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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?

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    Hint: If we take $e=0$ as the identity, then the inverse of, say, $\frac 14$ is $\frac 34$. – 2017-02-17
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    The inverse isn't the normal inverse. You need to find something $a'$ such that $a+a'=0$. There is no reason that $a'<0$. – 2017-02-17
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    Show if $a \in G$ then $1-a \in G$ (provided a is not the identity element). Then show $1-a$ is the inverse element for $a$ ... & you are done. – 2017-02-17
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    You are correct Donald: choose $0$ as the Identity element and $\forall a \in G,a \neq e=0, a^{-1}=1-a \in G$, since $0 \lt a \lt 1 \Rightarrow 0 \lt 1-a \lt 1$. This is because (I just learned that :) the Identity element is self-inverse. – 2017-02-17

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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

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.

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    Perhaps I missed something, but how can this help the OP if he still can't prove $\;(G,\circ)\;$ is a group to begin with? – 2017-02-17
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    @DonAntonio What's the point in proving the group structure to begin with? Isn't $\mathbb{Q}/\mathbb{Z}$ a group? – 2017-02-17
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    I don't think the asker is familiar with quotient groups. – 2017-02-17
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    I am just trying to understand the answer and I am kinda new to group theory so... Anyway, the first section of the answer must be proving that the $G$ is indeed a group, but I fail to find an $f$ and set $Y$ that does the required job... – 2017-02-17
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    @egreg Even if the OP knows that, how that helps him? That there's a map from $\;G\;$ to $\;\Bbb Q/\Bbb Z\;$ does **not** mean $\;G\;$ is *also* a group, and if you call that map "isomorphism" is because you know *already* it is group. This is a cyclic reasoning, I think. – 2017-02-17
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    @DonAntonio I think you are able to prove the assertion at the top, where the only assumption on $(X,\circ)$ is that it is a set with an operation. – 2017-02-17
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    @egreg I certainly am: I've used this precise exercise with my students several times... It is the OP the one who's having problems with this, as he states in his question. – 2017-02-18
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    Well, I tried to find it but didn't succeed: I am sure that we have to choose $X=Y=\{r \in \mathbb Q\;|\;0 \le r \lt 1\}$ and that $\circ$ is the operation as defined above, but for the $f,*$ what? I tried $f(x)=x$ and the $* \Rightarrow +$ but it is not possible. The $-1$ in the result of the operation $a \circ b$ kills me. – 2017-02-19
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    As for the second (the extra) part, I don't even understand (intuitively) how such a map will provide the answer! You don't have to answer this completely but it would really help to know the trick with the $f$ function on the first question and to really understand what is $\mathbb Q/ \mathbb Z$ - is it the class of (rational generally) remainders of the euclidean division of every rational with an integer? – 2017-02-19
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    @JohnZobolas You just have to prove the map is bijective and that $f(a\circ b)=f(a)+f(b)$. – 2017-02-19
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    I surely don't understand how without knowing $f$ I can prove the latter. Let's just say that I know what is $\mathbb Q/ \mathbb Z$. The map is bijective means that it is one-to-one and onto. For one-to-one let's take the kernel: only zero should be mapped to zero: $\phi (r)=0 \Rightarrow r+\mathbb Z = 0$, but it seems that with $r=0$ I can have the whole $\mathbb Z$ as the result? – 2017-02-19