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: Disprove (I) using trace as a second conjugation invariant.
If two matrices $P,Q \in GL_2(\mathbb{R})$ are conjugate, they share the same characteristic polynomial, so both $\det P = \det Q$ and $\operatorname{tr} P = \operatorname{tr} Q$ must hold. Take\[P = \begin{pmatrix}1&0\\0&1\end{pmatrix},\qquad Q = \begin{pmatrix}2&0\\0&\tfrac12\end{pmatrix}\]Both have $\det = 1$, but $\operatorname{tr} P = 2$ while $\operatorname{tr} Q = 2.5$. Since conjugate matrices must have equal trace, $P$ and $Q$ are not conjugate despite sharing the same determinant. This falsifies (I).

Step 2: Recompute the $A_5$ class sizes using centralizer orders.
For $x \in A_5$, class size $= |A_5|/|C_{A_5}(x)|$.\[x = e:\ |C(x)|=60 \Rightarrow \text{size } 1\]\[x = (12)(34):\ |C(x)|=4 \Rightarrow \text{size } 60/4 = 15\]\[x = (123):\ |C(x)|=3 \Rightarrow \text{size } 60/3 = 20\]\[x = (12345):\ |C(x)|=5 \Rightarrow \text{size } 60/5 = 12 \ (\times 2 \text{ classes})\]Adding gives $1+15+20+12+12=60$, which does not match $1,6,10,15,28$ given in (II). So (II) is also false.

Step 3: Conclusion.
Both statements fail to hold.\[\boxed{\text{Neither (I) nor (II)}}\]