Recently I saw this theorem :

Assume that $A\leq B\leq C$ with $A\cong\mathbb{Z}^n\cong C$. Then $B\cong\mathbb{Z}^n$.

I wonder whether it is possible to generalise this result to arbitrary finitely generated abelian groups.

What I mean is this : If $A\leq B\leq C$ are finitely generated abelian groups with $A\cong C$ then $B\cong C$.

Using the fundamental theorem of finitely generated abelian groups it seems like it is true but I could not prove it.

Edit : If we remove the condition that $C$ is abelian then we know that the claim is not true for free groups. There are subgroups of $F_2$ which is not isomorphic to $F_2$ but contain a subgroup which is isomorphic to $F_2$. But I also do not know an example of a solvable group $C$ with there exists $A\leq B\leq C, A\cong C$ but $B\not\cong C$.

  • $\begingroup$ We have $rank(B)=rank(A)=rank(C)$, see here, under properties. Then $B\cong C$. $\endgroup$ Dec 10, 2016 at 17:17
  • $\begingroup$ I see, using the properties of rank, the result is clear. Can you say something about the case when the groups are finitely generated solvable groups? By the way, if the result is valid for finitely generated abelian groups why is the theorem presented for only finitely generated free abelian groups? $\endgroup$
    – Levent
    Dec 10, 2016 at 17:22

1 Answer 1


For the case of solvable group, let $C$ be the wreath product of two copies of the integers. So this is a split extension of a direct sum of countably many copies $Z_i$ of the integers ($i$ running over the integers), by a copy of the integers, spanned by $g$, say, and written multiplicatively, with $g$ conjugating $Z_i$ to $Z_{i+1}$. This is a $2$-generated metabelian, hence solvable, group.

Now let $B$ be the subgroup generated by the $Z_i$ and $g^2$. Clearly it is not isomorphic to $C$, as there are two orbits of the $Z_i$ under the span of $g^ 2$, corresponding to $i$ even and $i$ odd. It is a $3$-generated group.

Finally, let $A$ be the subgroup of $B$ generated by the $Z_i$, for $i$ even, and $g^ 2$. This is isomorphic to $C$.

  • $\begingroup$ @Levent, thanks. $\endgroup$ Dec 12, 2016 at 9:37
  • $\begingroup$ What is it for? :) $\endgroup$
    – Levent
    Dec 18, 2016 at 15:42
  • 1
    $\begingroup$ :-) Your appreciation, which is becoming quite a scarce resource anywhere. $\endgroup$ Dec 18, 2016 at 15:44

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