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I’m studying for my exam in algebra and I’m trying to show that $$(\mathbb{Z}/36\mathbb{Z})^*$$ Is isomorphic to the product of two groups $A$ and $B$ with 2 and 6 elements respectively. I was thinking since $$\mathbb{Z}/36\mathbb{Z}\cong \mathbb{Z}/4\mathbb{Z} \times \mathbb{Z}/9\mathbb{Z} $$ Then maybe $$(\mathbb{Z}/36\mathbb{Z})^* \cong (\mathbb{Z}/4\mathbb{Z})^* \times (\mathbb{Z}/9\mathbb{Z})^* $$ It appears to be true in this case, but I would like to show the above statement in general, if it’s true, but I haven’t been able to show that there is a homomorphism between the two

Snildt
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    For any two rings $R,S$, we have $(R\times S)^{\ast}=R^{\ast}\times S^{\ast}$. Furthermore, an isomorphism of rings $T\cong R\times S$ induces an isomorphism $T^{\ast}\cong(R\times S)^{\ast}$. Putting these two facts together yields what you want. Which of the two (or perhaps both) is the one not clear to you? – Thorgott Dec 27 '21 at 15:37
  • Start from this: a pair $(u,v)$ is invertible (in the ring product) if and only if both $u$ and $v$ are invertible (in their corresponding rings). –  Dec 27 '21 at 15:57
  • @Thorgott I was not familiar with any of these results, but I suppose the second one follows more or less directly from how inverses behave under a homomorphism? – Snildt Dec 27 '21 at 17:55
  • Yes, that sounds correct. – Thorgott Dec 27 '21 at 18:25

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