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i am new in the site. Next week, i will take final exam on group theory. Today i wached i video on youtube: https://www.youtube.com/watch?v=QOxyxsCdkz0

In video, it is proved that $Z^*_{st} \cong Z^*_{s} \oplus Z^*_{t}$, here $\oplus$ means external direct product, and $\cong$ denotes isomorphism ans s and t are coprime

Realize that $Z^*_{st} $ sometimes denoted as $U(st)$

The youtuber prove it by defining function $f:x\to (x \pmod{s}, x \pmod{t})$

They proved the injectivity by $$f(x)=f(y) \Rightarrow (x \pmod{s}, x \pmod{t}) = (y \pmod{s}, y \pmod{t})$$

So, $$x \pmod{s}= y\pmod{s}$$ $$x \pmod{t}= y\pmod{t}$$ $$x \equiv y\pmod{s},x \equiv y\pmod{t}$$ because $(s,t)=1$ $$x \equiv y\pmod{st}$$

It is also onto because their size are equal.

Operation preserving: $f(x) \oplus f(y)= (x \pmod{s}, x \pmod{t}) \oplus (y \pmod{s}, y \pmod{t}) = (xy \pmod{s}, xy \pmod{t})=f(xy)$

I believe that my professor will ask its more general version such that $Z^*_{n_1n_2,...,n_k} \cong Z^*_{n_1} \oplus Z^*_{n_2} \oplus...\oplus Z^*_{n_k}$, because he did not prove it in class and said it is exercise for you

how can i prove the general version ? I found a method using induction in site,but it was open question and there were not any answer ? Can you help me for general version?

Edition: after Arturo Magidin hint about additive group:

Theorem: External direct product $G_1 \times ... \times G_n$ of finite number of finite cyclic groups is cyclic iff the order of groups are relatively prime.

If $n_i,n_j$ are relatively prime when $i \neq j$, then $Z_{n_1} \times ... \times Z_{n_k} $ is cyclic group, because by previous theorem.

Theorem: two cyclic group of same order is isomorphic to each other.

We know that $|Z_{n_1} \times ... \times Z_{n_k}|=|Z_{n_1n_2...n_k}|$ where $n_i,n_j$ are relatively prime when $i \neq j$

So, $ Z_{n_1n_2...n_k} \cong Z_{n_1} \times ... \times Z_{n_k} $

However, i could not understand how to apply restriction for $U(n)$ version

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    $\begingroup$ Yes, rings make this simple. The rings $\mathbb{Z}_{st}$ and $\mathbb{Z}_s \times \mathbb{Z}_t$ are isomorphic in this case, and so their respective groups of units are, too. But yes, the Chinese Remainder Theorem makes this pretty automatic. $\endgroup$
    – Randall
    Commented Sep 25 at 18:13
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    $\begingroup$ There are several proofs of this on this site. Did you look for them? $\endgroup$
    – Arturo Magidin
    Commented Sep 25 at 18:34
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    $\begingroup$ What you have above is a proof that $Z_{st}$ bijects with $Z_s\times Z_t$; it does not include the argument that the image of an element of $Z_{st}^*$ will land in $Z_s^*\times Z_t^*$, though that is not hard to do. You can use induction and this very same argument to show that $Z_{n_1\cdots n_k}\cong Z_{n_1}\times\cdots\times Z_{n_k}$, and then show that the restriction to the multiplicative group and induction to prove the general case. $\endgroup$
    – Arturo Magidin
    Commented Sep 25 at 18:38
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    $\begingroup$ Your argument that the function is a morphism does not, in and of itself, prove that if $x$ is relatively prime to $st$, then it is relatively prime to $s$ and to $t$, which is required to prove that the element lands in the appropriate place. If you can do it for two, and you know that the function also defines an isomorphism of additive groups, then you apply induction, like I said. I will not go review a slide-show elsewhere. $\endgroup$
    – Arturo Magidin
    Commented Sep 25 at 18:54
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    $\begingroup$ @ArturoMagidin sorry, for the last, they are cyclic groups. Theorem: two cyclic group of same order is isomorphic to each other." i wont ping you further $\endgroup$
    – user1409514
    Commented Sep 25 at 19:24

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