Question

Which of the following statements is/are correct?
(I) In \(GL_2(\mathbb{R})\), matrices with the same determinant always belong to the same conjugacy class.
(II) The class equation of \(A_5\) (the alternating group on 5 elements) is \(60=1+6+10+15+28\).

• A. Only (I)
• B. Only (II)
• C. Both (I) and (II)
• D. Neither (I) nor (II)

Hint:

Try a specific counterexample in \(GL_2(\mathbb{R})\) for (I), and recall the actual conjugacy class sizes of \(A_5\) for (II).

Answer 1

The Correct Option is D

Step 1: Test statement (I) with a counterexample.
Consider\[A = \begin{pmatrix}1&0\\0&1\end{pmatrix}, \qquad B = \begin{pmatrix}1&1\\0&1\end{pmatrix}\]Both have determinant 1. But \(A\) is the identity, so its conjugacy class is \(\{A\}\) alone. Since \(B \neq A\), \(B\) cannot lie in the same class as \(A\), even though \(\det A = \det B\). So statement (I) is false.

Step 2: Compute the actual conjugacy classes of \(A_5\).
\(|A_5| = 60\). The classes correspond to cycle types of even permutations of 5 symbols:\[\text{identity}: 1,\quad \text{double transpositions}: 15,\quad \text{3-cycles}: 20,\quad \text{5-cycles}: 24\]The 24 five-cycles split into two conjugacy classes of size 12 each inside \(A_5\) (the centralizer of a 5-cycle in \(S_5\), of order 5, lies entirely in \(A_5\), so the class does not fuse). This gives\[60 = 1+15+20+12+12\]

Step 3: Compare with the given equation.
The statement gives \(60=1+6+10+15+28\), which numerically sums to 60 but does not match the true class sizes \(1,15,20,12,12\). So statement (II) is false.

Step 4: Conclusion.
Neither (I) nor (II) is correct.\[\boxed{\text{Neither (I) nor (II)}}\]