Proving $\mathbb{Q} \times \mathbb{Z_2} \ncong \mathbb{Q}$ How do I prove $\mathbb{Q} \times \mathbb{Z_2} \ncong \mathbb{Q}$? I know that they are not isomorphic because for each element in $\mathbb{Q}$, say $\frac{a}{b}$, there are two corresponding elements $(\frac{a}{b}, 0), (\frac{a}{b}, 1) \in \mathbb{Q} \times \mathbb{Z_2}$, so the cardinality of $\mathbb{Q} \times \mathbb{Z_2}$ should be precisely double than that of $\mathbb{Q}$.
How can I show this formally though? You can prove that two infinite groups have the same cardinality by finding a bijection between them, but how can you prove that a bijection does not exist between two infinite sets?
 A: Your cardinality argument is wrong since finitely many product of at most countable set is at most countable. (Especially the product of finite set and countable set is countable.) Moreover, there are nonismorphic groups $A$ and $B$ which satisfy $A\times B\cong A$. (e.g. $A=\Bbb{Q}^\Bbb{N}$, $B=\Bbb{Q}$)
Despite of this, you can prove that $\Bbb{Q}\times \Bbb{Z}/2\Bbb{Z}$ is not isomorphic to $\Bbb{Q}$ because former one has an element of order 2 but later one does not.
A: To expound upon the good answer by tetori, suppose that there were an isomorphism $\varphi: \Bbb{Q} \times \Bbb{Z}_2 \to \Bbb{Q}$.  Consider the element $(0, 1) \in \Bbb{Q} \times \Bbb{Z}_2$ and its image $x = \varphi(0,1)$.  Calculate:
\begin{align}
2x &= x + x \\
&= \varphi(0,1) + \varphi(0,1) \\
&= \varphi\bigl( (0,1) + (0,1) \bigr) \tag{$\varphi$ is a homomorphism} \\
&= \varphi(0,0) \\
&= 0.
\end{align}
So you have the equation $2x = 0$ in $\Bbb{Q}$, which only has the solution $x=0$.  But $\varphi(0,0) = 0 = \varphi(0,1)$ shows that $\varphi $ is not injective; hence, it's not an isomorphism.
