Problem of the Day | 21-05-2025 | Abstract Algebra - Herstein Chapter 2 - Problem 1 - Determine if the following set is a group with the given operation

Sachchidanand Prasad

21-05-2025

Problem: Determine if the following sets $G$ with the operation indicated form a group. If not, point out which of the group axioms fail.

$G=$ set of all integers, $a * b=a-b$.
$G=$ set of all integers, $a * b=a+b+a b$.
$G=$ set of nonnegative integers, $a * b=a+b$.
$G=$ set of all rational numbers $\neq-1, a * b=a+b+a b$.
$G=$ set of all rational numbers with denominator divisible by 5 (written so that numerator and denominator are relatively prime), $a * b=$ $a+b$.
$G$ a set having more than one element, $a * b=a$ for all $a, b \in G$.

Solution: We recall that $G(,*)$ is a group if it satisfies the following properties:

$*$ is closed, that is, given any $a, b \in G,\ a * b \in G$;
Associativity, that is, given any $a,b,c \in G$, $(a*b)*c = a *(b*c)$;
Existence of identity, that is, there exists $e\in G$ such that for any $a\ in G$ $e*a = a = a*e$;
Existence of inverse, that is, given any $a \in G$, there exists $b \in G$ such that $a*b = e = b*a$.

For the given problem, we will check if any of the above axioms fail.

$(\mathbb{Z} , *)$, where $a * b = a - b$. This is NOT a group as it does not satisfy the associativity rule. For example, take $a = 3, b = 2, c = 1$, then \begin{align*} (a*b)*c = (3-2) - 1 = 0 \quad \text{ but } a*(b*c) = 3 - (2 - 1) = 2. \end{align*}

$(\mathbb{Q} , *)$, where $a*b = a + b + ab$. Clearly the $\mathbb{Q} $ is closed under the given operation. We will check the other axioms.
1. Associativity: For any $a,b,c \in \mathbb{Q} $, \begin{align*} (a * b) * c & = (a + b + ab) * c \\ & = a + b + ab + c + (a + b + ab)c \\ & = a + b + c + ab + ac + bc + abc \\[2ex] a * (b * c) & = a * ( b + c + bc) \\ & = a+ b + c + bc + a(b + c + bc) \\ & = a + b + c + bc + ab + ac + abc. \end{align*} Thus, the operation is associative.
2. Existence of identity: If $e$ is the identity, then for any $a \in \mathbb{Q} $, \begin{align*} a * e = a & \implies a + e + ae = a \\ & \implies e + ae = 0 \\ & \implies e = 0. \end{align*} This shows that $e = 0$ is the identity of the group.
3. Existence of inverse: Let $a \in \mathbb{Q} $. Then \begin{align*} a * b = 0 & \implies a + b + ab = 0 \implies b = -\frac{a}{1+a}, \end{align*} but the above one is not weill defined for $a = -1$. Also, note that $a = -1$, \[ -1 * b = -1 + b + b (-1) = -1. \]

$G$ is NOT a group as inverse of $1$ is $-1$ which does not belong to the set $G$.

The set $G$ can be written as \[ G = \left\{ \frac{p}{q} \in \mathbb{Q} : 5 \mid q \text{ and } \gcd (p,q) = 1\right\}. \] The operation is given by $a = \dfrac{p_1}{q_1}$ and $b = \dfrac{p_2}{q_2}$, then $a * b = a + b$. This is \textbf{NOT} a group. For example, take $a = \dfrac{4}{5}$ and $b = \dfrac{6}{5}$. Then \[ a * b = a + b = \frac{4}{5} + \frac{6}{5} = \frac{10}{5} = \frac{2}{1}, \] and hence $G$ is not closed under $*$.

This is NOT a group. For example, take $G = \{a,b\}$. Then one of them must be identity, say $a$. Applying the given operation, we have \begin{align*} a * a = a,\ b * b = b,\ a * a = a \text{ and } b * a = b. \end{align*} Since $a \neq b$, so $a * b \neq b * a$ and hence $a$ is not the identity. Hence $G$ is not a group.

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