Group theory, subgroup of index 2

Question

I have a group $G $ which is an abelian group such that $\mathbb R $ is a subgroup of index $2$. Can I conclude that $G $ is isomorphic to $\mathbb Z /2\mathbb Z \times \mathbb R $ ?

I am not sure, since it is not true for example with $\mathbb Z /4\mathbb Z$ and $\mathbb Z /2\mathbb Z$ instead of $G $ and $\mathbb R $.

Answer 1

Yes, You can conclude that $G=\mathbb{R}\times \mathbb{Z}/2\mathbb{Z}$.

Let $\eta:G\to G/\mathbb{R}=\mathbb{Z}/2\mathbb{Z}=\langle b\rangle $ be a standart homomorphism.

Let $a$ be the fixed preimage of $b$ in $G$.

Therefore $2a=x$ is a rational number. Denote by $c=a-x/2$ (we can write $x/2$ since $x$ is rational number). Therefore $2c=2a-x=0$ (i.e. $\langle c\rangle=\mathbb{Z}/2\mathbb{Z}$ is a subgroup of $G$) and $c$ is another preimage of $b$ under the homomorphism $\eta$.

Now we have $G=\langle \mathbb{R}, c\rangle$ and $\mathbb{R}\cap \langle c\rangle=0$, i.e. $G=\mathbb{R}\times\mathbb{Z}/2\mathbb{Z}$.

The same proof does not work for $\mathbb{Z}/4\mathbb{Z}$ since in $\mathbb{Z}/4\mathbb{Z}$ you cannot devide by $2$.