这是一份manchester曼切斯特大学 MATH32001作业代写的成功案例
问题 1.
since by ${ }^{*}$ note above $m_{i} n_{i}=|a|$. Thus
$$
2(|G|-1)=\sum_{i=1}^{k}\left(|G|-m_{i}\right) .
$$
证明 .
Divide by $|G|$ to get:
$$
2-2 /|G|=\sum_{i=1}^{k}\left(1-\frac{1}{n_{i}}\right)
$$
(Thus $\left.\sum_{i=1}^{k}\left(1-\frac{1}{n_{i}}\right)<2 .\right)$ We may assume $|G|>1$. Thus
$$
1 \leq 2-2 /|G|<2 \text {. }
$$
Now each $n_{i} \geq 2$; hence $1 / 2 \leqq 1-\frac{1}{n_{i}}<1$.
MATH32001 COURSE NOTES ：
Let $x \in A_{1}^{\prime} \cap A_{2}^{\prime}$. Then $x \in A_{1}^{\prime}$ means $x=\left(a_{1}, 0\right)$ and $x \in A_{2}^{\prime}$ means $x=\left(0, a_{2}\right)$. Thus
$$
\left(0, a_{2}\right)=\left(a_{1}, 0\right) .
$$
This means that $a_{1}=a_{2}=0$. Thus
$$
x=(0,0),
$$
the zero of $A_{1} \oplus A_{2}$. Hence is satisfied.