Let $F(S)$ be a free group with finite rank, $G$ a group with order $n$.

I want to know if the universal property can help determine the number of homomorphisms $F(S)\rightarrow G$. Say $H$ is the number of certain homomorphisms.

By the universal property, let $\varphi:S\rightarrow G$ be a set map, then there exists a unique homomorphism $\Phi(s_1^{a_1}s_2^{a_2}\cdots)=\varphi(s_1)^{a_1}\varphi(s_2)^{a_2}\cdots$.

There are $n^{|S|}$ set maps $\varphi$, hence $H\geq n^{|S|}$ (Is this true even when $G$ is not simple? If this is not always true, then when?).

(When) Can we say that $H=n^{|S|}$?

In the worst case I holp it at least works for $G=\mathbb{Z}/2\mathbb{Z}$.

-- Thanks


As $\text{Hom}(F(S),G)$ is in natural correspondence with $\text{Map}(S,G)$, they are equinumerous: there are $|G|^{|S|}$ homomorphisms from $F(S)$ to $G$.


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