1)prove that every finite subgroup of Q/Z is cyclic.
1) G={m1/n1 + Z,..., mk/nk + Z} finite simple of Q/Z where gcd(mi,ni)=1, set n=lcm(n1,...nk)
i don't know how to show that G is <1/n + Z>
2) invertible :b/a 11 doesnt divide 1 irred: 11k/b
3) Z/10Z ={(1),...(9)}
(2)=(4)
inver: (1)^(-1)=1, (2)^(-1)=8, (3)^(-1)=7, (4)^(-1)=4,(5)^(-1)=5,(6)^(-1)=6,(7)^(-1)=3,(8)^(-1)=2,(9)^(-1)=9
Let $\displaystyle G\le\mathbb{Q}\slash\mathbb{Z}$ be a finite group. Let us agree to write elements in $\displaystyle \mathbb{Q}\slash\mathbb{Z}$ as $\displaystyle \frac{\alpha}{\beta}+\mathbb{Z}\,,\,\,\alpha\,,\,\ beta\in\mathbb{Z}\,,\,\,\beta\neq 0\,,\,\,(\alpha,\beta)=1$ .
$\displaystyle \mbox{Let }\frac{a}{n}+\mathbb{Z}\in G\mbox{ be such that }\forall\,\,\frac{\alpha}{\beta}+\mathbb{Z}\,,\,\, n\geq \beta$. Now prove the following:
1) $\displaystyle \exists\,\,x\,,\,y\in\mathbb{Z}\,\,\,s.t.\,\,\,ax+ ny=1$
2) From the above it follows that $\displaystyle \frac{1}{n}+\mathbb{Z}\in G$
3) If there's an element $\displaystyle \frac{r}{s}+\mathbb{Z}\in G$ s.t. $\displaystyle \frac{r}{s}+\mathbb{Z}\notin \left<\frac{1}{n}+\mathbb{Z}\right>$ , then we get a contradiction to the maximality of n
4) It follows that $\displaystyle G=\left<\frac{1}{n}+\mathbb{Z}\right>$
Tonio
Ps. BTW, from the above it almost follows, adding some minor details, that $\displaystyle \mathbb{Q}\slash\mathbb{Z}$ has a unique (cyclic, of course) subgroup of order $\displaystyle n$ for every $\displaystyle n\in\mathbb{N}$