Solution. Replace $ n$ with $ 0$. Then $ S$ is replaced with $ \mathbb{Z}\slash n\mathbb{Z}$. Also, the condition is that if $ x,y\in A$ (reduced modulo $ n$), then $ (x + y)\bmod n\in A$. This is equivalent to saying that the set $ A$ is closed under addition modulo $ n$. Because $ A$ is finite, closure is sufficient to obtain $ A\leq\mathbb{Z}\slash n\mathbb{Z}$. By Lagrange's Theorem, $ |A|$ divides $ |\mathbb{Z}\slash n\mathbb{Z}| = n$, as desired. $ \Box$

boxedexe wrote: By Lagrange's Theorem, $ |A|$ divides $ |\mathbb{Z}\slash n\mathbb{Z}| = n$, as desired. $ \Box$ Can anyone explain to me what "Lagrange's Theorem" is?I have some difficulties when reading it on Google.This sounds strange to me.If you have any books about this theorem,please send me.Thanks.

How do we know there are inverses in $ A$ so it forms a group?

To answer the first question: the order of a subgroup divides the order of the group. To answer the second question: if we have $ x \in A$, add it successively to itself $ n - 2$ times to get $ (n - 1)x \equiv - x \pmod n$. In fact, since we have a cyclic group, that solution is serious overkill: let $ x$ be the smallest element of $ A$, and $ m$ be maximal such that $ mx \leq n$. Then $ A$ contains all of $ x, 2x, 3x, \ldots, mx$. If $ mx < n$ then $ (m + 1)x > n$ and so $ (m + 1)x - n \in A$; but then $ (m + 1)x - n < x$, a condtradiction with our choice of $ x$. Thus $ mx = n$, and $ |A| = m |n$.

Lagrange's Theorem: Let $ H\leq G$, where $ G$ is a finite group. Then the order of $ H$ divides the order of $ G$ and the number of left cosets of $ H$ in $ G$ equals $ \left|G\right|/\left|H\right|$.

SUPERMAN2 wrote: boxedexe wrote: By Lagrange's Theorem, $ |A|$ divides $ |\mathbb{Z}\slash n\mathbb{Z}| = n$, as desired. $ \Box$ Can anyone explain to me what "Lagrange's Theorem" is?I have some difficulties when reading it on Google.This sounds strange to me.If you have any books about this theorem,please send me.Thanks. I know you are Vietnamese,so I will advise you a book:"Đại số đại cương" by Hoàng Xuân Sính.It contains this theorem.

If an abstract algebra book doesn't contain Lagrange's theorem, it's incomplete. It's a pretty central theorem.