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Math Help - Isomorphism

  1. #1
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    Isomorphism

    Hey guys,
    I need to prove that there exists 2 groups with order 4 which are not isomorphic.
    I am short on examples. I am just surfacing this topic. Can you please help me out?
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  2. #2
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    Re: Isomorphism

    Quote Originally Posted by mathrld View Post
    Hey guys,
    I need to prove that there exists 2 groups with order 4 which are not isomorphic.
    I am short on examples. I am just surfacing this topic. Can you please help me out?
    Look at the Kline Four Group.
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  3. #3
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    Re: Isomorphism

    start from first principles.

    we have G = {e,a,b,c}.

    again, the products involving e are boring. ignore them. we need to focus on a,b and c.

    let's start with a. a*a has to be some member of the group.

    a*a cannot be a (we've been over this before). that leaves us with 3 choices:

    a*a = e
    a*a = b
    a*a = c

    let's look at the first choice: a*a = e (this means a = a-1). so far, so good.

    if we adopt this (for the time being), then we can look at a*b.

    a*b = e is not possible (for that implies a*a = a*b and thus a = b).
    a*b = a is not possible (for that implies b = e).
    a*b = b is not possible (for that implies a = e).

    so...if a*a = e, it must be the case that a*b = c. hold that thought!

    now, i claim that we can always find one of a,b or c such that either:

    a*a = e
    b*b = e, or
    c*c = e.

    why? look at inverses:

    we know e-1 = e...boring!

    well, if a*a = e, then we're done, a is it's own inverse. so suppose not. then a-1 is either b or c. if b is a's inverse, the a is b's inverse. in that case, c has to be its own inverse (it can't be e,a or b, whose inverses are already spoken for). on the other hand, if c is a's inverse, then a is c's inverse, so b has to be it's own inverse. so no matter how we slice it, at least one of a,b, or c has to be its own inverse.

    now, for us, a,b and c are just symbols representing the elements of G. let's re-label them, if we need to, so that a is an element that is its own inverse.

    this means we can write G = {e,a,b,a*b}. let's write out all 16 products, filling in the ones we can figure out:

    e*e = e
    e*a = a
    e*b = b
    e*(a*b) = a*b
    a*e = a
    a*a = e <---by the above discussion
    a*b = a*b (by definition)
    a*(a*b) = (a*a)*b = e*b = b
    b*e = b
    b*a = ...?
    b*b = ....?
    b*(a*b) = ...?
    (a*b)*e = a*b
    (a*b)*a = ...?
    (a*b)*b = ...?
    (a*b)*(a*b) =...?

    as you can see, we already know 10 of the 16. let's work on the remaining 6.

    b*a = e implies b = a-1. but a = a-1, and a and b are different. so this can't happen.
    b*a = a implies b = e. bad choice.
    b*a = b implies a = e. another bad choice.

    so b*a has to be a*b. this means (a*b)*a = a*(b*a) = a*(a*b) = (a*a)*b = e*b = b. 2 down, 4 to go.

    when we try to evaluate b*b, we hit a dead-end. that's too bad, because if we knew what b*b was, we could also figure out (a*b)*b = a*(b*b), as well as:

    b*(a*b) = (b*a)*b = (a*b)*b = a*(b*b).

    that would also take care of:

    (a*b)*(a*b) = (b*a)*(a*b) = b*(a*a)*b = b*e*b = b*b.

    so it all boils down to what b*b is.

    b*b = b isn't possible (that would mean b = e, been there, done that. it doesn't work).
    b*b = a*b also isn't possible (that would mean b = a, not gonna happen).

    so we are left to decide between: b*b = e, and b*b = a. you can play with this a bit yourself, but i'll just "skip ahead": either one works.

    suppose b*b = a. this is the same as saying: b2 = a. then a*b = b2*b = b3, and we get:

    G = {e,b,b2,b3}.

    note that this is the same as saying: 4 is the smallest positive number n, for which bn = e.

    on the other hand, suppose b*b = e. then (a*b)*(a*b) = b*b = e, too...every element of G is its own inverse! in this case, we can't "simplify" G any further:

    G = {e,a,b,a*b} is the best we can do.

    ********

    ok, perhaps you see that perhaps two different groups of order 4 might be possible...but does "one of each" actually occur?

    well, yes. a group of the "first kind" is:

    {0,1,2,3} under addition modulo 4 (in this case, "1" plays the role of "b"). in this group a*b is written a+b, and rather than bn, we write nb. this is just notation, but it often throws people off:

    nb does NOT mean "n times b", it means "b+b+...+b" (n times) the *same way*

    bn means "b*b*...*b" (n times). the group {0,1,2,3} is called "the cyclic group of order 4" (Z4, when it is written additively, and C4, when multiplicative notation is used).

    a group of the second kind is:

    {(0,0),(1,0),(0,1),(1,1)} where (a,b) + (a',b') = (a+a' (mod 2), b+b' (mod 2)). this group is called "the klein 4-group", or V, or Z2xZ2, when it is written additively.
    Thanks from mathrld
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    Re: Isomorphism

    Wow Thanks. I really appreciate it. You respond it great detail that its easy to follow. I am taking online math courses so I kinda learn from diff problems and examples. I really appreciate it
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