, it's not e, that just leaves 3 to check. what is the order of x?

    x^2 = e, so x is of order 2. let's keep looking.....

    y^2 = x. this looks promising....

    y^3 = (y^2)y = xy = z. so y is not of order 1,2 or 3...what's left? just to be sure, let's check:

    y^4 = (y^3)y = zy = e, so yep, y is of order 4.

    a homomorphism has to map identity to identity:

    so e--->0 is already chosen for us.

    an isomorphism has to send elements to images of the same order, so we have two choices for y:

    y-->1, or y-->3. let's pick y--->1. we can now write T as {e,y,y^2,y^3} (since we know y^2 = x, and y^3 = z)

    now, convince yourself that sending y^k <---> k (mod 4) is an isomorphism.

    the bijection part is easy...why is it a homomorphism (can you think of a way to prove it "in general" without verifying all 16 possible products in T)?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2011
    Posts
    28

    Re: Show a group is isomorphic. Table included!

    Quote Originally Posted by Deveno View Post
    Z4 is cyclic of order 4. in Z4, 1 is an element of order 4 which generates the group (3 is also a generator).

    if T and Z4 are to be isomorphic, T has to have an element of order 4.

    well, it's not e, that just leaves 3 to check. what is the order of x?

    x^2 = e, so x is of order 2. let's keep looking.....

    y^2 = x. this looks promising....

    y^3 = (y^2)y = xy = z. so y is not of order 1,2 or 3...what's left? just to be sure, let's check:

    y^4 = (y^3)y = zy = e, so yep, y is of order 4.

    a homomorphism has to map identity to identity:

    so e--->0 is already chosen for us.

    an isomorphism has to send elements to images of the same order, so we have two choices for y:

    y-->1, or y-->3. let's pick y--->1. we can now write T as {e,y,y^2,y^3} (since we know y^2 = x, and y^3 = z)

    now, convince yourself that sending y^k <---> k (mod 4) is an isomorphism.

    the bijection part is easy...why is it a homomorphism (can you think of a way to prove it "in general" without verifying all 16 possible products in T)?

    Can this be proven by figuring the table out?
    T' =

    lllll | 0 | 1 | 2 | 3 |
    l 0 | 0 | 1 | 2 | 3 |
    l 1 | 1 | 2 | 3 | 0 |
    l 2 | 2 | 3 | 0 | 1 |
    l 3 | 3 | 0 | 1 | 2 |


    A homomorphism between T and T' is any function, f, satisfying f(a*b) = f(a)*f(b) for all a,b in T

    For this table (T'), f(2*3) = f(1) vs f(2) * f(3) = f(1), right?

    which is the same for T which would be:
    f(y*z) = f(x) vs f(y) * f(z) = f(x), right?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,328
    Thanks
    702

    Re: Show a group is isomorphic. Table included!

    yes, you CAN compare tables....but that approach will prove useless with any group of large size (even a group of order 12 takes more time to construct a table than most people are willing to do).

    but behold! here is my "magic proof"

    define φ(y^k) = k (mod 4). we already have seen this is a bijection.

    φ((y^k)(y^m)) = φ(y^(k+m)) = k+m (mod 4) = k (mod 4) + m (mod 4) = φ(y^k) + φ(y^m), no muss, no fuss, φ is a homomorphism.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Oct 2011
    Posts
    28

    Re: Show a group is isomorphic. Table included!

    Quote Originally Posted by Deveno View Post
    yes, you CAN compare tables....but that approach will prove useless with any group of large size (even a group of order 12 takes more time to construct a table than most people are willing to do).

    but behold! here is my "magic proof"

    define φ(y^k) = k (mod 4). we already have seen this is a bijection.

    φ((y^k)(y^m)) = φ(y^(k+m)) = k+m (mod 4) = k (mod 4) + m (mod 4) = φ(y^k) + φ(y^m), no muss, no fuss, φ is a homomorphism.


    Wow! That proof is delightful. Thank you! Certainly too cool to be in my textbook.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: December 7th 2011, 10:58 AM
  2. Replies: 7
    Last Post: November 16th 2011, 11:39 PM
  3. isomorphic group
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: May 5th 2011, 10:51 PM
  4. Replies: 8
    Last Post: December 10th 2010, 03:42 AM
  5. Logic problem help (picture/table included)
    Posted in the Algebra Forum
    Replies: 2
    Last Post: January 3rd 2010, 01:53 PM

Search Tags


/mathhelpforum @mathhelpforum