1
$\begingroup$

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

$\endgroup$
2
$\begingroup$

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 $G$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.