GATE CS 2024 Set 2 Q53 — Monoids and Groups

Let $G = Z_2 \times Z_3 \times Z_4$ be the group under component-wise addition modulo $n$. An element $g = (a, b, c) \in G$ is its own inverse if and only if:

where $e = (0, 0, 0)$ is the identity element of $G$. This component-wise condition translates to:

1. $2a \equiv 0 \pmod 2$ for $a \in Z_2$
2. $2b \equiv 0 \pmod 3$ for $b \in Z_3$
3. $2c \equiv 0 \pmod 4$ for $c \in Z_4$

Let us find the number of solutions for each component:

• For $a \in Z_2 = \{0, 1\}$ :
  • $2(0) = 0 \equiv 0 \pmod 2$
  • $2(1) = 2 \equiv 0 \pmod 2$
  Thus, there are $2$ solutions: $\{0, 1\}$.
• For $b \in Z_3 = \{0, 1, 2\}$ :
  • $2(0) = 0 \equiv 0 \pmod 3$
  • $2(1) = 2 \not\equiv 0 \pmod 3$
  • $2(2) = 4 \equiv 1 \not\equiv 0 \pmod 3$
  Thus, there is only $1$ solution: $\{0\}$.
• For $c \in Z_4 = \{0, 1, 2, 3\}$ :
  • $2(0) = 0 \equiv 0 \pmod 4$
  • $2(1) = 2 \not\equiv 0 \pmod 4$
  • $2(2) = 4 \equiv 0 \pmod 4$
  • $2(3) = 6 \equiv 2 \not\equiv 0 \pmod 4$
  Thus, there are $2$ solutions: $\{0, 2\}$.

Using the product rule, the total number of elements in $Z_2 \times Z_3 \times Z_4$ that are their own inverses is:

The $4$ elements are: $(0, 0, 0)$, $(0, 0, 2)$, $(1, 0, 0)$, and $(1, 0, 2)$.