I need some help with the following question.
Prove that group $\mathbb{Q}\times Z_2$ is not isomorphic to $\mathbb{Q}.$
My proof:
Let $a,b \in \mathbb{Q}$ and let $\phi$ be isomorphism.We have $\phi(a,x)\phi((b,x)=\phi((a,x)(b,x))=\phi(ab,1)=\phi(a,1)\phi(b,1)$
But $\phi(a,x)\phi((b,x)\neq \phi(a,1)\phi(b,1).$ Thus $\phi$ isn't isomorphism.
Is proof true?
Best Answer
You seem to be under the impression that the operation of your group is component-wise multiplication. This is not possible, as $\Bbb Q \times \Bbb Z_2$ is not a group under that operation, as $(0,0)$ has no multiplicative inverse.
Instead, the proper operation should be component-wise addition, with the addition of the second coordinate modulo $2$.
Note that if we had such an isomorphism, $\phi$, its kernel is necessarily $\{(0,0)\}$. However:
$\phi((0,1)) + \phi((0,1)) = \phi((0,1) + (0,1)) = \phi((0,0)) = 0$. If we set:
$\phi((0,1)) = q$, the above tells us that $q + q = 0 \implies q = 0$ (since $q$ is rational).
Hence, $(0,1) \in \text{ker }\phi$, contradicting our assumption that $\phi$ is an isomorphism (since all isomorphisms are injective).