Given $N= \langle n_i|r_j \rangle$ and $G/N= \langle g_k|s_l \rangle$, how do we prove $G$ has a finite presentation?

We know that $G$ is f.g. by $\{n_i,g_k\}$ (I am being sloppy about directly lifting the generators from $G/N$ to $G$ of course) and we know the relations $r_j$ will hold in $G$ since $N$ is a subgroup of $G$. How do we get the other relations?

I imagine there has to be something with conjugates of elements in $N$, i.e. some relations like $g_kn_ig_k^{-1}=\ldots$ but I can't quite see what they should be.

  • $\begingroup$ That's right, the other relations are of the form $g_kn_ig_k^{-1} = w_{k,i}(n_i)$, one for each $i$ and $k$, for suitable words $w_{k,i}(n_i)$ in the $n_i$, and also $s_l = w_l(n_i)$, one for each $s_l$. $\endgroup$ – Derek Holt May 29 '14 at 10:56
  • $\begingroup$ @DerekHolt sorry, i think i'm getting confused by the notation, when you write $w_{k,i}(n_i)$, this means $w_{k,i}(n_1,n_2,\ldots)$ ? i.e. a word specific word in $N$ based on the $n_i,g_k$. How do I determine these two word functions $w_{k,i}$ and $w_l$? What do you mean by suitable choice? $\endgroup$ – user154035 May 29 '14 at 11:45
  • $\begingroup$ Related: math.stackexchange.com/questions/375589/… $\endgroup$ – user1729 May 29 '14 at 12:31
  • $\begingroup$ @user1729 I already have the finite generation part done. I am asking more about the relations. $\endgroup$ – user154035 May 29 '14 at 12:37
  • $\begingroup$ @user154035 I know, but I figured that if someone comes along later then they might be stuck at that bit. $\endgroup$ – user1729 May 29 '14 at 12:41

Suppose that the group $G$ has a normal subgroup $N$, and that we have presentations $\langle\, Y \mid S\, \rangle$ of $N$ and $\langle\, \overline{X} \mid \overline{R}\, \rangle$ of $G/N$ on generating sets $Y$ and $\overline{X}$, respectively. Here we shall describe a general recipe for constructing a presentation of $G$ as an extension of $N$ by $G/N$.

For each $\overline{x} \in \overline{X}$, choose $x \in G$ with $xN = \overline{x}$, and let $$X := \{\, x \mid \overline{x} \in \overline{X}\,\}.$$ Then, for any word $\overline{w} \in (\overline{X} \cup \overline{X}^{-1})^*$, we can define $w \in (X \cup X^{-1})^*$ with $wN = \overline{w}$, by substituting $x$ or $x^{-1}$ for each $\overline{x}$ or $\overline{x}^{-1}$ occurring in $\overline{w}$.

In particular, for each $\overline{r} \in \overline{R}$ there is a corresponding word $r$, and then $\overline{r} = 1_{G/N}$ implies that $r \in N$, so in the group $G$ we have $r =_G w_r$, for some word $w_r \in (Y \cup Y^{-1})^*$. Let $R$ be the set $\{\, rw_r^{-1} \mid \overline{r} \in \overline{R}\,\}$.

For each $y \in Y$ and $x \in X$, we have $x^{-1}yx \in N$, so $x^{-1}yx =_G w_{xy}$ for some word $w_{xy} \in (Y \cup Y^{-1})^*$. Let $T$ be the set $\{\, x^{-1}yxw_{xy}^{-1} \mid x \in X,\,y \in Y \,\}$.

Proposition. With the above notation, $\langle\, X \cup Y \mid R \cup S \cup T\,\rangle$ is a presentation of $G$.

Proof.Let $F$ be the group defined by the above presentation. To avoid confusion, let us denote the generators of $F$ mapping onto $x \in X$ or $y \in Y$ by $\hat{x}$ and $\hat{y}$, respectively. We have chosen the sets $R,S,T$ to be words that evaluate to the identity in $G$, so the mapping $\hat{x} \to x$, $\hat{y} \to y$ induces a homomorphism $\theta:F \to G$. In fact $\theta$ is an epimorphism, because clearly $G$ is generated by $X \cup Y$.

Let $K$ be the subgroup $\langle\, \hat{y} \mid y \in Y\, \rangle$ of $F$. Then $\theta(K) = N$. Since, by assumption, $\langle\, Y \mid S\, \rangle$ is a presentation of $N$, and the relators of $S$ are also relators of $K$, it follows that the map $y \to \hat{y}$ induces a homomorphism $N \to K$. But this homomorphism is then an inverse to $\theta_K$, so $\theta_K:K \to N$ is an isomorphism.

We now wish to assert that the relators in $T$ imply that $K \unlhd F$. This is not quite as clear as it may at first sight appear! For it to be clear, we would need to know that $\hat{x} \hat{y} \hat{x}^{-1} \in K$ for all $x \in X, y \in Y$, in addition to $\hat{x}^{-1} \hat{y} \hat{x} \in K$. But the fact that $N \unlhd G$ tells us that each $x \in X$ induces an automorphism of $N$ by conjugation, which implies that, for each $x \in X$, we have $N = \langle\, w_{xy} \mid y \in Y\, \rangle$. Since $N$ is isomorphic to $K$ via $\theta_K$, the corresponding statement is true in $K$ for each $\hat{x}$. So the fact that $\hat{x}^{-1}$ conjugates each word $w_{xy}$ into $K$ implies the desired property $\hat{x} \hat{y} \hat{x}^{-1} \in K$ for all $x \in X, y \in Y$.

So we do indeed have $K \unlhd F$. Now, by a similar argument to the one above for $\theta_K$, the fact that $\langle\, \overline{X} \mid \overline{R}\, \rangle$ is a presentation of $G/N$ implies that the induced homomorphism $\theta_{F/K}:F/K \to G/N$ is an isomorphism. So, if $g \in \ker(\theta)$, then $gK \in \ker (\theta_{F/K})$ implies that $g \in K$, but then $g \in \ker(\theta_K) = 1$, so $\theta$ is an isomorphism, which proves the result.


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