# Groups

• Jul 18th 2009, 04:45 PM
mms
Groups
Show that if $
f:G \to H

$
is a homomorphism, $
a \in G

$

and $
f(a)

$
has a finite order in H then
$
\left| a \right|

$
is infinite or $
\left| {f(a)} \right|

$
divides $
\left| a \right|

$

thank you.
• Jul 18th 2009, 09:43 PM
NonCommAlg
Quote:

Originally Posted by mms
Show that if $
f:G \to H

$
is a homomorphism, $
a \in G

$

and $
f(a)

$
has a finite order in H then
$
\left| a \right|

$
is infinite or $
\left| {f(a)} \right|

$
divides $
\left| a \right|

$

thank you.

if $|a|=n < \infty,$ then $1=f(1)=f(a^n)=(f(a))^n$ and thus $|f(a)| \mid n.$
• 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 $a$ is finite, then it's divisible by the order of $f(a).$ (see your original question again!)

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