I got the set: $G=\{p/q\in \Bbb Q : (p,q)=1$, with $q$ odd number $\}$ and the binary operation $a*b:=a+b$. And I say that $(G.*)$ isn't a group because it doesn't have an identity.
My proof is: We know that $G$ is a subset of rational numbers, then $e$ must be $0$, but since $0\not\in G$ then $G$ doesn't have an identity. Is this correct?
In general could it happen that if we've got a subset of a group, this could have a different identity?
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You're half-correct: If $G$ were a group, then it would be a subset of $\mathbb{Q}$ with the same operation as $(\mathbb{Q}, +)$, so the only possible identity would be $0$.
But since $0 = \frac 0 1$ and $\gcd(0, 1) = 1$ is odd, this doesn't actually give a counterexample.
More generally, if $G$ is a group, and $H$ is a subset of $G$ with the same operation, and $H$ is a group under this operation, then the identities of $H$ and $G$ coincide.
If $\frac{p_1}{q_1},\frac{p_2}{q_2}\in G$ then $\frac{p_1}{q_1}+\frac{p_2}{q_2}=\frac{p_1q_2+p_2q_1}{q_1q_2}$. Now the numerator may be even but the denominator isn't. So when the common primes are cancelled out we cannot have an even denominator. So $G$ is closed. Clearly all inverses are in $G$ as well following which it is a group.