Let $G=(\mathbb{Q}-\{0\},*)$ and $H=\{\frac{a}{b}\mid a,b\text{ are odd integers}\}$. Show $H$ is a normal subgroup of $G$. Let $G=(\mathbb{Q}-\{0\},*)$ and $H=\{\frac{a}{b}\mid a,b\text{ are odd integers}\}$.


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*Show $H$ is a normal subgroup of $G$.

*Show that $G/H \cong (\mathbb{Z},+)$


I know that there are multiple definitions for normal subgroup and I am having a hard time to develop the proof for these particular sets. 
For part 2. I need help developing a function from $G/H \to (\mathbb{Z},+)$.
 A: 1) $G$ is abelian, so any subgroup is normal.
2) Show that every element of $G/H$ is of the form $2^kH$ with $k\in \mathbb{Z}$; the map $2^k H \to k$ then gives you the required isomorphism.
A: Subgroups of abelian groups are always normal. One way to show normality is to show that for any $g\in G$, and $h\in H$, that $ghg^{-1}\in H$. But this follows immediately , since $ghg^{-1}=gg^{-1}h=h\in H$, since multiplication is commutative.
A: $\textbf{Show that $G/H \cong (\mathbb{Z},+)$.}$
$\textbf{Proof:}$ We can construct a function $f:G/H \to (\mathbb{Z},+)$ to be able to show that $G/H \cong (\mathbb{Z},+)$. However we can notice that $f$ can be written as $f:2^kH \to k$. Why?
Note that $\mathbb{Q}=\{\frac{r}{s} | r,s \in \mathbb{Z}, s\neq 0\}$. So $\mathbb{Q}-\{0\}=\{\frac{r}{s} | r,s \in \mathbb{Z}, r,s\neq 0\}$.
Let $\frac{r}{s} \in G/H$, we can write $r$ and $s$ into their prime factorization. 
(i.e. $\frac{r}{s}=\frac{2^{n_1}3^{n_2}\cdots p_m^{n_m}}{2^{t_1}3^{t_2}\cdots p_m^{t_m}}=\frac{2^{n_1}}{2^{t_1}} \cdot \frac{3^{n_2}\cdots p_m^{n_m}}{3^{t_2}\cdots p_m^{t_m}}$) 
Note that the only prime number that is not odd is 2. So we can break the fraction above to be the product of two fractions, and since the elements of $H$ are fractions with odd numerators and denominators then $\frac{r}{s}=2^{n_1-t_1}H$
We eventually want to use the conclusion of $G/\,Ker() \cong \,Im()$ So we need to show that $H=\,Ker()$ and $(\mathbb{Z},+)=\,Im()$. So the $\,Ker()=\{k=0\}$ because that means the fraction $\frac{r}{s}$ is only comprised of odd integers in both the numerator and denominator. So we can see that $\,Im()=(\mathbb{Z},+)$. So $f(2^kH)=k$.
To finally use the conclusion above, We need to show that (1) $f$ is a homomorphism, and (2) onto.$$f(2^{k_1}H2^{k_2}H)=f(2^{k_1+k_2}H)=k_1+k_2=f(2^{k_1}H)+f(2^{k_2})$$ Hence $f$ is a homomorphism. And lastly, we need to show that this homomorphism is onto (i.e. surjective). For all $k$ in $(\mathbb{Z},+)$, $f(2^kH)=k$ so $f$ is surjective. Since $f$ is an onto homomorphism we have $G/H \cong (\mathbb{Z},+)$
