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I looked up for a similar question but I haven't found any. If there is any, I can delete the question.

How can I show that the following theorem?

Let $G$ be a group, $S$ a set of generators for $G$, and $G'$ another group. Let $f:G \rightarrow G'$ be a map. If there exists a homomorphism $\bar{f}$ of $G$ into $G'$ whose restriction to $S$ is $f$, then there is only one.

So I assumed that there are more than one $\bar{f}$, took $f_1$ and $f_2$. I know that $f_1(S) = f_2(S) = f(S).$ I took an element $g \in S$, write it as $g=x_1^{\epsilon_1}\dots x_n^{\epsilon_n}$ where each $x_i \in S$ and $\epsilon_i =\pm1$ for each $i\in \{1,\dots,n\}$ since $S$ is a set of generators for $G$.

Then, I said

$f_1(g)=f_1(x_1^{\epsilon_1}\dots x_n^{\epsilon_n})= f_1(x_1)^{\epsilon_1}\dots f_1(x_n)^{\epsilon_n} = $

$f_2(g)=f_2(x_1^{\epsilon_1}\dots x_n^{\epsilon_n}) = f_2(x_1)^{\epsilon_1}\dots f_2(x_n)^{\epsilon_n}$ but I stuck here. Should I pick $g$ from $G$? I did not pick it from $G$ because I can not guarantee that $f_1(g) = f_2(g)$ in that case.

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In effect you did pick $g\in G$: every $g\in G$ can be written in the form $x_1^{\epsilon_1}\ldots x_n^{\epsilon_n}$ with each $x_k\in S$ and $\epsilon_k\in\{-1,1\}$. Then

$$f_1(g)=\prod_{k=1}^nf_1\left(x_k^{\epsilon_k}\right)=\prod_{k=1}^n\big(f_1(x_k)\big)^{\epsilon_k}=\prod_{k=1}^n\big(f_2(x_k)\big)^{\epsilon_k}=\prod_{k=1}^nf_2\left(x_k^{\epsilon_k}\right)=f_2(g)\;.$$

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  • $\begingroup$ How can we write $\prod_{k=1}^n\big(f_1(x_k)\big)^{\epsilon_k}=\prod_{k=1}^n\big(f_2(x_k)\big)^{\epsilon_k}$? Is it because of $S$ is a generator set? $\endgroup$ – Ninja Nov 27 '16 at 4:11
  • $\begingroup$ @Ninja: Because by hypothesis $f_1\upharpoonright S=f_2\upharpoonright S$, so $f_1(x)=f_2(x)$ for every $x\in S$. $\endgroup$ – Brian M. Scott Nov 27 '16 at 4:13
  • $\begingroup$ I see, thank you very much! $\endgroup$ – Ninja Nov 27 '16 at 4:15
  • $\begingroup$ @Ninja: You’re very welcome! $\endgroup$ – Brian M. Scott Nov 27 '16 at 4:17
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You are essentially done. You just have to recall that $f_1$ and $f_2$ agree on $S$, hence they agree on all the $x_i$. Therefore $f_1(g) = f_2(g)$.

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  • $\begingroup$ So since $S$ is a generator set, then $f_1(S) = f_2(S) \implies f_1(G) =f_2(G)$. Then we are done. Is that all we have to write? $\endgroup$ – Ninja Nov 27 '16 at 4:09
  • $\begingroup$ Write everything you wrote in your original post, then conclude that $f_1(g) = f_2(g)$ since $f_1$ and $f_2$ agree on the $x_i$ $\endgroup$ – Ethan Alwaise Nov 27 '16 at 4:10
  • $\begingroup$ I am sorry, does agreeing on something mean $f_1$ and $f_2$ have the same image? $\endgroup$ – Ninja Nov 27 '16 at 4:13
  • $\begingroup$ Yeah I meant $f_1(x_i) = f_2(x_i)$ for each $x_i$. $\endgroup$ – Ethan Alwaise Nov 27 '16 at 4:13
  • $\begingroup$ I understood. Thank you for all the help! $\endgroup$ – Ninja Nov 27 '16 at 4:16

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