# Groups with a dense chain of (isomorphic) subgroups

Are there any interesting non-Abelian examples of groups which are equal to the union of a dense chain of normal subgroups, with each subgroup isomorphic to the original group. That is, a group $G$ and set of normal subgroups $\{ H_{i}: i \in \mathbb{Q} \}$ such that:

1) $G=\bigcup_{i\in \mathbb{Q}} H_i$;

2) $H_i \subset H_j$ if and only if $i<j$];

3) For each $i$ we have $H_i\cong G$.

4) $G/H_i\cong G$.

One (although Abelian) example is $\bigoplus_{i\in \mathbb{Q}} \mathbb{Z}_2$; however this is isomorphic to $\bigoplus_{i\in \mathbb{N}} \mathbb{Z}_2$ (I believe), which wouldn't work...

The problem arises from a series of surjective (group) homomorphisms $\psi_{i}:G\rightarrow G_i$ ($i\in \mathbb{Q}$) of a group $G$ such that 1) -4) hold for normal subgroups $\{\text{Ker } \psi_i:i\in \mathbb{Q}\}$.

Note: $G$ will clearly not be hopfian or co-hopfian.

• $\bigoplus_{i\in \mathbb{N}} \mathbb{Z}_2$ is indeed isomorphic to $\bigoplus_{i\in \mathbb{Q}} \mathbb{Z}_2$, but the isomorphism is not given by the canonical inclusion, but by any bijection between $\mathbb{N}$ and $\mathbb{Q}$. – Captain Lama Apr 11 '16 at 16:17
• Indeed this is true; but are there any examples of G satisfying 1)-4) which are not isomorphic to a group without a dense chain of normal subgroups? – Tom Q Apr 12 '16 at 9:03
• Tautologically yes (all of them) : for instance $\oplus_{i\in \mathbb{N}} \mathbb{Z}_2$ has a dense chain of normal subgroups. It's just not easily visible on this decomposition. What I'm trying to say is that your conditions 1)-4) are clearly invariant by isomorphisms. – Captain Lama Apr 12 '16 at 9:23
• Thank you, I can see (roughly) how the dense chains would form. Are there any non-abelian examples? – Tom Q Apr 12 '16 at 11:06
• I think $\bigoplus_{i\in \mathbb{Q}} K$ for some non-abelian $K$ should work just as fine as your example. – Captain Lama Apr 12 '16 at 11:08