GATE CS 1994 Q1.10 — Monoids and Groups

A group $(G, \circ)$ is abelian (or commutative) if the group operation commutes for all elements in the group, i.e.,

$$g \circ h = h \circ g \quad \text{for all } g, h \in G$$

Let us evaluate the expression $(g \circ h)^2$ using the properties of an abelian group:

1. By definition of squaring: $$(g \circ h)^2 = (g \circ h) \circ (g \circ h)$$
2. By associativity of the group operation: $$(g \circ h)^2 = g \circ (h \circ g) \circ h$$
3. Since $G$ is abelian, we can swap $h \circ g$ with $g \circ h$: $$(g \circ h)^2 = g \circ (g \circ h) \circ h$$
4. Regrouping using associativity: $$(g \circ h)^2 = (g \circ g) \circ (h \circ h) = g^2 \circ h^2$$

Thus, Option C is always true.

Counterexamples for other options:

• Option A: For the abelian group $(\mathbb{Z}_3, +)$, the element $1$ has inverse $2 \neq 1$. Hence, $g = g^{-1}$ does not hold for all elements.
• Option B: In any group, $g = g^2 \implies g \circ e = g \circ g \implies g = e$. This is only true for the identity element, not for every element in a non-trivial abelian group.
• Option D: The set of integers under addition, $(\mathbb{Z}, +)$, is an abelian group of infinite order. Thus, $G$ does not have to be of finite order.