# Determine whether the statement is true: “If $g\in G$ has finite order $|g|=m$ then $f(g)$ has order $m$ in $H$.”

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 $$m$$.
• Yes, the counterexample is a solution to the original problem. But I guess they also wanted to show that the order of $f(g)$ in $H$ must divide $m$. – Mark Apr 23 at 22:17