Question

Let $\mathit{G}$be an abelian group of odd order. If role="math" localid="1654348008954" ${\mathit{a}}_{\mathbf{1}}\mathbf{,}{\mathit{a}}_{\mathbf{2}}\mathbf{,}{\mathit{a}}_{\mathbf{3}}\mathbf{,}\mathbf{\dots }\mathbf{,}{\mathit{a}}_{\mathbf{n}}$ are the distinct elements of $\mathit{G}$(one of which is the identity $\mathit{e}$), prove that role="math" localid="1654348057092" ${\mathit{a}}_{\mathbf{1}}{\mathit{a}}_{\mathbf{2}}{\mathit{a}}_{\mathbf{3}}\mathbf{\dots }{\mathit{a}}_{\mathbf{n}}\mathbf{=}\mathit{e}$ .

Short answer

We proved that the product$a_0{a}_1{a}_2\dots a_n=e$

Step-by-step solution

To show the product of two elements is identity

Let $G$be an abelian group of odd order.

Let $G=\{a_0,a_1,a_2,\dots ,a_n\}$, where one of the $a_i\text{'}s$ is identity $e$, for $0\le i\le n$.

We have to show that role="math" localid="1654348550369" $a_0{a}_1{a}_2\dots a_n=e$.

Since the order of $G$ is odd, then by Lagrange’s Theorem, there is no element of order 2 in $G$.

$\therefore$for every element $b\in G\backslash \left\{e\right\},\exists b^{-1}\in G$, which is unique, such that $b{b}^{-1}=e$and $b\ne b^{-1}$.

$\therefore$there exists a unique $j_i\in \{0,1,2,\dots ,n\}$ such that

$a_i{a}_{j_i}=e$ …… for every $0\le i\le n$

To prove  a1a2a3 … an= e

Now, from step 1, the product:

$$\begin{gathered} a_0{a}_1{a}_2\dots a_n=a_0{a}_{j_0}{a}_1{a}_{j_1}{a}_2{a}_{j_2}\dots a_n{a}_{j_n} \\ =e\cdot e\cdot e\cdot \dots \cdot e \\ =e \end{gathered}$$

Hence, $a_0{a}_1{a}_2\dots a_n=e$.