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While revising I came across this question:

Let $G$ be the group of complex numbers $\{1, i, -1, -i\}$ under multiplication.

Is it true that for every such homomorphism, there is an integer $k$ such that the homomorphism has the form $z\mapsto z^k$ ?

The answer is true, but how does one prove it?

I thought of considering four cases of $f(i)=1,i,-1,-i$, since $i$ is a generator.

Sincere thanks for any help.

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The title does not really reflect the question. –  lhf Jul 2 '12 at 10:26
    
You say "every such homomorphism", but you haven't mentioned any. What homomorphisms do you wish to ask about? –  Chris Eagle Jul 2 '12 at 10:51
    
thanks for the comments, just to clarify I am referring to homomorphisms of G into itself –  yoyostein Jul 3 '12 at 6:07
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1 Answer 1

up vote 6 down vote accepted

I'm assuming the question means: let $f:G\to G$ be an endomorphism. Prove that $f(z)=z^k$ for some $k\in\mathbb{Z}$.

This is trivial. Since $G$ is a cyclic group (with $i$ as a generator, for example), the homomorphism is completely defined by what the generator is sent to (since all elements are some power of the generator). But whatever it is sent to, it is sent to an element of $w\in G$, but all of those can be written as some power of $i$ (that is, $w=i^k$ for some $k\in\mathbb{Z}$). This directly implies that $f(z)=z^k$: let $z\in G$, then $z=i^n$ for some $n\in\mathbb{Z}$, and then $f(z)=f(i^n)=f(i)^n$ since $f$ is a homomorphism, and then $f(i)^n=w^n=i^{kn}=z^k$ by definition.

If $f$ were required to be an automorphism (ie. bijective), then necessarily we would have $k$ odd.

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Thanks, this is perfect. –  yoyostein Jul 3 '12 at 6:09
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