Consider the subset of $SL_3(\Bbb Z)$ consisting of matrices of the form

$$\begin{pmatrix} 1 & a & c\\ 0 & 1 & b \\ 0 & 0 & 1 \end{pmatrix}$$

for $a,b,c\in\Bbb Z$.

I need to show that it admits the presentation:

$$\langle A, B, C\mid AC=CA, BC=CB, ABA^{-1}B^{-1}=C\rangle.$$

I am not sure what exactly I need to show. I tried to follow the definition given in class, but didn't really understand it.

Definition 1.36: Let $G$ be a group and $S$ a generating set. Let $R\subseteq F(S)$. Denote by $\pi$ the morphism $F(S)\to G$. We say that $G$ admits the presentation $\langle S\mid R\rangle$ if $R$ normally generates $\ker \pi$, that is, if $\ker\pi$ is the smallest normal subgroup containing $R$.

Remark 1.38: We can also build a group with any presentation we choose: given a set $S$ and a set $R$ of words in $S$, the quotient $F(S)/\langle\langle R\rangle\rangle$ obviously admits the presentation $\langle S\mid R\rangle$. Note that this is a way to specify a group algebraically, not geometrically.

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    $\begingroup$ Here's the steps. (1) Choose the correct matrices $A,B,C$ of the form given. (2) Prove the three equations given in the presentation. (3) Prove that any equation in the letters $A,B,C$ and their inverses that holds in the group may be derived formally from the three equations given. How far have you gotten? $\endgroup$ – Lee Mosher Dec 5 '16 at 13:46
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    $\begingroup$ That's the hard part! For example, I'm sure your choice of $A,B,C$ satisfies the equation $A C B C^{-1} A^{-1} B^{-1} = C$; and you can see that this equation is a consequence of the 2nd and 3rd equations in the presentation, by making a substitution. You have to prove that for every one of the infinitely many equations in the symbols $A,B,C,A^{-1},B^{-1},C^{-1}$ that holds in the group, that equation is a consequence of the three equations given, by a sequence of substitutions. $\endgroup$ – Lee Mosher Dec 5 '16 at 14:35
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    $\begingroup$ The way I would do this is to show, using the relations of the presentation, that every element in the group defined by the presentation can be represented by a word of the form $A^iB^jC^k$ for $i,j,k \in {\mathbb Z}$. Then show that these elements correspond to distinct elements of the Heisenberg group. So we have an isomorphism between the two groups. $\endgroup$ – Derek Holt Dec 5 '16 at 16:37
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    $\begingroup$ OK So I Set X={A,B,C} and consider F(X) of X There is a homomorphisim phi:F(X)->H Do I need to show now that R={ACA^-1C^-1,BCB^-1C^-1,....} is a normal subgroup of Ker(phi) $\endgroup$ – matan Dec 5 '16 at 19:02
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    $\begingroup$ Or use that the Heisenberg group is semidirect product of $Z^2$ and $Z$, where the rank 2 subgroup is normal. $\endgroup$ – Moishe Kohan Dec 5 '16 at 20:31

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