GATE CS 2003 Q7 — Monoids and Groups

To determine the algebraic structure of $(\Sigma^*, \cdot)$, where $\Sigma = \{0, 1\}$ and $\cdot$ is the string concatenation operator, let us verify the group axioms:

1. Closure: For any two strings $x, y \in \Sigma^*$, their concatenation $x \cdot y$ is also a string over $\Sigma$, so $x \cdot y \in \Sigma^*$.
2. Associativity: String concatenation is associative. For any $x, y, z \in \Sigma^*$, we have $(x \cdot y) \cdot z = x \cdot (y \cdot z)$.
3. Identity Element: The empty string $\epsilon \in \Sigma^*$ acts as the identity element because for any $x \in \Sigma^*$, $$x \cdot \epsilon = \epsilon \cdot x = x$$ Thus, a unique identity element exists (which is both a left and right identity).
4. Inverse Element: For any non-empty string $x \in \Sigma^*$ (such as $x = 0$), there is no string $y \in \Sigma^*$ such that: $$x \cdot y = \epsilon$$ Since the length of the concatenated string $|x \cdot y| = |x| + |y| \ge 1$, it can never equal the length of the empty string ($|\epsilon| = 0$). Thus, non-empty strings do not have an inverse.

Since the inverse axiom fails, $(\Sigma^*, \cdot)$ is a monoid but does not form a group.

Analysis of Options:

• A is correct: It does not form a group because non-empty strings lack inverses.
• B is incorrect: It is not a group.
• C is incorrect: The empty string $\epsilon$ serves as the right (and left) identity.
• D is incorrect: Removing the empty string $\epsilon$ removes the identity element, leaving it as a semigroup, which still cannot be a group.