Show that if $H$ is any group and $h\in H$ and $|h|=n$ then there is a homomorphism from $\mathbb{Z}_n=\langle x\rangle$ to H such that $x\mapsto h$

Question

My attempt:

$h^n=1$ (given). Therefore $\{1,h^{2},h^{3},\ldots,h^{n-1}\}$ are distinct elements ....(1).

$f(x)=h$ is a mapping. $f(x^{a})=h^{a}$ and $f(x^{a})=h^{b}$ where $0\leq a,b\lt n$ where $a$ and $b$ are not equal. This cannot happen as $h^{a}$ not equal to$ h^{b}$ by(1).

Hence it is well defined. $f(x^{a+b})=h^{a}.h^{b}=f(x^{a}).f(x^{b})$ it is a homomorphism.

Is this right ?I am confused on well definedness

Answer 1

Given $|h|=n$, therefore $\{1,h,h^2,\dots, h^{n-1}\}$ are distinct. We define the map $f:\mathbb{Z}\rightarrow H$ by $ z \rightarrow h^z$, clearly a homomorphism. Furthermore, ${\rm Ker}(f)=n\mathbb{Z}$, and by the first isomorphism theorem we have $\mathbb{Z}/n \cong \{1,h,h^2,...,h^{n-1}\}$, as required.

Answer 2

There are several problems with your presentation.

First, while $|h|=n$ does imply that $h^n=1$, it is false that $h^n=1$ implies that the elements $1,h,\ldots,h^{n-1}$ are distinct. That latter conclusion follows from $|h|=n$, not from $h^n=1$.

Second, $H$ is not necessarily equal to $\{1,h,\ldots,h^{n-1}\}$; though the latter is equal to $\langle h\rangle$.

Third: your assertion that $f(x^{a+b})=f(x^a)f(x^b)$ is unjustifed: it only works if $0\leq a+b\lt n$. What if $a+b\geq n$? That is precisely the crux of the matter, and you completely ignore it.