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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 $G$?

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    $\begingroup$ I am probably missing something but, since ${\rm Aut}(F_2)$ is inducing ${\rm GL}(2,{\mathbb Z})$ on $F_2/[F_2,F_2]$ isn't $N$ just equal to the exponent of $G/[G,G]$? $\endgroup$ – Derek Holt Sep 9 '15 at 8:42
  • $\begingroup$ @DerekHolt Hmm...can you explain that a little more? $\endgroup$ – Will Chen Sep 9 '15 at 20:28
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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.

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