Let $F\_2$ be the free group on two generators.


Let $U\le F\_2$ be a characteristic subgroup of finite index, and let $f : F\_2\rightarrow\mathbb{Z}^2$ be the abelianization map.


It's easy to check that $f(U)$ is characteristic in $\mathbb{Z}^2$, so it must be of the form $N\mathbb{Z}\times N\mathbb{Z}$ for some $N\ge 1$, and hence the induced quotient $\mathbb{Z}^2/f(U)$ must be of the form $(\mathbb{Z}/N\mathbb{Z})^2$.


I apologize if this is a basic question, but my general question is - Is there a way of determining the $N$ from the characteristic subgroup $U$?


Of course this is too vague - I suppose an underlying question was - what are the characteristic subgroups of $F\_2$? (There seems to be very little literature on this)


**A more precise question might be:**


Let $G$ be a finite group, and let $H\lhd F\_2$ with $F\_2/H \cong G$, and let $\{H\_i\}$ be the $Aut(F\_2)$-orbit $H$, then $U\_H := \bigcap\_i H\_i$ is characteristic in $F\_2$. Then we know that $f(U) = N\mathbb{Z}\times N\mathbb{Z}$. Is there a way of determining $N$ from $H$?


If we let $U\_G$ be the intersection of *all* $G$-defining subgroups of $F\_2$, then is there a way of determinine the $N$ associated to $f(U\_G)$ from $G$?



Abelianization of characteristic quotients of $F_2$
Note that $N$ is clearly the exponent of $F\_2/[F\_2,F\_2]U$.


For any normal subgroup $U$ of $F\_2$, the exponent of $F\_2/[F\_2,F\_2]U$ is the same as the exponent of $K/[K,K]$, where $K=F\_2/U$. Thus, $N$ is the exponent of $K/[K,K]$.


Now, $K$ is a subgroup of the Cartesian product $G\times G\times\dots$ such that the projection of $K$ to each factor is surjective. This implies that $K/[K,K]$ is a subgroup of
$G/[G,G]\times G/[G,G]\times\dots$ and the projection of $K/[K,K]$ to each factor is surjective. 


Thus, the exponent of $K/[K,K]$ (that is $N$) is equal to the exponent of $G/[G,G]$. 


So, [Derek is right](https://mathoverflow.net/questions/217776/abelianization-of-characteristic-quotients-of-f-2#comment536665_217776). 


