Show that cyclic group of finite order $n$ is isomorphic to $(\mathbb Z_n,+)$
Actually, the answer to this question is available in MSE but I am not getting it,and one answer is not in the English language.Please kindly give a proof in easy language. Thank you.\
Here is a solution I don't understand: Proving that any finite cyclic group of order $n$ is isomorphic to $<Z_{n},+_{n}>$
Let $G$ be a cyclic group of order $n$ generated by $g$.
The map $\phi: k \mapsto g^k$ is a surjective group homomorphism $\mathbb Z \to G$ whose kernel is $n\mathbb Z$.
Therefore, $G = \operatorname{im} \phi \cong \mathbb Z / \ker \phi = \mathbb Z / n\mathbb Z$.
In a group, we define "powers" recursively, that is, we set, for $g \in G$:
$g^0 = e_G$ (this is a definition, it's best not to worry about what a "zeroth power" is). Think of it as a seed value for this "power thing".
The recursive step is this:
$g^{k+1} = g^k\ast g$. This is another defintion, so to evaluate say, $g^4$, you have to do something like this:
$g^4 = g^3 \ast g\\ g^3 = g^2 \ast g\\ g^2 = g \ast g$
and then evaluate backwards to find, in turn, $g\ast g, (g\ast g)\ast g,$ and finally:
$g^4 = ((g\ast g)\ast g)\ast g$.
Note that so far, we've only defined $g^k$ for $k \in \Bbb N$, and so we have to also define $g^{-k}$ for $k \in \Bbb N$, to get "negative powers". We do this, by setting:
$g^{-k} = (g^k)^{-1}$ (which we can do, since inverses exist in a group).
Having done this, we can use induction (the grandfather of all recursive processes) to prove things like:
$g^k\ast g^m = g^{k+m}$ (I really recommend you try to carry out this proof, it's enlightening), for all $k,m \in \Bbb Z$.
Now note what is happening here: we've taken the "action" of $\ast$ in the "down" part of $g^k$, and moved it to a different "action" in the "up" part-that is, we've "translated" (for powers of $g$ in $G$) the group multiplication of $G$ into something we understand better in the exponents, addition of integers. Mathematicians like to call (for perfectly understandable reasons) something that takes a multiplication, and turns it into addition, an exponential. This is actually a really important idea (that occurs in many different branches of math).
Note that we can now define a map, for any $g \in G$ by: $\phi: \Bbb Z \to G$, by:
$\phi(k) = g^k$.
If $G$ is a finite cyclic group, with $g$ a generator, then this map is surjective (onto), since powers of $g$ are all that is in $G$, and thus everything in $G$ is some $\phi(k)$, for some integer $k$.
Now $\phi$ is not just "any" function, it is a homomorphism, since:
$\phi(k+m) = g^{k+m} = g^k\ast g^m = \phi(k)\ast \phi(m)$ (note this is a kind of "reverse" exponential; that is, it acts like a logarithm function). Now since $\Bbb Z$ is infinite, and $G$ is finite, it's pretty clear this function is "many-to-one", so $G$ is not isomorphic to the integers. But we're close.
We can partition $\Bbb Z$ by the pre-images of $\phi$, that is to say define an equivalence relation on $\Bbb Z$, by:
$k \sim_{\phi} k' \iff \phi(k) = \phi(k')$.
This induces a bijection $[\phi]: \Bbb Z/\phi \to G$, by:
$[\phi]([k]_{\phi}) = \phi(k)$. This is a well-defined function on $\Bbb Z/\phi$ because $\phi$ is constant on our equivalence classes, so it doesn't matter which element of the equivalence class of $k$ we pick (since they all have the same image).
I leave it to you to show that: $[k]_{\phi} = [k']_{\phi}$ (that is $\phi(k) = \phi(k')$, or $g^k = g^{k'}$) if and only if $k \equiv k'\text{ mod }n$, where $n = |g| = |G|$. Here is a hint: if $k \equiv k'\text{ mod }n$, then $k = k' +tn$, for some integer $t$, and thus $g^k = g^{k' + tn} = g^{k'}\ast g^{tn} = ?$
What this tells us, is that $\Bbb Z/\phi = \Bbb Z_n$ (so $[k]_{\phi} = [k]_n$, they are the same subset of $\Bbb Z$). So at least we have a bijection between $\Bbb Z_n$ and $G$.
Finally, the "easy part": we can define $+_n$ on $\Bbb Z_n$, by:
$[k]_n +_n [m]_n = [k+m]_n$. We'll take it as given you already know this is a well-defined binary operation on $\Bbb Z/\phi = \Bbb Z_n$. Now we show that $[\phi]$ is thus a homomorphism, and you're done:
$[\phi]([k]_n +_n [m]_n) = [\phi]([k+m]_n) = \phi(k+m)$
$ = g^{k+m} = g^k \ast g^m = \phi(k) \ast \phi(m) = [\phi]([k]_n) + [\phi]([m]_n)$.
Let $G$ be a cyclic and $|G|=n$. If $G=<g>$ then $G=\{e,g,g^2,...,g^{n-1}\}$.
The mapping $f:G\to \Bbb (Z_n,+)$ defined by $f(g^k)=k$ is an isomorphism.
The idea is simple, map a generator to a generator.
First just a note, I assume that by $\mathbb Z_n$ you mean $\mathbb Z/n\mathbb Z$ (The integers mod $n$).
Let $G$ be your cyclic group of order $n$. By definition there exists $x \in G$ so that $x$ generates $G$. From here you should show that we can defined an isomorphism from $G$ to $\mathbb Z/n\mathbb Z$ by sending $x$ to 1 (so $a*x$ will map to $a$). Check that this is a homomorphism and is injective. From there you automatically get surjectivity because the sets in question are finite of the same cardinality.
