1
$\begingroup$

Finding all homomorphisms from $\mathbb Z_4$ to $\mathbb Q^*$

Let there be homomorphism $f: \mathbb Z_4 \rightarrow \mathbb Q^*$.

So there are several options for $|\ker f|: 1, 2, 4$

(1) $|\ker f|= 1 \rightarrow |\operatorname{im}f|=4$

So $f$ is monomorphism. So naturally $f(0)=1$. The order of $2 \in \mathbb Z_4$ is $2$, so it has to be mapped to $-1$ (because 1 was already mapped). The order of $1, 3 \in \mathbb Z_4$ is 4, but we already have mapped elements of order 1 and 2, and there aren't other elements in $\mathbb Q^*$ of such order or of order 4. So there is no such homomorphism.

Is there a more clever way of explaining that?

(2) $|\ker f|= 2 \rightarrow |\operatorname{im}f|=2$

$f(d)=1$ if $d=0\pmod2$, else $f(d)=-1$

(3) $|\ker f|=4$ trivial homomorphism.

Is this accurate? Thank you for any assistance!

$\endgroup$
1
$\begingroup$

Everything you've put is correct, but a clearer way to explain it:

Note that $f(1_4)$ must have order dividing $4$. But the only elements of order $1$, $2$ or $4$ in $\mathbb{Q}^*$ are $1$ and $-1$. Also, $1_4$ generates $\mathbb{Z}_4$, so the image of $1_4$ determines the homomorphism completely. So either $f(1_4) = 1$, in which case we have the trivial homomorphism, or $f(1_4) = -1$, in which case we have the homomorphism you describe in (2).

$\endgroup$
  • 1
    $\begingroup$ I have changed some of the $1$s to $1_4$s (the $1$s from $\mathbb{Z}_4)$. There were so many $1$s it was a tad confusing. Feel free to reject the edit though :-) $\endgroup$ – user1729 May 6 '14 at 13:56
  • $\begingroup$ No, it improves the answer's clarity a lot, thankyou :) $\endgroup$ – Christopher May 6 '14 at 14:16
1
$\begingroup$

An homomorphism $f : \mathbb{Z}_{4} \to \mathbb{Q}^*$ is determined by the image $f(1)$ .

A necessary condition is $f(1)^{4} = 1$ so $f(1)$ is a rational root of $x^4 -1$, and so there are $2$ possibilities $$f(1) =1$$$$f(1) = -1$$ But the previous condition is also sufficient and so each of these $2$ choices corresponds to an admissible homomorphism $f : \mathbb{Z}_{4} \to \mathbb{Q}^*$

$\endgroup$
0
$\begingroup$

Since $\mathbb Z/4\mathbb Z$ is cyclic, any homomorphism from $\mathbb Z/4\mathbb Z$ into another group is determined where the generator goes. You are allowed to send the generator to any element of order dividing 4. Since the only elements of finite order in $\mathbb Q^*$ are $\pm 1$, these are the only options.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.