Hw6 Problem 2

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### Problem 2

Let $G=\langle a \rangle$ be a cyclic group and $G'$ a group isomorphic to $G$. IF $\phi: G \rightarrow G'$ is an isomorphism, show that for every $x\in G$, $\phi(x)$ is completely determined by $\phi(a)$. (i.e., if $\phi, \psi: G \rightarrow G'$ are both isomorphisms such that $\phi(a)=\psi(a)$, then $\phi(x)=\psi(x)$ for all $x \in G$.)

Solution

By the homomorphism property $\phi(ab) = \phi(a)\phi(b)$, we have $\phi(a^{n}) = (\phi(a))^{n}$ for all $a \in \mathbb{Z}^+$. By theorem 3.14 we know that $\phi(a^{0})=\phi(e)=e'$. The equation $e'=\phi(e)=\phi(aa^{-1})=\phi(a)\phi(a^{-1})$ shows that $\phi(a^{-1})=(\phi(a))^{-1}$. From this, we see that $\phi(a^{-n})=(\phi(a))^{-n}$ for all negative integers $n$. Because $G$ is cyclic with generator $a$, this means that for all $g=a^{n} \in G$,$\phi(g)=\phi(a^{n})=[\phi(a)]^{n}$ is completely determined by the value $\phi(a)$.

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