Let $G$ and $K$ be groups and $f:G\rightarrow K$ be a group
homomorphism. Suppose $g\in G$ has finite order $n$. Prove that
$\mathrm{ord}(f(g))$ is finite and divides $n$.
How can I fix this proof?
We know $g^n=e$. Since $f$ is a homomorphism, then $[f(g)]^n = f(g^n)=f(e)=e$. Let $k\mid n$ and $j\mid n$. Then we can choose the smallest $k$ such that $[f(g)^k]^j=[f(g)]^n$. Thus, $\mathrm{ord}(f(g))=k$ where $k\mid n$.
This logic seems shaky. How can I improve it? Thanks!
Best Answer
I like to see the order of an element $g\in G$ as the cardinality of $$ \langle g\rangle=\{g^n:n\in\mathbb{Z}\} $$ (a natural number in case it's finite, infinity otherwise).
If you know this characterization, then you can observe that $f$ induces a surjective homomorphism $\tilde{f}\colon\langle g\rangle\to\langle f(g)\rangle$ and so, by the homomorphism theorem $$ \langle f(g)\rangle\cong \langle g\rangle/\ker\tilde{f} $$ which implies your assert, by Lagrange's theorem.
Now let's prove the characterization of the order. Consider the unique homomorphism $\varepsilon_g\colon \mathbb{Z}\to G$ such that $\varepsilon_g(1)=g$, that is, $\varepsilon_g(n)=g^n$. Then the image of $\varepsilon_g$ is precisely $\langle g\rangle$ and so $$ \langle g\rangle\cong \mathbb{Z}/\ker\varepsilon_g $$ If $\ker\varepsilon_g=\{0\}$, then $\varepsilon_g$ is injective, so $g^n\ne1$ for $n>0$, which means that $g$ has infinite order and that $\langle g\rangle$ is infinite. Otherwise, $\ker\varepsilon_g=k\mathbb{Z}$ with $k>0$.
In this case $|\langle g\rangle|=|\mathbb{Z}/k\mathbb{Z}|$. Moreover $g^k=1$, as $k\in\ker\varepsilon_g$. To finish off, observe that, for $0<r<k$, $r\notin\ker\varepsilon_g$, so $g^r\ne1$. Thus $k$ is indeed the minimum exponent $m>0$ such that $g^m=1$.