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Let $\phi: G_{1} \rightarrow G_{2}$ be a homomorphism and $G_{1} = \langle a \rangle$ and $G_{2} = \langle b \rangle$ be infinite cyclic groups. Must we have $\phi(a) = b$?

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Why cant $\phi(a)=b^0=e$ Where e is the identity of $G_2$ – Amr Dec 4 '12 at 21:47
Prove that if $b$ is a generator, then so is $b^{-1}$. – Alexei Averchenko Dec 4 '12 at 21:47
@AlexeiAverchenko: What does that have to do with anything? – Chris Eagle Dec 4 '12 at 21:48
@ChrisEagle it shows that there are at least two isomorphisms between $G_1$ and $G_2$. – Alexei Averchenko Dec 4 '12 at 21:50
OH I think the question should have been about isommorphisms not homomorphisms – Amr Dec 4 '12 at 21:51
up vote 4 down vote accepted

No every $\phi:G_1\rightarrow G_2$ with $\phi(\alpha)=(\beta)^k$ is a monomorphism of groups for every $k\in\mathbb{Z}$ ( $k\neq 0$, if $k=0$ then it's the trivial homomorpishm)

If you have $k=1$ or $k=-1$ then $\phi$ is an isomorphism of groups.

If what bothers you is “how we can fit” one infinite cyclic group into another infinite cyclic group if we don't map one generator to the other. Then just remember that every subgroup of an infinite cyclic group is also infinite cyclic.

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No, of course not. For example, let $G_1=G_2=\Bbb{Z}, a=b=1$. Then for any integer $k\neq1$, $\phi(n)=kn$ is a counterexample.

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As a counterexample, consider the zero homomorphism, that takes any generator to zero.

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Maybe you meant isomorphism. But even then, for $G_1=G_2=\Bbb{Z}/\Bbb{Z/p}, a=b=1$, $\phi(n)=kn$ with any $k$ coprime to $p$ and $k$ not congruent $1 \mod p$ is a counterexample.

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