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I want to compute the presentation for the kernel of $f:G\to(\mathbb{Z},+)$ given by $f(a)=f(b)=0$ and $f(c)=1$, where $G=\langle a,b,c\mid aba=bab,bc=cb\rangle$. It is easy to see that $a^{c^i}:=c^iac^{-i}\in\ker(f)~\forall~i\in\mathbb{Z}$, so we can expect to have $\ker(f)$ generated by $\lbrace b,a^{c^i}\rbrace_{i\in\mathbb{Z}}$. Hence, I claim that the kernel is: $$\langle b,a^{c^{i}},i\in\mathbb{Z}\mid a^{c^{i}}ba^{c^{i}}=ba^{c^{i}}b,i\in\mathbb{Z}\rangle$$

But I dont know whether this is right or not. It seems obvious that the kernel is generated by those elements, since every element $w\in\ker(f)$ I've tried can be obviously written as a product of those elements, however I can't prove it.

Any help will be appreciated.

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  • $\begingroup$ @lhf this is clearly false, every element $w\in G$ where the sum of the powers of $c$ in $w$ is zero is in the kernel. $\endgroup$
    – Marcos
    Apr 17, 2023 at 11:09
  • $\begingroup$ Does $c^3 a c^{-2} a c^{-1}$ fit your claim in the question? $\endgroup$
    – lhf
    Apr 17, 2023 at 11:12
  • $\begingroup$ @lhf It does indeed $c^3ac^{-2}ac^{-1}=c^3ac^{-3}cac^{-1}=a^{c^{-3}}a^{c}$ $\endgroup$
    – Marcos
    Apr 17, 2023 at 11:13
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    $\begingroup$ As you should be able to tell from the comments, it is not clear what you are asking. Please edit the question and make this clear. At the moment you have not even said what $i$ is in your displayed equation. $\endgroup$
    – Derek Holt
    Apr 17, 2023 at 12:45
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    $\begingroup$ You can prove your claim using the Reidemeister-Schreier algorithm. $\endgroup$
    – Derek Holt
    Apr 17, 2023 at 12:56

1 Answer 1

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As I said in my comments, you can do this using the Reidemeister-Schreier algorithm for computing presentations of subgroups of groups defined by a presentation. Let $K = \ker f$ be the subgroup.

To apply the algorithm, we first need a Schreier transversal $T$ of $K$ in $G$, which means a prefix-closed set of words in the group generators that map onto a set of (right) coset representatives of the subgroup. The obvious choice of $T$ in this case is $T = \{c^i : i \in {\mathbb Z} \}$. Formally, we define $c_0 := 1$ (the identity element) and define $c_{i+1} := c_ic$ for $i \ge 0$ and $c_{-i+1} = c_{-i}c^{-1}$ for $ i \ge 0$. Then of course $c_i$ is equal in $G$ to $c^i$.

Now $Kc_ia = Kc_i$ and $Kc_ib=Kc_i$ for all $i \in {\mathbb Z}$, so there are elements $a_i,b_i \in K$ with $c_ia = a_ic_i$ and $c_ib = b_ic_i$ for all $i \in {\mathbb Z}$, and by a well-known result of Schreier, we have $K = \langle a_i,b_i \mid i \in {\mathbb Z}\rangle$. As elements of $G$, we have $a_i = c^iac^{-i}$ and $b_i = c^ibc^{-i}$.

Now we can use the Reidemeister method to calculate a set of defining relations of $K$ on these generators. To do this we apply each relation to each $c_i$ on the right.

For the first relation $aba=bab$, $c_iaba = c_ibab$ yields the relation $a_ib_ia_i = b_ia_ib_i$ of $K$ for each $i$.

For the second relation $bc=cb$, $c_ibc = c_icb$ yields $b_i=b_{i+1}$ for all $i$ so, so after replacing each $b_i$ by $b_0$, we get $$ K \cong \langle a_i\,(i \in {\mathbb Z}),\,b_0 \mid a_ib_0a_i=b_0a_ib_0\,(i \in {\mathbb Z}) \rangle$$ and, since $a_i = c^iac^{-i}$ and $b_0=b$, this is equivalent to your conjectured presentation.

Or, if you want to be formal, the map $a_i \mapsto c^i ac^{-i}\,(i \in {\mathbb Z})$, $b_0 \mapsto b$ induces an isomorphism from group defined by the above presentation onto the subgroup $K = \ker f$ of $G$.

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