-1

Why $|\mathbb{Z} / (p_1^{\alpha_1} \dotsm p_k^{\alpha_k}) \mathbb{Z}| = |\mathbb{Z} / p_1^{\alpha_1} \mathbb{Z}| \times \dotsb \times |\mathbb{Z} / p_k^{\alpha_k} \mathbb{Z}|,$ could anyone explain this to me please?

Bill Dubuque
  • 272,048
weird
  • 29
  • 1
  • 11
  • 1
    How many elements do you think $\mathbb Z/n\mathbb Z$ has? It would be a little harder if you were asking about the multiplication units $\left(\mathbb Z/n\mathbb Z\right)^{\times}.$ – Thomas Andrews Dec 08 '22 at 14:03
  • By here in the linked dupe we have $|\Bbb Z/n\Bbb Z| = n,$ from which the sought result follows immediately. – Bill Dubuque Dec 08 '22 at 15:19

2 Answers2

1

For $n\in \mathbb{Z}\backslash\{0\}$, because of $|\mathbb{Z}/n\mathbb{Z}|=n$, your equality is trivial.

By the way $p_i^{\alpha_i}$ and $p_j^{\alpha_j}$ is coprime ($i\neq j$), by Chinese remainder theorem we get group isomorphic $\mathbb{Z}/(p_1^{\alpha_1}\cdots p_k^{\alpha_k})\mathbb{Z}\cong \mathbb{Z}/p_1^{\alpha_1}\mathbb{Z}\times \cdots \times \mathbb{Z}/p_k^{\alpha_k}\mathbb{Z}$.

Yos
  • 1,924
  • 7
  • 14
1

The equation $$\mathbb{Z}/M \mathbb{Z} = M$$ is true for any positive intrger $M$.

So by this, on the one hand [with $M=p_1^{\alpha_1}\cdots p_n^{\alpha_n}$]: $$|\mathbb{Z}/(p_1^{\alpha_1}\cdots p_n^{\alpha_n}\mathbb{Z})| = p_1^{\alpha_1}\cdots p_n^{\alpha_n}.$$

Also by this, on the other hand, for each $i=1,2,\ldots, n$ [with $M=p_i^{\alpha_i}$]:

$$|\mathbb{Z}/(p_i^{\alpha_i}\mathbb{Z})| = p_i^{\alpha_i}.$$ This yields,

$$|\mathbb{Z}/(p_1^{\alpha_1}\mathbb{Z})|\times \cdots \times |\mathbb{Z}/(p_n^{\alpha_n}\mathbb{Z})|$$ $$= p_1^{\alpha_1}\times \cdots \times p_n^{\alpha_n}.$$ So putting all this together:$$|\mathbb{Z}/p_1^{\alpha_1}\mathbb{Z}| \times \cdots \times |\mathbb{Z}/p_n^{\alpha_n}| $$ $$=p_1^{\alpha_1}\times \cdots \times p_n^{\alpha_n}$$ $$=|\mathbb{Z}/(p_1^{\alpha_1}\times \cdots \times p_n^{\alpha_n})|,$$ which is what you want.

Mike
  • 20,434