Question

Let \(Z_n\) be the group of integers \(\{0, 1, 2, \dots, n - 1\}\) with addition modulo \(n\) as the group operation. The number of elements in the group \(Z_2 \times Z_3 \times Z_4\) that are their own inverses is ............

Hint:

To find elements that are their own inverses in a group, solve \(2x \equiv 0\) modulo the group order for each component.

Answer 1

Step 1: Define the property of an element being its own inverse. An element \(x\) in a group is its own inverse if: \[ x + x \equiv 0 \, (\text{mod } n). \] This simplifies to: \[ 2x \equiv 0 \, (\text{mod } n). \] Step 2: Analyze each group. \(Z_2\): The elements are \(\{0, 1\}\). \[ 2x \equiv 0 \, (\text{mod } 2) \implies x = 0, 1. \] \(Z_3\): The elements are \(\{0, 1, 2\}\). \[ 2x \equiv 0 \, (\text{mod } 3) \implies x = 0. \] \(Z_4\): The elements are \(\{0, 1, 2, 3\}\). \[ 2x \equiv 0 \, (\text{mod } 4) \implies x = 0, 2. \] Step 3: Combine the results for \(Z_2 \times Z_3 \times Z_4\). The total number of elements in \(Z_2 \times Z_3 \times Z_4\) is: \[ |Z_2 \times Z_3 \times Z_4| = 2 \times 3 \times 4 = 24. \]
For an element \((x_2, x_3, x_4) \in Z_2 \times Z_3 \times Z_4\) to be its own inverse: \(x_2 \in \{0, 1\}\) (\(2\) choices). \(x_3 \in \{0\}\) (\(1\) choice). \(x_4 \in \{0, 2\}\) (\(2\) choices). The total number of such elements is: \[ 2 \times 1 \times 2 = 4. \] Final Answer: \[ \boxed{4} \]