Groups

• Jul 18th 2009, 04:45 PM
mms
Groups
Show that if $\displaystyle f:G \to H$ is a homomorphism, $\displaystyle a \in G$
and $\displaystyle f(a)$ has a finite order in H then
$\displaystyle \left| a \right|$ is infinite or $\displaystyle \left| {f(a)} \right|$ divides $\displaystyle \left| a \right|$
thank you.
• Jul 18th 2009, 09:43 PM
NonCommAlg
Quote:

Originally Posted by mms
Show that if $\displaystyle f:G \to H$ is a homomorphism, $\displaystyle a \in G$
and $\displaystyle f(a)$ has a finite order in H then
$\displaystyle \left| a \right|$ is infinite or $\displaystyle \left| {f(a)} \right|$ divides $\displaystyle \left| a \right|$
thank you.

if $\displaystyle |a|=n < \infty,$ then $\displaystyle 1=f(1)=f(a^n)=(f(a))^n$ and thus $\displaystyle |f(a)| \mid n.$
• Jul 19th 2009, 02:46 PM
mms
thank you, but i have a doubt, if f(a) has a finite order does that mean that a has a finite order?
• Jul 19th 2009, 03:05 PM
NonCommAlg
Quote:

Originally Posted by mms
thank you, but i have a doubt, if f(a) has a finite order does that mean that a has a finite order?

not necessarily, but if the order of $\displaystyle a$ is finite, then it's divisible by the order of $\displaystyle f(a).$ (see your original question again!)

example: define $\displaystyle f: \mathbb{Z} \longrightarrow \mathbb{Z}$ by $\displaystyle f(k)=0,$ for all $\displaystyle k \in \mathbb{Z}.$ then $\displaystyle f$ is a homomorphism. let $\displaystyle a=1.$ then $\displaystyle |f(a)|=|0|=1$ but $\displaystyle |a|$ is not finite.