Thread: Two more questions involving direct products

1. Two more questions involving direct products

1. Is Z x Z cyclic? [Here Z means (Z, +).]

2. Prove the Chinese remainder theorem: If $\displaystyle m_1,.....,m_n$ are positive integers such that $\displaystyle m_i$ and $\displaystyle m_j$ are relatively prime for i not equal to j, and $\displaystyle k_1,....,k_n$ are any integers, then there is an integer x such that x is congruent to $\displaystyle k_i(mod m_i)$ for 1 less than or equal to i less than or equal to n. (Hint: Consider the generator (1,1,....,1) for $\displaystyle Zm_1$ X......X $\displaystyle Zm_n.$)

2. Originally Posted by Janu42
1. Is Z x Z cyclic? [Here Z means (Z, +).]
no, however, it is generated by two elements, (1,0) and (0,1), i.e, <(1,0),(0,1)> = Z x Z

3. Originally Posted by Janu42
2. Prove the Chinese remainder theorem: If $\displaystyle m_1,.....,m_n$ are positive integers such that $\displaystyle m_i$ and $\displaystyle m_j$ are relatively prime for i not equal to j, and $\displaystyle k_1,....,k_n$ are any integers, then there is an integer x such that x is congruent to $\displaystyle k_i(mod m_i)$ for 1 less than or equal to i less than or equal to n. (Hint: Consider the generator (1,1,....,1) for $\displaystyle Zm_1$ X......X $\displaystyle Zm_n.$)
If $\displaystyle m_1,...,m_n$ are pairwise coprime then it means $\displaystyle G=\mathbb{Z}_{m_1}\times ... \times \mathbb{Z}_{m_n}$ is a cyclic group, furthermore, $\displaystyle ([1]_{m_1},...,[1]_{m_n})$ is a generator for this group. Let $\displaystyle k_1,...,k_n\in \mathbb{Z}$ and consider $\displaystyle ([k_1]_{m_1}, ... , [k_n]_{m_n}) \in G$. Therefore, there exists $\displaystyle x\in \mathbb{Z}$ so that $\displaystyle x([1]_{m_1},...,[1]_{m_n}) = ([k_1]_{m_1}, ... , [k_n]_{m_n})$, but that means $\displaystyle ([x]_{m_1}, ... , [x]_{m_n}) = ([k_1]_{m_1}, ... , [k_n]_{m_n})$. Therefore, $\displaystyle [x] = [k_j]_{m_j}$ for all $\displaystyle 1\leq j\leq n$. Thus, $\displaystyle x\equiv k_j(\bmod m_j)$.