Question

The order of the permutation \(\begin{pmatrix}1& 2& 3& 4& 5& 6\\2& 4& 6& 5& 1& 3\end{pmatrix}\) is

• A. \(1\)
• B. \(2\)
• C. \(4\)
• D. \(8\)

Hint:

The order of a permutation is the LCM of the lengths of its disjoint cycles.

Answer 1

The Correct Option is C

Concept:
The order of a permutation is the least positive integer \(n\) such that applying the permutation \(n\) times gives the identity permutation. To find the order, first write the permutation as a product of disjoint cycles. Then the order is the LCM of the lengths of the cycles.

Step 1: Write the mapping. The permutation is \[ 1\mapsto 2,\quad 2\mapsto 4,\quad 3\mapsto 6 \] \[ 4\mapsto 5,\quad 5\mapsto 1,\quad 6\mapsto 3 \]

Step 2: Form disjoint cycles. Start with \(1\): \[ 1\mapsto 2,\quad 2\mapsto 4,\quad 4\mapsto 5,\quad 5\mapsto 1 \] So we get the cycle \[ (1\ 2\ 4\ 5) \] Now start with \(3\): \[ 3\mapsto 6,\quad 6\mapsto 3 \] So we get the cycle \[ (3\ 6) \] Thus, \[ \begin{pmatrix}1& 2& 3& 4& 5& 6\\2& 4& 6& 5& 1& 3\end{pmatrix} = (1\ 2\ 4\ 5)(3\ 6) \]

Step 3: Find the order. The lengths of the cycles are \[ 4 \] and \[ 2 \] Therefore, the order is \[ \operatorname{LCM}(4,2)=4 \]

Step 4: Final answer. \[ \boxed{4} \]