Not an isomorphism

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

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 $\displaystyle \mathbb{Q}$ isn't cyclic and I don't see how $\displaystyle \mathbb{Q} \times Z_2$ would be either. Any thoughts?

Thanks!
-Dan

Dan,
Suppose $\displaystyle \phi:\mathbb{Q}\times\mathbb{Z}_2\rightarrow \mathbb{Q}$ is an isomorphism. Let x = (0,1). Then $\displaystyle 0=\phi(0)=\phi(2x)=2\phi(x)$. So in $\displaystyle \mathbb{Q},\,2\phi(x)=0\text{ and then }\phi(x)=0$
Since $\displaystyle \phi$ is an isomorphism, x=0, contradiction.

Quote:

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Originally Posted by johng

Dan,
Suppose $\displaystyle \phi:\mathbb{Q}\times\mathbb{Z}_2\rightarrow \mathbb{Q}$ is an isomorphism. Let x = (0,1). Then $\displaystyle 0=\phi(0)=\phi(2x)=2\phi(x)$. So in $\displaystyle \mathbb{Q},\,2\phi(x)=0\text{ and then }\phi(x)=0$
Since $\displaystyle \phi$ is an isomorphism, x=0, contradiction.

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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

IF the two were isomorphic so we had an isomorphism $\displaystyle \phi$, then $\displaystyle \phi(\langle (0,1) \rangle)$ would necessarily be a cyclic subgroup of $\displaystyle \mathbb{Q}$ 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 $\displaystyle \mathbb{Q} \times \mathbb{Z}_n$, for any positive integer n.

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Originally Posted by topsquark

Logically speaking the text should be looking for a proof using that concept.

-Dan

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(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

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)?

Quote:

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Originally Posted by Deveno

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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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. (Doh)

-Dan

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

for example:

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

Quote:

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Originally Posted by Deveno

The operation in Z₄ is ADDITION.

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Gaaaaaa! (Doh) Hey, when I screw up I like to do it right! Thanks for the catch and the extra problem. (Bow)

-Dan