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?
If  $\mathbb R^*$ and $\mathbb C^*$ be respectively multiplicative groups of non-zero real and complex number then whether $\operatorname{Hom}(\mathbb Z_6,R^*\oplus C^*)$  is  isomorphic to $\mathbb Z_{12}$? or $\mathbb Z_6$? or $\mathbb Z_6\oplus \mathbb Z_2$?
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This is an exmaple given by my professor, and his solution is:       The number of solutions to this equation satisfying the given   constraints is equal to the coefficient of $x^{21}$ in the expression   $$\left(\sum_{n=2}^5x^n\right)\left(\sum_{n=3}^6x^n\right)\left(\sum_{n=4}^7x^n\right)=x^9\left(\sum_{n=0}^3x^n\right)\left(\sum_{n=0}^3x^n\right)\left(\sum_{n=0}^3x^n\right).$$   This must be 3, since the coefficient of $x^8$ in $(1+x+x^2+x^3)^3$ is   3.   I understand actually nothing here. The expression of summation just pops up, and then comes the answer. What I only know that the idea of generating function is used here. Can someone help explain how the steps in the solution come? Why are z-values replaced with zero after in_memory use in Python? Best Pattern For Adding Commentary in align Environment? 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