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Math Help - Isomorphism and homomorphism properties

  1. #1
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    Isomorphism and homomorphism properties

    I have a bunch of properties of group homomorphism and isomorphism that I need to prove:
    So far, I'm stuck on a few properties. Hope someone can give me a hand.
    Let \phi:G \rightarrow H be a group homomorphism.
    1)If ord(g)=n then ord(\phi(g)) divides n
    In particular, if \phi is an isomorphism, then ord(\phi(g))=ord(g)
    2) If G is cyclic, then \phi(G) is cyclic. In particular, if \phi is an isomorphism and \phi(G) is cyclic, then G is cyclic.
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  2. #2
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    Hello,

    1) \phi(g)^n = \phi(g^n) = \phi(1_G)=1_H , so \mbox{ord}(\phi(g)) \le n.
    Let \mbox{ord}(\phi(g))=m. Suppose m doesn't devide n. Then n=km+q where  0<q<m. We have 1_H=\phi(g)^n = \phi(g)^{km+q} = \phi(g)^{km} \phi(g)^q = \phi(g)^q. But this contradicts our assumption that \mbox{ord}(\phi(g))=m (because q<m).

    Can you use this to prove the assertion about isomorphism?

    2) Let G be generated by an element g. Let's show that \phi(g) generates \phi(G). So be h \in \phi(G). This means h=\phi(k) for some k\in G. Since g generates G, there is some integer m that k=g^m. We have \phi(g)^m = \phi(g^m)= \phi(k) = h. Since h was arbitrary, \phi(g) generates \phi(G).
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  3. #3
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    Thank you so much for your help, Taluivren.
    I think I'm still stuck on the second part of 1). I tried to show n divides m where n=ord(g) and m=ord(\phi(g)), but I was unable to show this. I also tried to write n=mk since m divides n by previous result, but could not get anywhere.

    But I think I got the second part of 2). I proved that {\phi}^{-1}(x) generates G, where x is a generator of \phi(G).

    Now I'm stuck on another homomorphism property. Show that if H is a subgroup of G with ord(H)=n, then ord(\phi(H) divides n. Any help is very much appreciated.
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  4. #4
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    Consider

    1_H=(\phi(g))^m=\phi(g^m)
    But we know \phi is isomorphism so kernel is trivial. Hence g^m=1_G. But now n|m as n=ord(g) using the logic referred by Taluivren.

    For your last question use the fact \phi(H) is isomorphic to a H|(Kernel(\phi)\cap H)


    Quote Originally Posted by jackie View Post
    Thank you so much for your help, Taluivren.
    I think I'm still stuck on the second part of 1). I tried to show n divides m where n=ord(g) and m=ord(\phi(g)), but I was unable to show this. I also tried to write n=mk since m divides n by previous result, but could not get anywhere.

    But I think I got the second part of 2). I proved that {\phi}^{-1}(x) generates G, where x is a generator of \phi(G).

    Now I'm stuck on another homomorphism property. Show that if H is a subgroup of G with ord(H)=n, then ord(\phi(H) divides n. Any help is very much appreciated.
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  5. #5
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    Quote Originally Posted by aman_cc View Post
    Consider

    1_H=(\phi(g))^m=\phi(g^m)
    But we know \phi is isomorphism so kernel is trivial. Hence g^m=1_G. But now n|m as n=ord(g) using the logic referred by Taluivren.

    For your last question use the fact \phi(H) is isomorphic to a H|(Kernel(\phi)\cap H)
    Thank you so much for your help, aman_cc. Is there a more elementary way to approach my last question without using quotient group.
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  6. #6
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    Isomorphism means there is an inverse which is also a homomorphism going the other way. Apply the same result to see that both of their orders must divide each other and are therefore equal
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  7. #7
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    Quote Originally Posted by jackie View Post
    Thank you so much for your help, aman_cc. Is there a more elementary way to approach my last question without using quotient group.
    Not too sure, my wild guess would be no. You could start from the basic - define a relation on H. a\sim b if \phi(a)=\phi(b). This is an equivalence relation. Show that each equivalence class has same number of elements. (This is equal to number of elements in H where \phi(h)=1)

    But essentially I am using same logic used to derive the quotient group equation.
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