I am reading the book: Geometric Group Theory by
Cornelia Druţu and Michael Kapovich With an Appendix by Bogdan Nica. [https://www.math.ucdavis.edu/~kapovich/EPR/ggt.pdf](https://www.math.ucdavis.edu/%7Ekapovich/EPR/ggt.pdf). On page 125 we have:


\*Def: We say that two groups $G\_{1}$ and $G\_{2}$ are virtually isomorphic (abbreviated as VI) if there exist finite-index subgroups $H\_{i} \subset G\_{i}$ and finite normal subgroups $F\_{i}\triangleleft H\_{i}$, i = 1, 2, so that the quotients $H\_{1}/F\_{1}$ and $H\_{2}/F\_{2}$ are isomorphic.


The authors proved, on page 125 Proposition 5.15, that the virtual isomorphism relation is an equivalence relation between groups.


The proof goes as follows:


* The normal closure of the union of two finite normal subgroups is a finite subgroup. Suppose now that $G\_{1}$ is VI to $G\_{2}$ and $G\_{2}$ is VI to $G\_{3}$. Then we have
$F\_{i} \triangleleft H\_{i} < G\_{i}$, $|G\_{i} :H\_{i}|<\infty$, $|F\_{i}|<\infty$, $i=1,2,3$, and
$F\_{2}’\triangleleft H\_{2}’< G\_{2}$, $|G\_{2} : H\_{2}’| < \infty$, $| F\_{2}’ | < \infty$, so that
$H\_{1}/F\_{1}\cong H\_{2}/F\_{2}$ and $H\_{2}’/F\_{2}’\cong H\_{3}/F\_{3}$.
Now $H\_{2}’’=H\_{2}\cap H\_{2}‘$ has finite index in $G\_{2}$. Let $K\_{2}$ be the normal closure in $H\_{2}’’$ of the union of the finite subgroups $H\_{2}’’\cap F\_{2}$ and $H\_{2}’’\cap F\_{2}’$.
**From** now on things become unclear to me.


We have quotient maps:
$f\_{i}:H\_{2}’’\rightarrow C\_{i}=f\_{i}(H\_{2}’’)\leq H\_{i}/F\_{i}$, $i=1,3$, with finite kernels and cokernels. (I am not sure how these maps are defined.) The subgroups $E\_{i} \mathrel{:=} f\_{i}(K\_{2})$ are finite and normal in $C\_{i}$, $i = 1,3$. (This is obvious to me.) We let $H\_{i}’,F\_{i}’ \subset H\_{i}$ denote the preimages of $C\_{i}$ and $E\_{i}$ under the quotient maps $H\_{i} \rightarrow H\_{i}/F\_{i}$, $i = 1,3$. Then $|F\_{2}’| < \infty$, $|G\_{i} : H\_{i}’| < \infty$, $i = 1,3$.
Lastly (this is completely unclear to me),
$H\_{i}’/F\_{i}’\cong C\_{i}/E\_{i}\cong H\_{2}’’/K\_{2}$, $i=1,3$.


My attempt to convince myself:
Consider the isomorphisms
$g\_{1}:H\_{2}/F\_{2}\rightarrow H\_{1}/F\_{1}$ and $g\_{3}:H\_{2}’/F\_{2}’\rightarrow H\_{3}/F\_{3}$. Then $g\_{1}$ induces a surjective homomorphism $f\_{1}:H\_{2}\rightarrow H\_{1}/F\_{1}$ given by $f\_{1}(x)=g\_{1}(xF\_{2})$; similarly $g\_{2}$ induces a surjective homomorphism $f\_{3}:H\_{2}’\rightarrow H\_{3}/F\_{3}$ given by $f\_{3}(x)=g\_{3}(xF\_{2}’)$. Restricting to $H\_{2}’’$, we obtain the quotient maps:
$f\_{i}:H\_{2}’’\rightarrow C\_{i}=f\_{i}(H\_{2}’’)\leq H\_{i}/F\_{i}$, $i=1,3$. However, I cannot justify the isomorphisms
$H\_{i}’/F\_{i}’\cong C\_{i}/E\_{i}\cong H\_{2}’’/K\_{2}$, $i=1,3$.



virtual isomorphism of groups is an equivalence relation
OK, here is a proof (I just fixed it). Let $F\_i\triangleleft H\_i<\_{f.i.} G\_i$, $i=1,2$, $F'\_i\triangleleft H\_i'<\_{f.i.} G\_i, i=2,3$ and $f\colon H\_1\to H\_2/F\_2, g\colon H\_2' \to H\_3'/F\_3'$ be surjective homomorphisms with kernels $F\_1, F\_2'$. Let 
 $H=H\_2\cap H\_2' <\_{f.i} G\_2$. Then $H/(H\cap F\_2')\cong HF\_2'/F\_2'<\_{f.i.} H\_2'/F\_2'<\_{f.i.} H\_3'/F\_3'$. 


Let $P$ be $f^{-1}((H\cap F\_2')/(H\cap F\_2'\cap F\_2)$. Then $P$ is normal in $U=f^{-1}(H/F\_2)$ which is of f.i. in $G\_1$. Therefore $U/P\cong H/(H\cap F\_2'\cap F\_2)$ which is of f.i. in $H\_3'/F\_3'$ (see above), the proof is complete. The proof uses two standard isomorphism theorems in group theory (the second and the third, I think).


This is not the only proof available. As I was
 told by M. Kapovich, it is Proposition 5.15 in the book by Drutu and Kapovich. "For the proof, we first prove Lemma 5.16 and then finish the proof of the proposition." 


