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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?
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  2. #2
    Super Member PaulRS's Avatar
    Oct 2007
    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
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