Let $(G, ∗)$ be a finite abelian group and $G=\{a_1,a_2,...,a_n\}$. Let $a_1∗a_2∗···∗a_n=x$. Show that $x∗x=e$, where $e$ is the identity of $(G, ∗)$.

Question

Let $(G, ∗)$ be a finite abelian group and $G = \{a_1, a_2, . . . , a_n\}$. Let $a_1 ∗a_2 ∗· · · ∗a_n = x$. Show that $x ∗ x = e$, where $e$ is the identity of $(G, ∗)$.

Please give me head start on how to proceed. I am completely blank.

Answer 1

Let $x=a_1\ast\dots \ast a_n$. Since $(G, \ast)$ is a group, each $a_i$ has an inverse $a_{j_i}$ in $G$ with respect to $\ast$ for some $j_i$. Because inverses are unique, we have a bijection between $\{a_i\mid i\in\{1, \dots, n\}\}$ and $\{a_{j_i}\mid j_i\in\{1, \dots, n\}\}$ given by taking inverses. Thus the product that is $x$ is equal to the product of all the inverses $a_{j_i}$ of each $a_i$. But $(G, \ast)$ is abelian. Thus we can reason as follows:

$$x^2=(a_1\ast\dots\ast a_n)\ast\color{red}{(a_1\ast\dots \ast a_n)},$$

so we can move each of the inverses $\color{red}{a_{j_i}}$ to the left until they are paired off with their respective $a_i$, like this:

$$\begin{align} x^2&=(a_1\ast\dots\ast a_n)\ast\color{red}{(a_{j_1}\ast\dots \ast a_{j_n})}\\ &=(a_1\ast \color{red}{a_{j_1}})\ast\dots\ast(a_n\ast \color{red}{a_{j_n}}), \end{align}$$

meaning that $x^2$ is the product of pairs of elements with their inverses, which are each equal to the identity $e$. Hence $x^2=e.$

Answer 2

Hint:

Among the factors in the product $a_1\times a_2\times\cdots\times a_n$ is each element and its inverse,

unless the element is its own inverse. An element times its inverse is the identity.

Answer 3

In any group $a^{-1}=b^{-1} \iff a=b$. Hence the set $G=\{a_1, a_2, . . . , a_n\}=\{a_1^{-1}, a_2^{-1}, . . . , a_n^{-1}\}$.

Therefore since $(G,*)$ is abelian $x^{-1}=(a_1 ∗a_2 ∗· · · ∗a_n)^{-1}=a_n^{-1} ∗a_{n-1}^{-1} ∗· · · ∗a_1^{-1}=a_1 ∗a_2 ∗· · · ∗a_n=x$.