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 $G$?

  • 2
    $\begingroup$ 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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.