Question

If $\mathit{G}$ is a group under the stated operation, prove it; if not, give a counterexample:

(a) .$\mathit{G}\mathbf{=}\mathit{\mathbb{Q}}\mathbf{;}\mathit{a}\mathbf{\ast }\mathit{b}\mathbf{=}\mathit{a}\mathbf{+}\mathit{b}\mathbf{+}\mathbf{3}$

Short answer

It is proved that$(G,\ast )$ forms a group

Step-by-step solution

Group

A non-empty set $\mathit{G}$ is said to form a group under binary composition $\mathbf{\ast }$ if,

• $\mathit{G}$is closed under$\mathbf{\ast }$ .
• $\mathbf{\ast }$is associative.
• There exists an element $\mathit{e}$in $\mathit{G}$ such that $\mathit{e}\mathbf{\ast }\mathit{a}\mathbf{=}\mathit{a}\mathbf{\ast }\mathit{e}\mathbf{=}\mathit{a}$ for all $\mathit{a}\mathbf{\in }\mathit{G}$.
• For each element$\mathit{a}\mathbf{\in }\mathit{G}$ ,there exists an element${\mathit{a}}^{\mathbf{\text{'}}}$ in $\mathit{G}$ such that,${\mathit{a}}^{\mathbf{\text{'}}}\mathbf{\ast }\mathit{a}\mathbf{=}\mathit{a}\mathbf{\ast }{\mathit{a}}^{\mathbf{\text{'}}}\mathbf{=}\mathit{e}$

.

Proof

1.Let ,$a,b\in G$then, $a\ast b=a+b+3\in \mathbb{Q}\Rightarrow a\ast b\in G$, which shows that $G$ is closed under $\ast$

2.Let $a,b,c\in G$ then,

$\begin{array}{l}(a\ast b)\ast c=(a+b+3)\ast c\\ =a+b+3+c+3\\ =a+(b+c+3)+3\\ =a+(b\ast c)+3\\ =a\ast (b\ast c)\end{array}$

which shows that $\ast$ is associative.

3. Now,

$\begin{array}{l}a\ast (-3)=a+(-3)+3=a;\\ (-3)\ast a=a+(-3)+3=a\end{array}$.

which proves that $(-3)$ is the identity element of .

4. Let $a\in G$ be arbitrary. Then,

$\begin{array}{l}a\ast (-6-a)=a+(-6-a)+3=-3;\\ (-6-a)\ast a=(-6-a)+a+3=-3.\end{array}$

This shows that $(-6-a)$ is the inverse of $a$. So, every element in$G$ has an inverse.

This proves that$(G,\ast )$ forms a group under the binary operation$\ast$ .