Question

Let \( A_6 \) be the group of even permutations of 6 distinct symbols. Then the number of elements of order 6 in \( A_6 \) is ..................

Hint:

The number of elements of order 6 in \( A_6 \) is equal to the number of 6-cycles, which is a specific type of permutation.

Answer 1

Correct Answer: 0

1. Element Order in Permutation Groups

The order of a permutation is the Least Common Multiple (LCM) of the lengths of the cycles in its cycle decomposition.

We are looking for permutations in $A_6$ whose cycle structure has an $\text{LCM}$ of 6.

Possible cycle structures in $S_6$ whose $\text{LCM} = 6$ are:

Cycle Structure	Partition of 6	Order (LCM)	Parity (Must be Even)
$(6)$	$6$	6	Odd
$(3)(2)(1)$	$3+2+1$	$\text{lcm}(3, 2, 1) = 6$	$\text{sign} = (-1)^{3-1} \cdot (-1)^{2-1} = (+1) \cdot (-1) = \text{Odd}$

2. Analysis of Parity

A permutation is in the alternating group $A_n$ if and only if it is an even permutation.

The sign of a $k$-cycle is $(-1)^{k-1}$.

The sign of a permutation is the product of the signs of its disjoint cycles.

Case 1: Cycle of length 6, $(\mathbf{a_1 \ a_2 \ a_3 \ a_4 \ a_5 \ a_6})$

Order: 6

Sign: $(-1)^{6-1} = (-1)^5 = -1$ (Odd permutation)

Result: $\mathbf{A \notin A_6}$

Case 2: Cycles of length 3 and 2, $(\mathbf{a_1 \ a_2 \ a_3})(\mathbf{a_4 \ a_5})$

Order: $\text{lcm}(3, 2) = 6$

Sign: $(\text{sign of 3-cycle}) \cdot (\text{sign of 2-cycle}) = (-1)^{3-1} \cdot (-1)^{2-1} = (+1) \cdot (-1) = -1$ (Odd permutation)

Result: $\mathbf{A \notin A_6}$

Conclusion

Since all possible cycle structures that yield an order of 6 result in an odd permutation, there are no elements of order 6 in the alternating group $A_6$.

Final Answer

The number of elements of order 6 in $A_6$ is $\mathbf{0}$.