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### Problem 9

(+) Let $G$ be a group that is generated by the set $\{a_i|i\in I\}$ where $I$ is some indexing set and $a_i\in G$ for all $i\in I$. Let $\phi:G\rightarrow G'$ and $\mu: G\rightarrow G'$ be two homomorphisms from $G$ into a group $G'$ such that $\phi(a_i)=\mu(a_i)$ for all $i\in I$.Prove that $\phi=\mu$. (This shows that any homomorphism of $G$ is determined by its action on the generators of $G$.)

**Solution**

Let $G=\{a_1^{n_1}a_2^{n_2}\dots a_i^{n_i}|i\in I,n_i\in\mathbb Z^+\cup\{0\}\}$.(Since $G$ is generated by $\{a_i|i\in I\}$).

Because $\phi: G\rightarrow G'$ and $\mu:G\rightarrow G'$ are two homomorphisms,then

$\forall x\in G$,we have

$\phi(x)=\phi(a_1^{n_1}a_2^{n_2}\dots a_i^{n_i})$

$=\phi(a_1^{n_1})\phi(a_2^{n_2})\dots \phi(a_i^{n_i})$

$=[\phi(a_1)]^{n_1}[\phi(a_2)]^{n_2}\dots [\phi(a_i)]^{n_i}$

and

$\mu(x)=\mu(a_1^{n_1}a_2^{n_2}\dots a_i^{n_i})$

$=\mu(a_1^{n_1})\mu(a_2^{n_2})\dots \mu(a_i^{n_i})$

$=[\mu(a_1)]^{n_1}[\mu(a_2)]^{n_2}\dots [\mu(a_i)]^{n_i}$

by the homomorphism property.

Because $\phi(a_i)=\mu(a_i)$ for all $i\in I$,

$\phi(x)=\phi(a_1^{n_1}a_2^{n_2}\dots a_i^{n_i})$=$\mu(x)=\mu(a_1^{n_1}a_2^{n_2}\dots a_i^{n_i})$

That is,$\forall x\in G$, $\phi(x)=\mu(x)$.

Hence,$\phi=\mu$.