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Thread: Group isomorphism

  1. #1
    Member javax's Avatar
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    Group isomorphism

    Hello. Need some fast help on this:

    Let $\displaystyle \phi G, \star)\rightarrow(H, \triangle) $ be an isomorphism between these two groups. If $\displaystyle a' \in G$ is the inverse element of $\displaystyle a \in G$, then prove that $\displaystyle \phi(a')\in H$ is the inverse element of $\displaystyle \phi(a) \in H$.

    I tried like this:
    Since $\displaystyle \phi$ is isomorphism, that means it is a one to one map thus
    $\displaystyle \phi(a)=\phi(b)$ only when $\displaystyle a = b$ and since $\displaystyle \phi $ is isomorphism, $\displaystyle \phi(a')$ must be the inverse of $\displaystyle \phi(a)$.

    Do you think my prove is enough, if not please elaborate.

    Thanks for your time.
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  2. #2
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    Quote Originally Posted by javax View Post
    Hello. Need some fast help on this:

    Let $\displaystyle \phi G, \star)\rightarrow(H, \triangle) $ be an isomorphism between these two groups. If $\displaystyle a' \in G$ is the inverse element of $\displaystyle a \in G$, then prove that $\displaystyle \phi(a')\in H$ is the inverse element of $\displaystyle \phi(a) \in H$.

    I tried like this:
    Since $\displaystyle \phi$ is isomorphism, that means it is a one to one map thus
    $\displaystyle \phi(a)=\phi(b)$ only when $\displaystyle a = b$ and since $\displaystyle \phi $ is isomorphism, $\displaystyle \phi(a')$ must be the inverse of $\displaystyle \phi(a)$.


    I can't see any explanation or justification here...in fact, the argument is true for ANY group homomorphism, isomorphism or not, and the proof is boringly simple:

    $\displaystyle \phi(a)\phi(a')=\phi(aa')=\phi(e_G)=e_H$

    Tonio

    Do you think my prove is enough, if not please elaborate.

    Thanks for your time.
    .
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