Past exam of the mathematics course of the University of Cambridge / 2016 / iii / Paper 104 / 5 / d

For $M>a\ge 0$ and $N>b\ge 0$, defineThe normal form theorem for a free product giveswhere $X,Y,T$ map to $x^M,y^N,t(a,b)$. To see injectivity, expand an alternating word in $⟨x^M,y^N⟩$ and powers of $t(a,b)$. The conjugating $x,y$ factors commute with the lattice syllables, so each intervening nonidentity lattice syllable stays nonidentity in the expanded free-product normal form.

Its intersection with $T_K$ is the free group with basisThis follows by moving all $x^M,y^N$ factors to the right; the zero-exponent lattice remainder characterizes membership in the kernel.

The required isomorphisms are specified on the three generators byandEach source and target has the same presentation $⟨X,Y,T\mid [X,Y]=1⟩$, and each displayed assignment identifies its three abstract generators. Hence these are genuine isomorphisms, not merely maps of generating sets.

On the kernel basis they give exactly the machine transitions:For example, rewrite the first input as $x^{-mu}{y}^{-mv}t(a_i,b_i)x^{mu}{y}^{mv}$ and apply ${\varphi }_i$ to each factor. These identities also hold for negative $u,v$, since they are identities in the groups.