Results 1 to 2 of 2

Thread: İsomorphism

  1. #1
    Junior Member mathemanyak's Avatar
    Jul 2008


    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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member PaulRS's Avatar
    Oct 2007
    We'll prove that f is bijective and that f(ab)=f(a)f(b)

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

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

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

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

    Now obviously <br />
f\left( {ab} \right) = \left( {ab} \right)^n  = a^n b^n  = f\left( a \right)f\left( b \right)<br />
thus it is an Isomorphism
    Follow Math Help Forum on Facebook and Google+

Search Tags

/mathhelpforum @mathhelpforum