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: $$\boxed{4}$$