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Let be $\mathbb{R^*}$ and $\mathbb{C^*}$ the multiplicative groups of non-zero real numbers and non-zero complex numbers respectively. Is $\operatorname{Hom}(\mathbb{Z_6},\mathbb{R^*}\oplus\mathbb{C^*})\cong\mathbb{Z_6}\oplus \mathbb{Z_2}? $ Why?

Jyrki Lahtonen
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user
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1 Answers1

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A homomorphism from $\mathbb{Z}/6\mathbb{Z}$ into any group $G$ is the same as an element $g \in G$ such that $g^6$ is the identity element of $G$ (look at where $1 \in \mathbb{Z}/6\mathbb{Z}$ goes). Can you write down the elements of $\mathbb{R}^{\times} \oplus \mathbb{C}^{\times}$ with this property?

Justin Campbell
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