# Way to Tietze's Transformation Theorem

During our knot-theory lecture we have talking about the following theorem:

Given two finite presentations of the same group, one can be obtained from the other by a finite sequence of Tietze transformations.

This theorem is also given in the book from Gilbert & Porter: Knots and Surfaces. Here they are given a proof with some exercises (I also will write down them here), but I don't come further. Can someone help me to solve the exercises to go further with the proof?! Here the questions:

• Let $P=(X:R)$ and $Q=(X:R\cup S)$ be presentations of the groups $G$ and $H$ resp. with $f:P\rightarrow Q$ given by $f=id:F(X)\rightarrow F(X)$. Show that f is a presentationmapping. What is the corresponding homomorphism $\theta:G\rightarrow H$? When is $\theta$ an isomorphism?

From my point of view this is trivial, true?! the corresponding $\theta$ (it exists) is just an inclusion map and it is an isomorphism if $S$ is a consequence from $R$ ?!

• Let $Y$ be a set, $y\in Y$ and $X=Y-\{y\}$. Suppose $P=(X:R)$ and $Q=(Y:S)$ presentations and $f:F(X)\rightarrow F(Y)$ the homomorphism induced by the inclusion of $X$ into $Y$, then $f(R)\subseteq S$. Describe the corresponding group homomorphism $\theta$. Now suppose $S=f(R)\cup\{yw^{-1}\}$ ($w$ a word in the image of $f$). Prove that $\theta$ is an isomorphism. Hint: Define $g:F(Y)\rightarrow F(X)$ by $g(x)=x$ if $x\in X\subset Y$, g(y)=w. Prove that $g$ a presentationsmapping from $Q$ to $P$ and the corresponding homomorphism is the inverse of $\theta$.

• Let $P=(X:R)$ presentation of a group $G$ and let $\phi:F(X)\rightarrow G$ a epimorphism corresponding to the presentation. Suppose $Z$ a finite set disjoint from $X$. Pick homomorphism $\theta:F(X\cup Z)\rightarrow F(X)$ such tha $\theta(x)=x\ \forall x\in X$. Try to describe a set of normal generators for $Ker(\phi\theta)$. Of course $R\subset Ker(\phi\theta)$ but what other elements do we need? The case $Z=\{y\}$ is a good place to start (look up to the previous exercise)
• What do you mean by a presentation mapping? For the first exercise, it is not true that $\theta$ is an inclusion mapping. In general it is not injective. But you are correct in saying that it is an isomorphism if and only if the relators in $S$ are consequences of those in $R$; that is, if $S \subseteq \langle R^{F(X)} \rangle$. – Derek Holt Mar 27 '13 at 18:25