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When $r$ and $s$ are relatively prime we have the ring isomorphism $\mathbb{Z}/rs \cong \mathbb{Z}/r \times \mathbb{Z}/s$

Given a prime factorization of $n$ where $n = p_1^{k_1} \cdots p_n^{k_n}$

Show that for groups of units invertible under multiplication:

$(\mathbb{Z}/n)^\times \cong (\mathbb{Z}/p_1^{k_1})^\times \times \cdots \times (\mathbb{Z}/p_n^{k_n})^\times$

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    $\begingroup$ Hint: Since you know $\mathbb{Z}/rs\simeq\mathbb{Z}/r\times\mathbb{Z}/s$ as rings, then you know that the units of one map to units to the other. This shows $\mathbb{Z}/rs^\times\simeq\mathbb{Z}/r^\times\times\mathbb{Z}/s^\times$ (there is something small to be checked here, namely that the units of $(\mathbb{Z}/r\times\mathbb{Z}/s)$ are $\mathbb{Z}/r^\times\mathbb{Z}/s^\times$). Then, use induction. $\endgroup$ – Michael Burr May 12 '15 at 20:52
  • $\begingroup$ Thank you! Can you get me started on the induction proof of that last point. Thanks! $\endgroup$ – clay May 12 '15 at 20:55
  • $\begingroup$ Induct on the number of distinct prime factors. The case where there is one distinct prime factor is easy ... $\endgroup$ – Michael Burr May 12 '15 at 21:07
  • $\begingroup$ In the inductive step I need to show $\left(\mathbb{Z}/r \times \mathbb{Z}/s\right)^\times = \left(\mathbb{Z}/r\right)^\times \times \left(\mathbb{Z}/s\right)^\times$ for relatively prime $r$ and $s$. How can I do that? $\endgroup$ – clay May 12 '15 at 22:49
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Base case: You've already done it.

$\mathbb{Z}/n^{\times} = \left(\mathbb{Z}/\prod_{i=1}^{j} p_i ^{k_j}\right)^{\times}= \bigg(\mathbb{Z}/\left(\prod_{i=1}^{j-1} p_i^{k_i}\right)\cdot p_j^{k_j}\bigg)^{\times} \cong \bigoplus_{i=1}^j \left(\mathbb{Z}/p_i^{k_i}\right)^{\times}$

The product of the $p_i^{k_i}$ for $1 \leq i \leq j-1$ is relatively prime with $p_j^{k_j}$.

The induction hypothesis would be to suppose it is true up to $j-1$, then we show above that up to $j$ resolves to the base case.

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