Results 1 to 11 of 11

Math Help - isomorphism between two groups problem..

  1. #1
    Junior Member
    Joined
    Jun 2010
    From
    Skopje
    Posts
    54

    isomorphism between two groups problem..

    Hi i got problem cant find isomorphic function between this groups:

    G=[{1, -1, i, -i}, *] and [Z4, +4]

    first group is defined with normal multiplication, second is cyclic Z4 group with mod 4 addition, here is what i thought of:

    since i need to convert * op to + op has to do something with Log function cause the groups must be closed under the op. and the function that i must find has to be homomorphic: kinda f(x) = Log(x), so for check on closed property i get x and y E in G and f(x*y)=Log(x) + Log(y), the problem is when i need to get f(-1) or f(-i) since under the log there cant be negative or else i get another complex number but in Z4 there's no complex numbers.. so im confused here i cant find the function.. any help would be appreciated thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    Quote Originally Posted by goroner View Post
    Hi i got problem cant find isomorphic function between this groups:

    G=[{1, -1, i, -i}, *] and [Z4, +4]

    first group is defined with normal multiplication, second is cyclic Z4 group with mod 4 addition, here is what i thought of:

    since i need to convert * op to + op has to do something with Log function cause the groups must be closed under the op. and the function that i must find has to be homomorphic: kinda f(x) = Log(x), so for check on closed property i get x and y E in G and f(x*y)=Log(x) + Log(y), the problem is when i need to get f(-1) or f(-i) since under the log there cant be negative or else i get another complex number but in Z4 there's no complex numbers.. so im confused here i cant find the function.. any help would be appreciated thanks
    You are making this harder then it needs to be. The element 1 generates \mathbb{Z}_4 and i generates G.

    So let

    f(i)=1 Now isomorphism must respect the group operations so in G you have

    i\cdot i=-1 \iff f(i)+f(i)=f(-1) \iff 2=f(-1)

    The Key idea is that the mapping of the generator to generator determines the isomorphism. This also tells you that there is more that one isomorphism -i is also a generator of G.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jun 2010
    From
    Skopje
    Posts
    54
    hmm, so the funct is just f(i)=1?, how do i translate 1 with this funct? i dont understand you say in G you have i*i = -1 and then you use additive operation between f(i)+f(i) which equals f(-1) , didnt you mean f(i^2)=f(-1), and how come f(-1)=2? you messed me up bit now :P, from f(i)+f(i) ye you get 2 but why that equals f(-1) i dont understand..

    and cause -i is generator too it turns out that f(-i)=1, so from here f(-i)+f(-i) = f(i)+f(i) isnt this breaking the rule that the function must be bijection so that there's isomorphism between the groups..? more precise f(-i)=f(i) # ?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
    Joined
    Feb 2008
    From
    Yuma, AZ, USA
    Posts
    3,764
    Thanks
    78
    You need to understand that isomorphism preserve group operations. If you multiply two elements in G and then map to the integers mod4 it needs to be the same as mapping the two elements to the integers mod 4 and then adding them.

    In symbols let g_1,g_2 \in G and suppose that f(g_1)=a \quad f(g_2)=b note that  f(g_1),f(g_2) \in \mathbb{Z}_4 then by the definition of isomorphism we must have that

    f(g_1\cdot g_2)=f(g_1)+f(g_2)

    The multiplication is in G where the addition is in the integers mod 4.

    So as I said in my example map a generator to a generator to get

    f(i)=1 So I know what the image of i is.

    Now using f(g_1\cdot g_2)=f(g_1)+f(g_2)

    f(i\cdot i)=f(i)+f(i)=1+1=2 \iff f(-1)=2

    Now lets keep extending this

    f(i\cdot i\cdot i)=f(i)+f(i)+f(i)=1+1+1=3 \iff f(-i)=3


    f(i\cdot i\cdot i \cdot i)=f(i)+f(i)+f(i)+f(i)=1+1+1+1=0 \iff f(1)=0

    Now just for good measure what happens if we do this one more time

    f(i\cdot i\cdot i \cdot i\cdot i)=f(i)+f(i)+f(i)+f(i)+f(i)=1+1+1+1+1=1 \iff f(i)=1


    ahhh we are back where we started!


    Now we have the isomorphism

    f(i)=1,f(-1)=2,f(-i)=3,f(1)=0

    Notice that the identity of G was mapped to the identity of the integers mod 4

    You should try to find another isomorphism but this time send -i to 1 and see what happens.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jun 2010
    From
    Skopje
    Posts
    54
    wow looks easy , well thanks a lot mate, im not really long into abstract algebra and my brain is not working that smooth yet but i guess ill make it work with practice (im trying ), and ye that was my doubt about, i was trying to find just one function that will map all the set members from G into Z4, i thought of function by parts too but i was not really sure about it, anyway thanks for your help now is clear to me
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,273
    Thanks
    666
    since an isomorphism is invertible, we can just as easily consider an isomorphism from (Z4, +) --> ({i,-1,-i,1},*).

    so how about:

    \varphi(n) = e^{in\pi/2} ?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Jun 2010
    From
    Skopje
    Posts
    54
    that looks rather complicated to me, the first solution is fine but thanks anyway!
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,273
    Thanks
    666
    perhaps it will seem less complicated if we use that fact that:

    e^{iy} = cos(y) + isin(y).

    then we have: \varphi(n) = cos(n\pi/2)+isin(n\pi/2)

    i wrote it in exponential form, so that you could see that addition transforms into multiplication. this is, in fact, the isomorphism:

    0<-->1 = 1+0i
    1<-->i = 0+i
    2<-->-1 = -1+0i
    3<-->-i = 0-i

    a slight variation of this isomorphism can be used to find an isomorphism between Zn (for any natural number n) and vertices of a regular n-gon inscribed within

    the unit circle in the complex plane, with the first vertex at 1 = 1+0i = (1,0).
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by goroner View Post
    that looks rather complicated to me, the first solution is fine but thanks anyway!
    The first isomorphism might work, but Deveno's one is telling you what is going on. Which is much, much more important!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by Deveno View Post
    since an isomorphism is invertible, we can just as easily consider an isomorphism from (Z4, +) --> ({i,-1,-i,1},*).

    so how about:

    \varphi(n) = e^{in\pi/2} ?
    Or perhaps the maybe-less-scary-looking form \varphi(n)=i^n.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    MHF Contributor Swlabr's Avatar
    Joined
    May 2009
    Posts
    1,176
    Quote Originally Posted by Drexel28 View Post
    Or perhaps the maybe-less-scary-looking form \varphi(n)=i^n.
    ...but...rotating a circle!...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Isomorphism between groups...
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: November 5th 2010, 09:02 AM
  2. Isomorphism of 2 groups
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: January 23rd 2010, 09:24 AM
  3. determine up to isomorphism all groups of order 13^2*3^2
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 10th 2010, 01:12 AM
  4. Finding all groups of order n up to isomorphism
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 8th 2009, 02:55 AM
  5. Isomorphism of abelian groups
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 29th 2009, 08:51 AM

Search Tags


/mathhelpforum @mathhelpforum