Prove that $\mathbb{Q}\times \mathbb{Z_2}$ not isomorphic to $\mathbb{Q}$

Question

My attempt:I was thinking about the fact that $\mathbb{Q} \times \Bbb Z_2$ has twice the no of elements as compared to $\mathbb{Q}$ and thus there cannot be injectivity. Am I correct?

Answer 1

Consider the element $(0,1)$ of $\left( \mathbf{Q}, \mathbf{Z_2} \right) $. Its order is $2$ for the additive law we are working with because $(0,1) \neq (0,0)$ and $1 + 1 = 0$ in $\mathbf{Z_2} $.

But there is no element of order $2$ in the group $(\mathbf{Q}, +) $. Indeed, such a non-zero element $x$ would respect $2x = 0$ in $\mathbf{Q}$, which is impossible.

This is sufficient to claim the two groups are not isomorphic. If they were, we would find a group isomorphism $\Phi : \left( \mathbf{Q}, \mathbf{Z_2} \right) \rightarrow \mathbf{Q}$ and we would have $\Phi(0, 1) \neq 0$ by injectivity. But $ 2 \cdot \Phi(0, 1) = \Phi(2 \cdot (0, 1)) = \Phi(0, 0) = 0$. So we must also have $\Phi(0, 1) = 0$: contradiction.

To conclude, $\left( \left( \mathbf{Q}, \mathbf{Z_2} \right) , + \right)$ and $\left( \mathbf{Q} , + \right)$ are not isomorphic groups.

Answer 2

No, your proof doesn't work. Injectivity doen't work like that for infinite sets. In fact, $|\mathbb{Q}\times \mathbb{Z_2}| = |\mathbb{Q}|$ hence there is a bijection between those sets.

Here, we are probably talking about a ring-isomorphism.

Answer 3

This can be proven in several ways.

As already noted, $\Bbb Q$ has no element of order $2$ while $\Bbb Q\times\Bbb Z_2$ does.

Another approach is to observe that $\Bbb Q\times\Bbb Z_2$ has a subgroup of index $2$ while $\Bbb Q$ doesn't. Indeed if $H<\Bbb Q$ were a subgroup of index $2$ fix an element $q\notin H$. Since $q=2(q/2)$ also $q/2\notin H$. But then it would be $q-(q/2)=q/2\in H$, a contradiction.

A third possibility is to observe that $\Bbb Q$ is a divisible group (since $q/n\in\Bbb Q$ for every $q\in\Bbb Q$ and for every $0\neq n\in\Bbb Z$) while $\Bbb Q\times\Bbb Z_2$ is not, e.g. there's no $(a,b)$ such that $2(a,b)=(0,1)$.