Please Login to access more options.

Problem 50 (The Inverse In A Finite Group Is A Power Of The Element)

Let $G$ be finite group with $a\in G$. Prove that there exists a positive integer $k$ such that $a^k=a^{-1}$.

Hint: Consider the sequence $a, a^2, a^3, a^4, \ldots$. Why must you eventually have a repeated element? So if $a^i=a^j$ for some $j>i$, how can you use this to find $a^{-1}$?

Solution

Let $f : \mathbb{N} \rightarrow G$ be the sequence defined by $f(n) = a^n$. We know that since $G$ is a finite group then the image of $f$ is also finite; therefore, there must be repeated values in the codomain. This implies there exists $i, j \in \mathbb{N}$ where $j > i$ such that $a^j = a^i$.

Let $a^i, a^j \in G$ for some $i, j \in \mathbb{N}$ such that $a^j = a^i$ where $i = j - 1$. This means that $a^j = a^i = a^{j - 1}$ which is logically equivalent to $e = a^{-1}$ where $e$ is the identity in $G$. This implies that $a = a^{-1}$, and therefore $k = 1$ satisfies $a^k = a^{-1}$.

Let $a^i, a^j \in G$ for some $i, j \in \mathbb{N}$ such that $a^j = a^i$ where $j - 1 > i$. We know that since $G$ is a group $a^{-i} \in G$ as well (which is the repeated operation of $a^{-1}$ an $i$ number of times). Left multiplying $a^j = a^i$ by $a^{-i}$, and right multiplying by $a^{-1}$ yields $(a^{-i}a^j)\;a^{-1} = (a^{-i}a^i)\;a^{-1}$. Using the identity property of G we can reduce the right hand side to get $(a^{-i}a^i)\;a^{-1} = (e)\;a^{-1} = a^{-1}$. An important step to note is that since $i, j \in \mathbb{N}$ we may use the left hand side of $(a^{-i}a^j)\;a^{-1} = a^{-1}$ and compute $$ \begin{align} (a^{-i}a^j)\; a^{-1} &= ( \underbrace{ a^{-1}a^{-1} \cdots a^{-1} }_\text{$i$}\; \underbrace{ aa \cdots a }_\text{$j$} )\; a^{-1} \\ &= ( \underbrace{ aa \cdots a }_\text{$j - i$} )\; a^{-1} \\ &= a^{j - i - 1}. \end{align} $$ Therefore we have $a^{j - i -1} = a^{-1}$. Let $k = j - i -1$. It follows from our condition that $j - 1 > i$ that the greatest integer that $i$ may assume is $j - 2$, and therefore $k = j - i -1 \geq j - ( j - 2 ) - 1 = 1$. Thus $a^k = a^{-1}$ for $k$ a positive number.

Q.E.D.

Tags
  • When you are ready to submit this written work for grading, add the phrase [[!Submit]] to your page. This will tell me that you have completed the page (it's past rough draft form, and you believe it is in final draft form). Don't type [[!Submit]] on a rough draft.
  • If I put [[!NeedsWork]] on your page, then your job is to review what I've written, address any comments made, and then delete all the comments I made. When you have finished reviewing your work, leave [[!NeedsWork]] on your page and type [[!Submit]]. (Both tags will show up). This tells me you have addressed the comments.
  • I'll mark your work with [[!Complete]] after you have made appropriate revisions.