Question

Which of the following statements are correct:
A. Permutation group $S_n$ has order $n!$, $n > 1$.
B. Alternating group $A_n$ of degree $n$ has order $\frac{n!}{2}$, $n > 1$. C. Set of all even permutations forms a subgroup of $S_n$.
D. Every permutation in $S_n$, $n > 1$ is a product of 2-cycles.
E. $A_4$ has a subgroup of order 6.
Choose the correct answer from the options given below:

• A. A, B, C, D Only
• B. B, C, D, E Only
• C. A, B, C Only
• D. A, B, D Only

Hint:

Remember that $A_4$ (order 12) is the classical example showing that the converse of Lagrange's theorem does not hold, as it has no subgroup of order 6.

Answer 1

The Correct Option is A

Step 1: Concept:
This question checks basic concepts of symmetric groups $S_n$, alternating groups $A_n$, transpositions, and subgroups.

Step 2: Key Formula or Approach:
1. The symmetric group $S_n$ consists of all bijections on $n$ elements, so $|S_n| = n!$.
2. The alternating group $A_n$ is the kernel of the sign homomorphism $\text{sgn}: S_n \to \{1, -1\}$, so $|A_n| = \frac{|S_n|}{2} = \frac{n!}{2}$.
3. Every permutation can be written as a product of transpositions (2-cycles).

Step 3: Step-by-step Explanation:

• Statement A:
The order of the symmetric group $S_n$ is the number of distinct permutations of $n$ distinct symbols, which is $n!$. Hence, Statement A is correct.

• Statement B:
The alternating group $A_n$ consists of all even permutations in $S_n$. Since exactly half the permutations in $S_n$ ($n > 1$) are even, $|A_n| = \frac{n!}{2}$. Hence, Statement B is correct.

• Statement C:
The product of two even permutations is even, the identity is even, and the inverse of an even permutation is even. Thus, the set of even permutations forms a subgroup $A_n \le S_n$. Hence, Statement C is correct.

• Statement D:
Any cycle $(a_1 a_2 \dots a_k)$ can be decomposed into transpositions as $(a_1 a_k)(a_1 a_{k-1})\dots(a_1 a_2)$. Since every permutation is a product of disjoint cycles, every permutation in $S_n$ ($n > 1$) is a product of 2-cycles. Hence, Statement D is correct.

• Statement E:
The alternating group $A_4$ has order $|A_4| = \frac{4!}{2} = 12$.
By Lagrange's theorem, any subgroup must have order dividing 12.
However, $A_4$ has no subgroup of order 6 (it serves as a famous counterexample to the converse of Lagrange's theorem). Hence, Statement E is incorrect.

Step 4: Final Answer:
Statements A, B, C, and D are correct. Therefore, option (A) is the correct answer.