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**Drexel28** The notation is just notation. If you like better $\displaystyle \mathbb{Z}_{m_1}\times\cdots\times\mathbb{Z}_{m_k}$ or $\displaystyle \mathbb{Z}_{m_1}\oplus\cdots\oplus\mathbb{Z}_{m_k}$ in lieu of $\displaystyle \bigoplus_{j=1}^{k}\mathbb{Z}_{m_j}$. They all mean the same thing except that $\displaystyle \oplus$ is reserved for when the groups are abelian (btw $\displaystyle \bigoplus$ is an analogous index operator for $\displaystyle \oplus$ as that of $\displaystyle \sum$ for $\displaystyle +$).

Ok, so for the idea. You need a quick little lemma.

**Lemma:** Let $\displaystyle C_n$ denote a cyclic (this lemma will show *the* cyclic) group of order $\displaystyle n$. Then, $\displaystyle \mathbb{Z}_n\cong C_n$

**Proof:** Let $\displaystyle C_n$ be generated by $\displaystyle c$ and define $\displaystyle \theta:\mathbb{Z}_n\to C_n$ by $\displaystyle k\overset{\theta}{\longmapsto}c^k$.

Clearly $\displaystyle \theta$ is injective since $\displaystyle \theta(k)=\theta(m)\implies c^k=c^m$ now we may assume WLOG that $\displaystyle 0\leqslant m\leqslant k\leqslant n$ and so the above implies that $\displaystyle c^{k-m}=e$. But! $\displaystyle |c|=n\geqslant k-m$ and so $\displaystyle k-m$ can't be strictly positive and since it's non-negative it follows that $\displaystyle k-m=0\implies k=m$

Surjectivity is clear since $\displaystyle g\in C_n\implies g=c^k$ for some $\displaystyle 0\leqslant k\leqslant n$ and so $\displaystyle k\in\mathbb{Z}_n$ and $\displaystyle \theta(k)=c^k=g$

The homomorphism property follows since $\displaystyle \theta(m+k)=c^{m+k}=c^mc^k=\theta(m)\theta(k)$.

The conclusion follows. $\displaystyle \blacksquare$

Thus, we must only prove that $\displaystyle \bigoplus_{j=1}^{k}\mathbb{Z}_{m_j}$ is cyclic and it will follow from the lemma that it is equal to $\displaystyle \mathbb{Z}_{m_1\cdots m_k}$

To see this we will prove that $\displaystyle (\underbrace{1,\cdots,1}_{k\text{ times}})$ generates $\displaystyle \bigoplus_{j=1}^{k}\mathbb{Z}_{m_k}$. To do this note that $\displaystyle |(1,\cdots,1)|$ is the least positive number such that $\displaystyle (1,\cdots,1)^d=(d,\cdots,d)=(0,\cdots,0)\quad{\col or{red}(1)}$. But, notice that the first coordinate's order is $\displaystyle m_1$ and so it will only be $\displaystyle e$ when $\displaystyle d=a_1m_1$ where $\displaystyle a_1\in\mathbb{N}$ and $\displaystyle d$ is as in $\displaystyle \color{red}(1)$. Similarly, the second slot will only be $\displaystyle e$ when $\displaystyle d=a_2m_2$ and so on and so forth. Thus, $\displaystyle |(1,\cdots,1)|$ is the first occurrence of the multiples of $\displaystyle m_1,\cdots,m_k$ matching up. In other words $\displaystyle |(1,\cdots,1)|=\text{lcm}(m_1,\cdots,m_k)$. But, since $\displaystyle \text{gcd}(m_1,\cdots,m_k)=1$ and $\displaystyle \text{lcm}(m_1,\cdots,m_k)=\frac{m_1\cdots m_k}{\text{gcd}(m_1,\cdots,m_k)}$ it follows that $\displaystyle |(1,\cdots,1)|=\text{lcm}(m_1,\cdots,m_k)=m_1\cdot s m_k$. And since $\displaystyle \left|\bigoplus_{j=1}^{k}\mathbb{Z}_{m_k}\right|=m _1\cdots m_k$ it follows that $\displaystyle (1,\cdots,1)$ generates it. Thus, $\displaystyle \bigoplus_{j=1}^{k}\mathbb{Z}_{m_k}$ is a cyclic group of order $\displaystyle r=m_1\cdots m_k$ and thus $\displaystyle \bigoplus_{j=1}^{k}\mathbb{Z}_{m_j}\cong \mathbb{Z}_r$ by our lemma.