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How to find all homomorphism $\delta :V_4 \to \mathbb{C}^{*}$. Where $V_4$ is Kleins 4 group and $\mathbb{C}^{*}$ is multiplicative group of nonzero complex numbers.

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  • $\begingroup$ I am not able to think how to start $\endgroup$ – user253340 Aug 28 '15 at 5:23
  • $\begingroup$ How many elements of order 2 does $V_4$ have? How many elements of order 2 does $\mathbb{C}^*$ have? $\endgroup$ – Yeldarbskich Aug 28 '15 at 5:24
  • $\begingroup$ $ V_4$ has 3 and $\mathbb{C}^*$ has just one $\endgroup$ – user253340 Aug 28 '15 at 5:28
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Hint: look at the order of the generators for $V_4$. Warning: spoilers ahead.

Note that $V_4\cong \mathbb{Z}/2\mathbb{Z}\oplus \mathbb{Z}/2\mathbb{Z}$ which is generated by the set $\{(1,0),(0,1)\}$. So, in order to specify a homomorphism from $V_4$ to $\mathbb{C}^*$, we only need to say what happens to each of these elements, and then check to make certain they are consistent with a homomorphism. We know that $(1,0)$ and $(0,1)$ each have order $2$ in $V_4$, which means that if $\phi:V_4\rightarrow\mathbb{C}^*$ is a homomorphism, then $\phi((1,0))\cdot\phi((1,0))=1$, so $\phi((1,0))=\pm 1$, and similarly $\phi((0,1))=\pm 1$. So, we have only four options for a homomorphism: 1. $\phi((1,0))=\phi((0,1))=1$. 2. $\phi((1,0))=-\phi((0,1))=1$. 3. $-\phi((1,0))=\phi((0,1))=1$. 4. $\phi((1,0))=\phi((0,1))=-1$. I'll leave it to you to check that each of these four choices produces a distinct homomorphism.

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