Let $G=\{r \in \mathbb Q\;|\;0 \le r \lt 1\}$ be the set of non-negative rationals smaller than $1$. Define the operation:
$a \circ b=\begin{cases} a+b, & \text{if $a+b<1$} \\ a+b-1, & \text{if $a+b \ge 1$} \end{cases}$
I need to verify that $G$ is a group with this operation.
Closure and Associativity are just calculations and I proved them but when it comes to the Identity and Inverse element I have difficulties finding these:
- If I choose $e=0$ as the Identity element then: $a \circ 0=0 \circ a=a$, since $a+0 \Rightarrow a \lt 1,a \in G$ which is OK, but then the Inverse element of every $a \in G$ has to be $-a<0, -a \notin G$, since $a \circ (-a) = (-a) \circ a=a+(-a)=0=e$.
- Now, if you could choose $e=1$ ($1 \notin G$ so you cannot!) then it serves very nice as Identity element but there is no Inverse element that I could find.
Any ideas?
Extra: We need to show that for any finite set $\{g_1,...,g_n\}\subseteq G$ there is a proper subgroup $H \subset G,H \neq G$, such that $\{g_1,...,g_n\}\subseteq H$. If I choose the subgroup $H$ with operation the $\circ$ and set the $A=\{g_1,...,g_n\} \cup$$\{$any inverse elements of $g_i$ that are not already in the $g_i$ set$\}$ $\cup$ $\{$the Identity element if it is not in the ${g_i}$ set$\}$, then $H \subset G$ and $H \neq G$ since the $g_i$ set is finite (while the rationals are infinitely many in $[0,1)$) and of course $A \supseteq \{g_1,...,g_n\}$. Is this correct?
If $\circ$ is an operation on the set $X$ and $*$ is an operation on the set $Y$ so that $(Y,*)$ is a group and there is a bijection $f\colon X\to Y$ such that $f(a\circ b)=f(a)*f(b)$, for all $a,b\in X$, then $(X,\circ)$ is a group.
Consider the map $$ \varphi\colon G\to \mathbb{Q}/\mathbb{Z} $$ defined by $\varphi(r)=r+\mathbb{Z}$ and prove this is an isomorphism for the $\circ$ operation on $G$ and the standard residue class addition on $\mathbb{Q}/\mathbb{Z}$.
You can also prove directly the properties, but it's much longer and tedious.
If you have been able to prove associativity, then $0$ is clearly a neutral element, because for every $r\in G$, $0+r<1$, so $0\circ r=0+r=r$.
If $r=0$, the inverse is $0$. If $0<r<1$, the inverse is $1-r$. Indeed, $r+(1-r)\ge1$, so $$ r\circ(1-r)=r+(1-r)-1=0 $$
Your attempt at the second part is incorrect, but the idea is good. Any finitely generated subgroup of $G$ is finite. The proof is longer than you did.