# Show that the kernel of a homomorphism from $G_1*G_2$ to $G_1\times G_2$ is free on $S$

Show that the kernel of $$\omega: G_1*G_2 \to G_1 \times G_2$$ given by $$\omega(g_1g_2)= \iota_1(g_1) \cdot \iota_2(g_2) = (g_1,g_2)$$ is free on $$S=\{ gg'g^{-1}g'^{-1}, 1\neq g \in q_1(G_1), 1 \neq g'\in q_2(G_2)\}$$.

$$G_1 * G_2$$ denoting the free product of groups and $$G_1 \times G_2$$ denoting the direct product of groups.

I've proved in a previous exercise before that $$\ker\omega$$ is the subgroup of $$G_1*G_2$$ generated by $$S'= \{ gg'g^{-1}g'^{-1}, g \in q_1(G_1), g'\in q_2(G_2)\}$$.

To try to prove this, I tried proving the universal property of free groups held here. To do so, I defined $$j: S \to\ker\omega$$ given by $$j(gg'g^{-1}g'^{-1})=gg'g^{-1}g'^{-1}$$.

Then given whatever group $$G$$ and a function $$f: S\to G$$, I define $$\phi:\ker\omega \to G$$ such that $$\phi(j(gg'g^{-1}g'^{-1}))=f(gg'g^{-1}g'^{-1})$$ and then I tried to prove it is a homomorphism and that it is unique. But I stumbled upon some doubts and problems. Firstly, I am not sure if defining $$\phi$$ like that proves its existence and then I had trouble proving it was a homomorphism. I think I am taking the wrong path and I'm quite lost, any help or advice is appreciated, thanks.

• Use $\ker \omega$ for $\ker \omega$.
– Shaun
Commented Jan 13, 2022 at 21:17
• Try proving no nontrivial relation can hold among these, by showing that any reduced word in them cannot vanish. Commented Jan 14, 2022 at 3:26

You need to show that a non-empty word $$u_1u_2 \cdots u_k$$ with each $$u_i \in S \cup S^{-1}$$ and no $$u_i = u_{i+1}^{-1}$$ cannot reduce to the identity.
Reduction can only occur between $$u_i$$ and $$u_{i+1}$$ with $$u_i \in S$$ and $$u_{i+1} \in S^{-1}$$ or vice versa. In that case, the condition $$u_i \ne u_{i+1}^{-1}$$ means that the last of the four letters of $$u_i$$ could cancel with the first of those of $$u_{i+1}$$, but then no further cancellation can occur. So the middle two letters of $$u_i$$ cannot cancel, and hence the word cannot reduce to th identity.