Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 104 / 1

Because the generating sets are disjoint, the free product has the presentationThis is immediate from the universal property of a group presentation: a map out of this presented group is exactly a pair of homomorphisms from $G$ and $H$ to the target group.

A reduced word is either the empty word or a productin which every syllable $u_i$ is a nonidentity element of $G$ or $H$, and consecutive syllables belong to different factors.

Let $\mathcal{W}$ be the set of reduced words. Each $g\in G$ acts on the right of $\mathcal{W}$: if the last syllable lies in $H$, append $g$; if it lies in $G$, multiply it by $g$ and delete it when the product is the identity. Define the action of each $h\in H$ analogously. These rules give genuine actions of the two factors by permutations of $\mathcal{W}$, hence an action of the free group $F(X\bigsqcup Y)$. Every relation in $R\bigsqcup S$ acts trivially, so the action factors through the displayed presentation of $G\ast H$.

The element represented by a reduced word $w$ sends the empty word to $w$. Therefore two reduced words representing the same element induce the same permutation and have the same value on the empty word. They must be identical. This proves the normal form theorem for a free product.

Let $w$ be a nonempty reduced word. If its first syllable belongs to $G$, choose any nonidentity $h\in H$. The reduced form of $hw$ begins with an $H$-syllable, whereas that of $wh$ begins with the original $G$-syllable; reduction at the right end cannot change the first syllable. Hence $hw\ne wh$. The same argument with a nonidentity $g\in G$ applies when $w$ begins in $H$. No nonidentity element is therefore central, and

Choose distinct nonidentity elements $g_1,g_2\in G$ and $h_1,h_2\in H$, and putExpand a freely reduced word in $u^{\pm 1},v^{\pm 1}$. At a boundary where two syllables from one factor meet, their product is one of $g_1^{-1}{g}_2$, $g_2^{-1}{g}_1$, $h_1{h}_2^{-1}$, or $h_2{h}_1^{-1}$, all nonidentity by the choices above. Every other boundary already alternates between the factors. Thus the expansion reduces to a nonempty reduced word in $G\ast H$ and cannot represent the identity. The homomorphism from the rank-two free group sending its free generators to $u,v$ is injective, so