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 $m$.

  • $\begingroup$ Thank you so much! It's amazing how much clearer something can be when simply reworded! Could they have omitted the first part though? It seems to me like solely including the trivial counterexample should suffice. $\endgroup$ – John Arg Apr 23 at 22:11
  • 1
    $\begingroup$ 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 $m$. $\endgroup$ – Mark Apr 23 at 22:17

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