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I'm trying to find all units in the ring $\Bbb Z_{5} \times \Bbb Z_{8}$. I know that the units of $\Bbb Z_{5}$ are $\{ 1,2,3,4 \}$ and the units of $\Bbb Z_{8}$ are $\{ 1,3,5,7 \}$. Just wondering do I combine these two units or is there a whole other way to find this out?

Matt L.
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  • $(\mathbb Z_5\times \mathbb Z_8)^=\mathbb Z_5^\times \mathbb Z_8^*$. – Todd Jul 25 '20 at 20:11
  • Hello: many people have this habit of asking whatever is on their mind without giving a thought to the fact that answers might already exist. Try to get in the habit of searching before asking. We do, after all, have over a million questions now with comparably as many answers, so simple questions especially often already have answers on the site. – rschwieb Jul 27 '20 at 14:40
  • I did search before posting this question, the problem was that the "duplicate" questions didn't answer my question in a way that I understood, so I posted my own version of the question I was looking for an answer for and I received an answer that made sense to me. – Matt L. Jul 28 '20 at 04:27

2 Answers2

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Now that you know the units of each individual group, the units in $ \mathbb{Z}_5 \times \mathbb{Z}_8$ is simply formed by taking all possible combinations. For example, $(1,1), (1,3), (3,1)$ are all units in the product you're considering. Just be careful here - in $(3,1)$, $3 \in \mathbb{Z}_5$ but in $(1,3)$, $3 \in \mathbb{Z}_8$.

Prasanna Maddila
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The group of units functor respects products, since it has a left adjoint, the group ring construction.

Furthermore, $\Bbb Z_p^*\cong \Bbb Z_{p-1}$.

Hence we get $(\Bbb Z_5×\Bbb Z_8)^*\cong\Bbb Z_5^*×\Bbb Z_8^*\cong \Bbb Z_4×\Bbb Z_2×\Bbb Z_2$. Here I used the fact that $\Bbb Z_8^*\cong V_4:=\Bbb Z_2\times\Bbb Z_2$.