Prove that group $\mathbb{Q}\times Z_2$ is not isomorphic to $\mathbb{Q}$ 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?
 A: 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).
A: The group Q*Z2 has an element (0,1), which is of order two, but Q doesn't have an element of order two. That's why we can't define any isomorphism between them.
A: Another way to see that $\mathbf Q$ is not isomorphic to $\mathbf Q\times Z_2$ is to use the fact that $\mathbf Q$ is divisible while $\mathbf Q\times Z_2$ is not.
Definition. A group $G$ is divisible if for each $x\in G$ and each nonzero integer $k$, there is some $y\in G$ such that
$$
x = y^k \quad\text{or, written additively,}\quad x = ky.
$$
Intuitively, divisibility of a group is equivalent to the existence of integer roots of elements of the group.
Proposition. $\mathbf Q$ is divisible. (Check this.)
Put $G=\mathbf Q\times Z_2$. Negating the definition of divisibility, we want to show that there is some $x\in G$ and some nonzero integer $k$ such that for all $y\in G$, $x \ne ky$. Put $x = (2,1)$ and $k = 2$. For the sake of contradiction, suppose that $(q,z)\in G$ satisfies $(2,1)=2(q,z)$. Then,
$$
2(q,z)=(q,z)+(q,z)=(2q,2z).
$$
Evidently $q=1$ if there is any hope for this to work out. However, $2z\equiv 0\pmod 2$, for all $z\in Z_2$. Hence, the second coordinate can never be $1$, no matter which $z$ we pick in $Z_2$, so $G$ is not divisible. Since divisibility is preserved under isomorphisms, the claim is proved.
