# Restriction of homomorphism to finite index subgroup gives finite index subgroup of kernel

Suppose we have a surjective homomorphism $f:G \to H$ with $H$ a finitely generated free abelian group and that $\ker f$ contains a finite index subgroup $B$. Can we always find a finite index subgroup $A$ of $G$ such that $f$ retstricted to $A$ has kernel $B$?

I have can see that the kernel of $f$ restricted to $A$ equals $\ker f \cap A$, but this does not really help me any further.

reason why I ask this question: I am currently writing my masters thesis. A have a statement which says that a given group has a finite index subgroup which has property $\mathcal{P}$. I have shown that the kernel of $f$ has such a finite index subgroup with property $\mathcal{P}$, but the text says that the kernel therefore has property $\mathcal{P}$. I assume that the text implicitly used the above conjecture, by restricting $f$ to the finite index subgroup $A$, and the kernel would then be the group with property $\mathcal{P}$.

I hope this makes any sense...

EDIT: on proposal of @stewbasic, I include the proof in which I do not udnerstand a part. The set $PG(F_n)$ denotes the set of polynomially growing outer automorphisms, the set $UPG(F_n)$ the set of unipotent polynomially growing outer automorphism. These are outer automorphisms such that the abelianisation matrix is conjugated to a upper triangular matrix with ones on the diagonal.

lemma: Let $\mathcal{H}$ be a subgroup of $Out(F_n)$ that does not contain a free subgroup of rank $2$. Then there are finite index subgroup $\mathcal{H}_0$ of $\mathcal{H}$, a finitely generated abelian group $A$ and a map $\Phi: \mathcal{H}_0 \to A$ such that $\ker \Phi$ is UPG. Proof: Let $\mathcal{L} = \{ \Lambda_1, \ldots, \Lambda_k\}$ and $\mathcal{H}_0$ as in lemma 7.0.10. Define $\Phi = \sum PF_{\Lambda_i^+}: \mathcal{H}_0\to \mathbb{Z}^k$, where each $PF_{\Lambda_i^+}$ is as in corollary 3.3.1. By corollary 5.7.6. it suffices to show that $\ker \Phi$ is contained in $PG(F_n)$... (proof then shows that this is true and this is the last part of the proof)

The proof then continuous to show that the kernel is PG. corollary 7.5.6. however states that

if $\mathcal{O} \in PG(F_n)$ is contained in the kernel of the natural homomorphism $$Out(F_n) \mapsto GL(n, \mathbb{Z}) \to GL(n, \mathbb{Z}/3\mathbb{Z})$$ then $\mathcal{O} \in UPG(F_n)$. In particular, every subgroup of $PG(F_n)$ contains a finite index subgroup in $UPG(F_n)$

and the proof does not show that $\ker \Phi$ is in the kernel of this natural homomorphism.

• The only chance this will work is if you can set $A$ to be the normalizer of $B$. $B$ must be normal in $A$ and you want $A$ as large as possible. – Max Apr 29 '18 at 19:22
• @Max: it does not seem like this is the case in the paper. anyway, thanks for your suggestion. – Student Apr 29 '18 at 19:25
• So based on your question title, $B$ has finite index in ker $f$, correct? – Andrew Kelley May 1 '18 at 2:24
• It sounds like you want this result in order to complete a proof of some other result. I suggest asking about the latter result directly, in case the omitted steps are not as you assume. – stewbasic May 1 '18 at 2:58
• @AndrewKelley yes, $B$ has finite index in $\ker f$. – Student May 1 '18 at 6:43

No. Let $N$ be the set of maps from $\mathbb Z$ to $\mathbb Z_2$, which forms an Abelian group under the operation of pointwise addition; $(a+b)(n)=a(n)+b(n)$. Now use the action of $\mathbb Z$ on $N$ given by $$(n\cdot a)(m)=a(m-n)$$ to form the semidirect product $G=N\rtimes\mathbb Z$ (recall that $(a,n)(b,m)=(a+n\cdot b,n+m)$). For $n\in\mathbb Z$ consider the homomorphism $$eval_n:N\to\mathbb Z_2,\,a\mapsto a(n),$$ Let $B=\ker eval_0$. Note that $B$ is normal of finite index in $N$ (indeed $N/B\cong\mathbb Z_2$). Moreover $$eval_n(n\cdot a)=a(0)=eval_0 a,$$ so $n\cdot B=\ker eval_n$. Thus for $(a,n)\in G$, $$(a,n)B(a,n)^{-1}=\ker eval_n,$$ so the normalizer of $B$ in $G$ is exactly $N_G(B)=N$.

Finally let $H=G/N$ and $f:G\to H$ be the natural surjection. Note that $H\cong\mathbb Z$ and $\ker f=N$. If such a subgroup $A$ exists, then $A\subseteq N_G(B)=N$. But $N$ does not have finite index in $G$, a contradiction.

As for the specific result, the proof can be completed as follows. Let $\Psi:Out(F_n)\to GL(n,\mathbb Z_3)$ denote the map in Corollary 7.5.6. The Corollary states that $$PG(F_n)\cap\ker\Psi\subseteq UPG(F_n).$$ The proof of the lemma constructs $\mathcal H_0\le\mathcal H$ and $\Phi:\mathcal H_0\to A$ such that $[\mathcal H:\mathcal H_0]<\infty$ and $\ker\Phi\subseteq PG(F_n)$. Now the subgroup $\mathcal H_1=\mathcal H_0\cap\ker\Psi$ has the required properties. Indeed $[\mathcal H_0:\mathcal H_1]<\infty$ since the codomain of $\Psi$ is finite, ensuring $[\mathcal H:\mathcal H_1]<\infty$. Also $$\ker\Phi|_{\mathcal H_1}=\ker\Phi\cap\ker\Psi \subseteq PG(F_n)\cap\ker\Psi\subseteq UPG(F_n)$$ as required.

• In your definition of $eval_n$, did you mean $a \mapsto a(n)$? – Andrew Kelley May 1 '18 at 15:45
• Also, isn't your example basically the same as mine except that you are dealing with the unrestricted wreath product instead of the restricted wreath product? – Andrew Kelley May 1 '18 at 18:17
• @AndrewKelley Thanks for the correction. Yes, it is essentially the same as your answer, which I didn't see until after I wrote mine. – stewbasic May 1 '18 at 21:51
• I'll have to read this carefully when I'm home,but this looks exactly what I need! – Student May 2 '18 at 7:35

No, it is possible that no such finite index subgroup $A$ of $G$ exists.

First, notice the following about any such group $A$. (This is inspired by Max's comment.) Since $B$ will be the kernel of a homomorphism with domain $A$, it must be normal in $A$. Thus $A$ must be contained in $N_G(B)$, the normalizer of $B$. So if $N_G(B)$ does not have finite index in $G$ then neither does $A$, and hence, we will have a counterexample.

Let $G = \mathbb{Z}/2\mathbb{Z} \wr \mathbb{Z} = (\bigoplus_{i \in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z}) \rtimes \mathbb{Z}$, the lamplighter group. (G is a wreath product, a specific type of semidirect product.) We will represent an element of $G$ as a pair $(g,n)$, where $g \in \mathbb{F}_2[x,x^{-1}]$ is a Laurent polynomial, and $n \in \mathbb{Z}$. The group operation in $G$ is given by $$(g_1,n)(g_2,m) = (g_1 + x^ng_2, n+m).$$ Let $H = \mathbb{Z}$. Then there is an obvious surjection $f : G \to H$ given by $f((g,n)) = n$. So ker $f$ = $\bigoplus_{i \in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z}$.

Consider the following subgroup $B$ of ker $f$: $B = \langle x^n | n \neq 0 \rangle$. $B$ is the set of all Laurent polynomials whose coefficient of $x^0$ is 0. So $B$ has index 2 in ker $f$.

Claim: $N_G(B)$ = ker $f$, which has infinite index in $G$.

Proof: We know that ker $f$ = $\bigoplus_{i \in \mathbb{Z}} \mathbb{Z}/2\mathbb{Z}$ normalizes $B$ because ker $f$ is abelian. Hence $N_G(B) \supseteq$ ker $f$. To show the other containment, we will show that no element of $G$ normalizes $B$ unless it is in ker $f$. Let $(f_1,n) \in G$. Then $(f_1,n)^{-1} = (x^{-n}f_1,-n)$. We just need to show that $(f_1,n)$ does not normalize $B$ if $n \neq 0$. Let $(g,0) \in B$ (so again, $g$ is a Laurent polynomial). We will specify later what $g$ will be.

We have $(f_1,n)(g,0)(x^{-n}f_1,-n) = (f_1 + x^ng,n)(x^{-n}f_1,-n) = (f_1 + x^ng + f_1, 0) = (x^ng,0)$. Notice that if we take $g = x^{-n}$, then we have shown that $(f_1,n)$ conjugates $(g,0)$ to an element outside of $B$ (assuming $n\neq 0$). Therefore assuming $n\neq0$, we have shown that $(f_1,n) \notin N_G(B)$.

Final remark: Notice that in the above counterexample, ker $f$ is not finitely generated as a group. (Regardless, $G$ itself is finitely generated. In fact it is 2-generated.) I was led to the stated example because each of the other examples I tried where ker $f$ was finitely generated, I failed to produce a counterexample.