Question

Suppose$\mathit{G}$is a group with operation $\mathbf{\ast }$.Define a new operation $\#$on$\mathit{G}$by . $\mathit{a}\mathbf{\#}\mathit{b}\mathbf{=}\mathit{b}\mathbf{\ast }\mathit{a}$Prove that $\mathit{G}$is a group under$\mathbf{\#}$

Short answer

It is proved that$(G,\#)$ forms a group under the binary operation $\#$.

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, $a\#b=b\ast a\in G\Rightarrow a\#b\in G$,which shows that $G$is closed under$\#$

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

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

which shows that$\#$is associative.

2. Let$e\in G$be the identity of$\ast$, then$a\ast e=e\ast a=a,\forall a\in G$,.

.$\begin{array}{l}a\#e=e\ast a=a;\\ e\#a=a\ast e=a\end{array}$

3. Let $a^{\text{'}}\in G$be the inverse of$a$ in $G$with respect to$\ast$

Then, we have,$a\ast a^{\text{'}}=a^{\text{'}}\ast a=e$. So,

$\begin{array}{l}a\#a^{\text{'}}=a^{\text{'}}\ast a=e;\\ a^{\text{'}}\#a=a\ast a^{\text{'}}=e\end{array}$

This shows that$a^{\text{'}}$ is the inverse of$a$ . So, every element in $G$has an inverse under$\#$.

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