Thread: homomorphism

Show that if H is any group and $\displaystyle h$ is an element of $\displaystyle H$ with $\displaystyle h^n = 1$, then there is a unique homomorphism from $\displaystyle Z_n = <x>$ to $\displaystyle H$ such that $\displaystyle x$ --> $\displaystyle h$.

Please help. Thank you.

Originally Posted by dori1123

Show that if H is any group and $\displaystyle h$ is an element of $\displaystyle H$ with $\displaystyle h^n = 1$, then there is a unique homomorphism from $\displaystyle Z_n = <x>$ to $\displaystyle H$ such that $\displaystyle x$ --> $\displaystyle h$.

define $\displaystyle f: <x> \longrightarrow H$ by $\displaystyle f(x^k)=h^k, \ \forall k.$ it's obvious that $\displaystyle f$ is a homomorphism and $\displaystyle f(x)=h.$ you only need to prove that $\displaystyle f$ is basically well-defined:

if $\displaystyle x^k=1,$ then $\displaystyle k=mn,$ for some integer $\displaystyle m.$ thus: $\displaystyle f(x^k)=h^k=(h^n)^m=1.$ hence $\displaystyle f$ is well-defined. this proves the existence of such homomorphism.

now suppose $\displaystyle g: <x> \longrightarrow H$ is another homomorphism with $\displaystyle g(x)=h.$ then for all $\displaystyle k: \ g(x^k)=(g(x))^k=h^k.$ hence $\displaystyle g=f,$ which proves the uniqueness.