homomorphism

• September 28th 2008, 08:19 PM
dori1123
homomorphism
Show that if H is any group and $h$ is an element of $H$ with $h^n = 1$, then there is a unique homomorphism from $Z_n = $ to $H$ such that $x$ --> $h$.

Show that if H is any group and $h$ is an element of $H$ with $h^n = 1$, then there is a unique homomorphism from $Z_n = $ to $H$ such that $x$ --> $h$.
define $f: \longrightarrow H$ by $f(x^k)=h^k, \ \forall k.$ it's obvious that $f$ is a homomorphism and $f(x)=h.$ you only need to prove that $f$ is basically well-defined:
if $x^k=1,$ then $k=mn,$ for some integer $m.$ thus: $f(x^k)=h^k=(h^n)^m=1.$ hence $f$ is well-defined. this proves the existence of such homomorphism.
now suppose $g: \longrightarrow H$ is another homomorphism with $g(x)=h.$ then for all $k: \ g(x^k)=(g(x))^k=h^k.$ hence $g=f,$ which proves the uniqueness.