Question

Consider the group $G=\{A\in M_2\left(R\right):A{A}^T=I_2\}$ with respect to matrix multiplication. Let

$Z\left(G\right)=\{A\in G:AB=BA,\text{for all}B\in G\}\mathrm{.}$

Then, the cardinality of Z(G) is

1. 1
2. 2
3. 4
4. Infinite

Answer: 2

Solution :-

Given that $G=\{A\in M_2\left(R\right):A{A}^T=I_2\}$ . Here $G$ is subset of $G{l}_2\left(ℝ\right)=\{A_{2\times 2}:det\left(A\right)\ne 0\}$ .

Let’s see why $G$ is subset of $G{l}_2\left(ℝ\right)$ :

Given $A{A}^T=I_2\cdots \cdots \left(1\right)$ . Now, taking determinant both sides of eq(1), we get

So, $det\left(A\right)\ne 0$ and we know that any matrix of order 2 with non-zero determinant are elements of $G{l}_2\left(ℝ\right)$ .
Hence, $G$ is a subset of $G{l}_2\left(ℝ\right)$ .

Now, let’s see which elements of $G{l}_2\left(ℝ\right)$ can also contain in $G$ .

Let $A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]\in G\mathrm{.}$

Hence, any matrix which is in $G$ satisfies the above two conditions.

Now, We know that center of $G{l}_2\left(ℝ\right)\text{is}a\cdot I_2,\text{where}0\ne a\in ℝ$ . Means $Z\left(G{l}_2\left(ℝ\right)\right)=\{a{I}_2:0\ne a\in ℝ\}$ .

And $G$ is subset of $G{l}_2\left(ℝ\right)$ so center of $G{l}_2\left(ℝ\right)$ is also center of $G$ .

So, $\{\left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right]:0\ne a\in ℝ\}$ is center of $G$ . But all elements which are in the center of a group, all those elements must also be in the group.

So, $\left[\begin{array}{cc}a& 0\\ 0& a\end{array}\right]\in G\text{iff}{a}^2+0^2=1$ from above (condition 1). Hence, $a=\pm 1.$

Therefore, $Z\left(G\right)=\{a{I}_2:a=\pm 1\}$ = $\{\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right],\left[\begin{array}{cc}-1& 0\\ 0& -1\end{array}\right]\}$ .

So, the correct answer is option-2, $\to$ 2