If $X \subseteq G$ is set of all pairwise commutative elements of $G$, prove that $\langle X \rangle \leqslant G$ abelian I am trying to prove that if $X$ is a set of elements of a group $G$ such that each pair of elements of $X$ commute, then the subgroup $\langle X \rangle \leqslant G$ generated by $X$ is abelian.
FYI: I have not learned what a center or a centralizer is yet; therefore, I cannot use either of those things in my proof.
My attempt so far has been the following:

Let $X \subseteq G$ s.t. $\forall a, b \in X$, $ab=ba$. 
Now, the subgroup $\langle X \rangle$ is the unique minimal subgroup of $G$ containing all elements of $X$.
So, if $a,b \in X \cap \langle X \rangle$, there is nothing to prove.

My idea then was to take $c \in \langle X \rangle$ but not in $X$, and so $\forall a \in \langle X \rangle$, $ca \neq ac$, and derive some kind of contradiction from that. The same thing for the case where both $c, d \in \langle X \rangle$ but not in $X$. However, I cannot for the life of me figure out how to do that. 
Could somebody please tell me how I should prove this (not involving centralizers, centers, etc.)?  
 A: The subgroup $\left<X\right>$ is equal to all possible finite products of elements of $X$ and their inverses, i.e.
$$\left<X\right> = \{x_1^{a_1}x_2^{a_2}\cdots x_s^{a_2} : x_i \in G, s \geq 1, a_i = \pm 1\},$$
Note that the $x_i$ need not be distinct. Then just use the fact that the elements of $X$ are pairwise commutative to conclude that $\left<X\right>$ is abelian.
A: I. If $x\in G$ then $\{g\in G:xg=gx\}$ is a subgroup of $G.$
Proof. $$xe=x=ex;$$
$$xg_1=g_1x,\ xg_2=g_2x\implies xg_1g_2=g_1xg_2=g_1g_2x;$$
$$xg=gx\implies xg^{-1}=g^{-1}gxg^{-1}=g^{-1}xgg^{-1}=g^{-1}x.$$
II. If $x\in X,$ every element of $\langle X\rangle$ commutes with $x.$
Proof. Supose $x\in X.$ Then $H=\{g\in G:xg=gx\}$ is a subgroup of $G$ which contains $X.$ Since $\langle X\rangle$ is the smallest subgroup of $G$ containing $X,$ it follows that $\langle X\rangle\subseteq H,$ i.e., every element of $\langle X\rangle$ commutes with $x.$
III. If $a,b\in\langle X\rangle,$ then $ab=ba.$
Proof. Suppose $a,b\in\langle X\rangle.$ Let $J=\{g\in X:ag=ga\}.$ Since $J$ is a subgroup of $G$ which contains $X,$ we have $\langle X\rangle\subseteq J.$ Since $b\in\langle X\rangle,$ it follows that $b\in J,$ ie., $ab=ba.$
