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 $f$ is bijective and that $f(ab)=f(a)f(b)$

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

Now suppose $
a,b \in G
$
we have $
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 $
a \cdot b^{ - 1} \in G
$
and by (1) it follows that $
a \cdot b^{ - 1}=e$
thus $a=b$ this means it's injective

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

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

Now obviously $
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