Existence of Unique Homomorphism Implies Generating Set

Question

The following question is taken from "An Invitation to General Algebra and Universal Constructions", p. 23 Ex. 2.1.2 (available online here).

Let $G$ be a group and let $\{a,b,c\}\subseteq G$ such that for every group $F$ and every $\{x,y,z\}\subseteq F$ there is a unique homomorphism: \begin{array}{c c c l} h: & G & \rightarrow & F \\ & a & \mapsto & x \\ & b & \mapsto & y \\ & c & \mapsto & z \end{array}

Prove that $G$ is generated by $a,b,c$.

Thank you!

Answer 1

Let $H$ be the subgroup of $G$ generated by $a,b,c$. Then by assumption, there is a unique homomorphism $\phi\colon G\to H$ sending $a\mapsto a$, $b\mapsto b$, $c\mapsto c$. Together with the inclusion $\iota\colon H\to G$, we get a homomorphism $G\to G$ that sends $a\mapsto a$, $b\mapsto b$, $c\mapsto c$. Of course $\operatorname{id}_G$ is also such a homomorphism. By uniqueness, these are the same, hence have the same image, i.e. $H=G$.

Answer 2

Here's an idea to get you on the right track. Suppose some $g \in G$ cannot be formed from $a$, $b$, and $c$. Can you think of multiple homomorphisms from $G$ to itself that act as the identity on $a$, $b$, and $c$?