Isomorphic subgroups such that the quotient is isomorphic.

Question

Does there exists an infinite group $G$ which admits a normal subgroup $H$ such that $G,H,G/H$ are all isomorphic groups?

If I don't remember badly, some time ago I read that the additive group of $\mathbb{R}^2$ is isomorphic to $\mathbb{R}$. Is this true? If it is, then the answer to my question is affirmative considering $G=\mathbb{R}^2$ and as $H$ we can take the $x$-axis. Is this fact true (and why)?

I have other related questions: does such an example can be of countable cardinality? Does there exist an example in which every element is of finite order (or just there are some of them different from the unit)? Are there such examples in non-abelian groups?

Answer 1

There even exist nontrivial finitely generated groups $G$ isomorphic to $G\times G$, hence, taking the normal subgroup $H=G\times 1$ in $G\times G\cong G$, we obtain an isomorphism $G\cong G/N$. See

J. Jones, Direct products and the Hopf property, J. Austral. Math. Soc. 17 (1974), 174-196.

Such groups $G$ are necessarily nonabelian. I am not sure about finitely generated torsion groups, but, most likely, they exist as well.

It is unknown if there are nontrivial finitely presented groups $G$ isomorphic to $G\times G$; see

R. Hirshon, Misbehaved direct products. Expo. Math. 20 (2002), no. 4, 365–374.

Answer 2

Take $G=\mathbb{Z}^{\infty}\times\mathbb{Z}^{\infty}$.

Answer 3

Under pointwise addition the family of functions $\Bbb F_2^{X}$ from any set $X$ to the finite field $\Bbb F_2$ of two elements is an abelian group all of whose non-trivial elements have finite order 2.

Let $X=\Bbb N$ and let $2\Bbb N, 2\Bbb N+1$ be the set of even/odd positive integers respectively. Let $H$ be the $\Bbb F_2^{\Bbb N}$ subgroup of functions which vanish on $2\Bbb N+1$. Clearly, $$H\approx\Bbb F_2^{2\Bbb N}\approx\Bbb F_2^{2\Bbb N + 1}\approx\frac{\Bbb F_2^{\Bbb N}}{H}$$ and because $\Bbb N,2\Bbb N$ have the same cardinality we also have
$$\Bbb F_2^{\Bbb N}\approx\Bbb F_2^{2\Bbb N}$$

Here, we can replace $\Bbb N$ with any set $X$ and replace $2\Bbb N, 2\Bbb N+1$, respectively, with any two subsets of $X$ that partition $X$ and have the same cardinality as $X$.