That's a very strangely worded question, but I think it's probably meant to be a fill-in the blank multiple choice question which would more intelligibly be stated as:
Every subgroup of $(\mathbb{Q}, +)$ is ____________________.
A) cyclic and finitely generated, but not necessarily abelian or normal.
B) cyclic and abelian, but not necessarily finitely-generated or normal.
C) abelian and normal, but not necessarily cyclic or finitely-generated.
D) finitely generated and normal, but not necessarily cyclic or abelian.
Since $(\mathbb{Q}, +)$ is an abelian group, every subgroup is obviously abelian and normal, which rules out A, B, and D, so the answer is C. Of course you want to actually be able to put your hands on a non-finitely generated subgroup of $(\mathbb{Q}, +)$, so as anon says in the comments, consider the additive group of $\mathbb{Z}[\frac{1}{2}]$, i.e. the subgroup of $(\mathbb{Q}, +)$ consisting of elements where the denominator is a power of 2.
(Note that a finitely-generated subgroup of $(\mathbb{Q}, +)$ has an upper bound on its denominators, and is in fact cyclic.)
Since no one answered my question, I did some reading and found out that this is a very well known result. By using the concept of a "basis" of an abelian group, I did the following proof.
Suppose $G$ is a finite abelian group and $H\leq G$. Let $|G|=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}$ and $|H|=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$ where $p_1,...,p_k$ are distinct primes. By the elementary divisor decomposition, we have $G\cong G_1\oplus G_2\oplus\cdots\oplus G_k$ and $H\cong H_1\oplus H_2\oplus\cdots\oplus H_k$ where $G_i$ is a Sylow $p_i$-subgroup of $G$ and $H_i$ is a Sylow $p_i$-subgroup of $H$ for all $i=1,2,...,k$. Since $H_i\unlhd G_i$ for all $i=1,...,k$, $G/H\cong(G_1/H_1)\oplus(G_2/H_2)\oplus\cdots\oplus(G_k/H_k)$. So it suffices to show the result when $G$ is an abelian $p$-group. We proceed by induction.
Notice it is easier to think of $G$ as an additive group in stead of a direct product.
If $|G|=p$, then $H=1$ or $G$, so $G/H\cong1$ or $G$.
Suppose the result holds for all abelian $p$-groups of order less than $|G|$. Since $G$ is an abelian $p$ group, by the fundamental theorem of finite abelian groups, $G=\left<x_1\right>\oplus\left<x_2\right>\oplus\cdots\oplus\left<x_t\right>$ with $|x_i|=p^{\alpha_i}$ for all $i\in\{1,...,t\}$ and $\alpha_1\geq\alpha_2\geq\cdots\geq\alpha_t\geq1$. Notice that $x_1,x_2,...,x_t$ are linearly independent in the sense that $x_i$ cannot be write as a linear combination of $x_1,...,x_{i-1},x_{i+1},...,x_t$.
Case 1: $H=\left<g\right>$ with $|g|=p$. Since $|g|=p$, $g=m_1p^{\alpha_1-1}x_1+m_2p^{\alpha_2-1}x_2+\cdots+m_tp^{\alpha_t-1}x_t$ with $m_i\in\{0,1,...,p-1\}$ for all $i\in\{1,...,t\}$ (hence, $m_i=0$ or $(m_i,p)=1$). WLOG, assume $m_t\neq0$. Let $x_t'=m_1p^{\alpha_1-\alpha_t}x_1+m_2p^{\alpha_2-\alpha_t}x_2+\cdots+m_{t-1}p^{\alpha_{t-1}-\alpha_t}x_{t-1}+m_tx_t$. Notice that $|x_t'|=p^{\alpha_t}$. Since $\left<x_t\right>\cap(\left<x_1\right>\oplus\cdots\oplus\left<x_{t-1}\right>)=0$, $\left<x_t'\right>\cap\left<x_i\right>=0$ for all $i\in\{0,1,...,t-1\}$ otherwise $x_t'$ would be a linear combination of $x_1,...,x_{t-1}$. Hence, $G=\left<x_1\right>\oplus\left<x_2\right>\oplus\cdots\oplus\left<x_t'\right>$ (This is basically changing the basis of $G$). Since $\left<g\right>\leq\left<x_t'\right>$, $G/H\cong\left<x_1\right>\oplus\left<x_2\right>\oplus\cdots\oplus(\left<x_t'\right>/\left<g\right>)$. Since $\left<x_t'\right>$ is cyclic, $\left<x_t'\right>/\left<g\right>$ is isomorphic to a subgroup of $\left<x_t'\right>$. So $G/H$ is isomorphic to a subgroup of $G$.
Case 2: $|H|>p$. By Cauchy's theorem, there exists $g\in H$ such that $|g|=p$. By Case 1, $G/\left<g\right>$ is isomorphic to a subgroup of $G$. Now by the third isomorphism theorem, $G/H\cong(G/\left<g\right>)/(H/\left<g\right>)$. Since $|G/\left<g\right>|<|G|$, by the induction hypothesis, $(G/\left<g\right>)/(H/\left<g\right>)$ is isomorphic to a subgroup of $G/\left<g\right>$ which is isomorphic to a subgroup of $G$. Hence $G/H$ is isomorphic to a subgroup of $G$.
Best Answer
Hint: assume for contradiction that some $a, b \in G$ don't commute, and consider $H = \left< a, b \right> \leqslant G$.
PS It is also not hard to unpack the reasoning so that it does not use any external facts. From the assumption we get a group homomorphism $\varphi : G \to \mathbb{Q}$ such that $\ker \varphi = Z(G)$. Now take any $a, b \in G$. We will show that $ab = ba$. The subgroup
$$\varphi[ \left< a, b \right> ] = \left< \varphi(a), \varphi(b) \right> \leqslant \mathbb{Q}$$
is cyclic, so we can find $c \in \left< a, b \right>$ and $i, j \in \mathbb{Z}$ such that $\varphi(a) = i \cdot \varphi(c)$ and $\varphi(b) = j \cdot \varphi(c)$. It follows that $a \equiv c^i \pmod{Z(G)}$ and $b \equiv c^j \pmod{Z(G)}$ and hence $ab = ba$.