1. ## İsomorphism

Let G be a finite commutative group, Let n in Z with (|G|,n)=1. Show that the function f:G to G defined by f(g)=g^n, all g in G. Prove that it is isomorphism?

2. We'll prove that $\displaystyle f$ is bijective and that $\displaystyle f(ab)=f(a)f(b)$

Since $\displaystyle \left( {n,\left| G \right|} \right) \ne 1$ and $\displaystyle g \ne e$ then $\displaystyle g^n \ne e$ (1) where $\displaystyle e$ is the identity element. Because the order of $\displaystyle g$ divides $\displaystyle |G|$

Now suppose $\displaystyle a,b \in G$ we have $\displaystyle f\left( a \right) = f\left( b \right) \Leftrightarrow a^n = b^n \Leftrightarrow \left( {a \cdot b^{ - 1} } \right)^n = e$ but by definition of Group $\displaystyle a \cdot b^{ - 1} \in G$ and by (1) it follows that $\displaystyle a \cdot b^{ - 1}=e$ thus $\displaystyle a=b$ this means it's injective

Now let $\displaystyle G = \left\{ {a_1 ,...,a_{\left| G \right|} } \right\}$ where $\displaystyle a_i=a_j$ iff$\displaystyle i=j$ and $\displaystyle a_1=e$

Then the set $\displaystyle \left\{ {f(a_1) ,...,f(a_{\left| G \right|}) } \right\}$ has $\displaystyle |G|$ different elements in G, because f is injective, thus it follows that $\displaystyle G= \left\{ {f(a_1) ,...,f(a_{\left| G \right|}) } \right\}$ and so it's surjective. Thus f is bijective

Now obviously $\displaystyle f\left( {ab} \right) = \left( {ab} \right)^n = a^n b^n = f\left( a \right)f\left( b \right)$ thus it is an Isomorphism