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Math Help - Not an isomorphism

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    Not an isomorphism

    I need to show that \mathbb{Q} \times Z_2 is not isomorphic to \mathbb{Q}.

    Mind you, I'm in the section of my book dealing with cyclic groups. I was able to handle the other problems in this set by showing that one group is cyclic and the and the other isn't, etc. But \mathbb{Q} isn't cyclic and I don't see how \mathbb{Q} \times Z_2 would be either. Any thoughts?

    Thanks!
    -Dan
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    Re: Not an isomorphism

    Dan,
    Suppose \phi:\mathbb{Q}\times\mathbb{Z}_2\rightarrow \mathbb{Q} is an isomorphism. Let x = (0,1). Then 0=\phi(0)=\phi(2x)=2\phi(x). So in \mathbb{Q},\,2\phi(x)=0\text{ and then }\phi(x)=0
    Since \phi is an isomorphism, x=0, contradiction.
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    Re: Not an isomorphism

    Quote Originally Posted by johng View Post
    Dan,
    Suppose \phi:\mathbb{Q}\times\mathbb{Z}_2\rightarrow \mathbb{Q} is an isomorphism. Let x = (0,1). Then 0=\phi(0)=\phi(2x)=2\phi(x). So in \mathbb{Q},\,2\phi(x)=0\text{ and then }\phi(x)=0
    Since \phi is an isomorphism, x=0, contradiction.
    Thanks. That does show that they are not isomorphic, but what is this problem doing in a section for cyclic groups? Logically speaking the text should be looking for a proof using that concept.

    -Dan
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    Re: Not an isomorphism

    IF the two were isomorphic so we had an isomorphism \phi, then \phi(\langle (0,1) \rangle) would necessarily be a cyclic subgroup of \mathbb{Q} of order 2 under rational addition. Do the rational numbers possess such an additive subgroup (that is, is there ANY rational number q not equal to 0 with: q + q = 0)?

    In other words, one group has a finite cyclic subgroup, the other does not. A similar proof holds for \mathbb{Q} \times \mathbb{Z}_n, for any positive integer n.
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    Re: Not an isomorphism

    Quote Originally Posted by topsquark View Post
    Logically speaking the text should be looking for a proof using that concept.

    -Dan
    (groans) I finally got those last two steps at 4 AM this morning. I now see that this proof is dependent on cyclic group properties. In my defense the text I am using almost exclusively uses the multiplicative format for cyclic groups.

    -Dan
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    Re: Not an isomorphism

    Here is a fun exercise:

    The groups (Z4,+) and (Z5*, .) are isomorphic, but

    f(k (mod 4)) = k+1 (mod 5) is not an isomorphism. Why not, and what is a "better" formula for an isomorphism (there are two I am thinking of)?
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    Re: Not an isomorphism

    Quote Originally Posted by Deveno View Post
    Here is a fun exercise:

    The groups (Z4,+) and (Z5*, .) are isomorphic, but

    f(k (mod 4)) = k+1 (mod 5) is not an isomorphism. Why not, and what is a "better" formula for an isomorphism (there are two I am thinking of)?
    For starters f(k) isn't a homomorphism: For x, y in Z_4 we have
    f(xy) = xy + 1
    f(x)f(y) = (x + 1)(y + 1) = (xy + 1) + (x + y)

    I can't seem to construct a bijective function g(k). The best I can do using the homomorphism property is g(k) = g^k(1), a cyclic group with generator g(1). But g(1) can only take the values 2, 3, or 4 and none of them work.

    -Dan
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    Re: Not an isomorphism

    The operation in Z4 is ADDITION.

    However, it IS true f isn't a homomorphism:

    f(x + y) = x + y + 1
    f(x)f(y) = (x + 1)(y + 1) = xy + x + y + 1, and since 5 is prime, Z5 is an integral domain, so if x,y are non-zero, so is xy.

    The two possible isomorphisms are:

    f(k (mod 4)) = 2k (mod 5) and
    g(k (mod 4)) = 3k (mod 5)

    for example:

    f(k (mod 4) + m (mod 4)) = f(k+m (mod 4)) = 2k+m (mod 5) = (2k (mod 5))(2m (mod 5)) = f(k (mod 4))f(m (mod 4)).

    The group (Z5*, .) is also known as the group of units of Z5, or U(5). In this group, 2 and 3 are primitive elements (generators). This is in direct analogy to the primitive 4-th roots of unity, as both 2 and 3 are square roots of -1 (which is 4 in Z5):

    (x - 2)(x - 3) (mod 5) = x2 - (2 + 3)x + 6 (mod 5) = x2 + 1 (mod 5).
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    Re: Not an isomorphism

    Quote Originally Posted by Deveno View Post
    The operation in Z4 is ADDITION.
    Gaaaaaa! Hey, when I screw up I like to do it right! Thanks for the catch and the extra problem.

    -Dan
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