Today, I wanted to make a post for this question. There are some approach in which we can overcome the problem like this and this. According to my knowledge, I could solve the problem via the approach I had learned. Since I don't know what is this way called, so I refused to post an answer. I am very thankful if somebody tells me what is the method named? Here is some of the preliminaries but not the whole things because it is difficult for me to translate whole story (sorry).
Definition: Let $R=\{r_1,r_2,...,r_m\}$, $X=\{x_1,x_2,...,x_n\}$ and $G=\langle X| R\rangle$. And consider the abelian group $G/G'$. If $\alpha_{ij}$ be the sum of the powers of $x_j$ in relation $r_i$ so we can call the following matrix, the relation matrix of abelian group $G/G'$: $$M=\begin{pmatrix} \alpha_{11} & \alpha_{12} & ... & \alpha_{1n}\\ \alpha_{21} & \alpha_{22} & ... & \alpha_{2n}\\ \vdots &\vdots &\vdots &\vdots\\ \alpha_{m1} & \alpha_{m2} & ... & \alpha_{mn} \end{pmatrix}$$
Definition: Let $G=\langle X| R\rangle$, such that $|X|-|R|\le 0$. If we can make $M$ to have an standard diagonal form: $$D:=\begin{pmatrix} d_1 & 0 & ... & 0 & 0\\ 0 & d_2 & ... & 0 & 0\\ 0 & 0 & d_3 &0 & 0\\ \vdots &\vdots &\vdots &\vdots &d_k\\ 0 & 0 & 0 &0 & 0\\\vdots &\vdots &\vdots &\vdots &\vdots\\0 & 0 & 0 &0 & 0 \end{pmatrix}_{m\times n}$$ wherein $d_i\in\mathbb N\cup\{0\}$, by employing elementary row operations, then we can have $G/G'\cong\mathbb Z_{d_1}\times\mathbb Z_{d_2}\times...\mathbb Z_{d_k}$.
For example: Let $$G=\langle a,b\mid a^{2^{n-1}}=1, a^{2^{n-2}}=b^2, b^{-1}ab=a^{-1}\rangle$$ so $$G/G'=\langle a,b\mid a^{2^{n-1}}=1, a^{2^{n-2}}=b^2, a^2=1, [a,b]=1\rangle$$ Now we have $$M=\begin{pmatrix} 2^{n-1} & 0\\ 2^{n-2} & -2\\ 2 &0\\ \end{pmatrix}\xrightarrow{R_3\leftrightarrow R_1}\begin{pmatrix} 2 & 0\\ 2^{n-2} & -2\\ 2^{n-1} &0\\ \end{pmatrix}\to\begin{pmatrix} 2 & 0\\ 0 & -2\\ 0 &0\\ \end{pmatrix}$$ So $G/G'\cong\mathbb Z_{2}\times\mathbb Z_{2}$.
Thanks for your time reading my question. :)
