Determine whether the following statement is true or false: "Suppose $f : \{G, *\} \mapsto \{H, \circ \}$ is a homomorphism of groups, and let $f(G) = \{f(g)~|~g\in G\}$. If $g\in G$ has finite order $|g|=m$ then $f(g)$ has order $m$ in $H$."
The solution of the book is as follows:
The statement is false in general.
As $|g|$ = m, $g^m = e_G$
$\implies f(g^m) = f(e_G)$
$\implies f(g*g*g*....*g) = e_H$ where $g$ is repeated $m$ times on the left hand side
$\implies f(g) \circ f(g) \circ f(g) \circ ....\circ f(g) =e_H$ where $f(g)$ is repeated $m$ times on the left hand side
$\implies (f(g))^m = e_H$
Now here it seems to me like they are proving that the statement is true...
The solution then goes on to say:
$\implies$ order of $f(g)$ is a factor of $m$
Why is the order of $f(g)$ a factor of $m$ and not $m$ itself? Is that because the group is cyclic?
The next line is:
Also $f:\{G,*\} \mapsto \{G,*\}$ where $f(g)=e_G$ is a homomorphism and $|f(g)|=|e_G|=1$ for all $g \in G$.
So how do they get this new codomain of $f$? And how do they get $f(g)=e_G$?
Finally, it ends with:
But $|g| \not = 1$ for all $g \in G$ if $G \not = \{e_G\}$
Now though this line is evidently true, what is its relevance?
And overall, how does this solution prove anything?
Showing that $(f(g))^m=e_H$ is not enough to conclude that $m$ is the order of $f(g)$. That doesn't prove that $m$ is the least natural number $k$ such that $(f(g))^k=e_H$. So we only know that the order of $f(g)$ divides $m$, nothing more. And this is what they proved.
And then they just gave an example that shows that the order of $f(g)$ might not be equal to $m$. Define the trivial homomorphism $f:G\to G$ by $f(g)=e_G$ for all $g\in G$. This is just a definition. Take any element $g\ne e_G$ and let's say it has order $m$. Since $g$ is not the identity we know that $m>1$. However $|f(g)|=|e_G|=1$. So the order of $f(g)$ is not $m$.