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Describe all non-injective group homomorphisms from $\mathbb Z$ to $\mathbb Q^*$.

So $\mathbb Z=\langle 1\rangle$. So when $f(z) \neq 0, z \neq 0$ we will get a non-injective homomorphism.

What's the way of thinking about this when we speak of infinite groups? I usually solve these questions using the fact that the order an element's image divides the element's order, but $1$ has infinite order. What do now?


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Note that the neutral element of $\mathbb Q^*$ is $1$, not $0$. So what you need is $f(z)=1$ for some $z\neq 0$ to get a non-injective homomorphism. – Christoph May 7 '14 at 8:39
up vote 2 down vote accepted

As you noted, $\mathbb Z = \langle 1 \rangle$. So for any homomorphism $f:\mathbb Z \to \mathbb Q^*$ you have $f(k)=f(1)^k$ for all $k\in\mathbb Z$. Thus, picking the image of $1$ determines the homomorphism.

Now assume $f$ is not injective, so we have some $k\neq 0$ such that $f(k)=f(1)^k = 1$. Now the only roots of unity in $\mathbb Q^*$ are $1$ itself and $(-1)$. So either $f(1)=1$ so $f(k)=1$ for all $k$, or $f(1)=(-1)$ so $f(k)=(-1)^k$ for all $k$.

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If you send 1 to an element of infinite order it will be injective. So you should send it to an element of finite order in $\mathbf{Q}^*$. Such elements are not difficult to locate. Just look for rational numbers $x$ such thta $x^n=1$ for some positive integer $n$.

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Isn't -1 the only one? – dsfsf May 7 '14 at 8:37
$-1$ satisfies, so does $+1$. – P Vanchinathan May 7 '14 at 8:54

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