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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 $a^k=a^{-1}$.
Solution
Note that $a^0 = e$. Consider the sequence $a^1, a^2, a^3, \ldots$. This is an infinite sequence, but we know that $G$ is a finite group. This means that some elements in the sequence are repeated. Thus there exists $i, j \in \mathbb{N}$ such that $i > j$ and $a^i = a^j$. Note that if $a^i = a^j$, then $a^{i-1} = a^{j-1}$ because the inverse of an element in a group is unique. Let $i, j \in \mathbb{N}$ such that $i > j$ and $a^i = a^j$. If we multiply both sides of $a^i = a^j$ on the right by $(a^{-1})^{i}$ (we multiply by $a^{-1}$ on the right $i$ times) we get $a^{i-i} = a^{j-i}$. Since $a^{i-i}$ is clearly equal to the identity, we get $e = a^{i-j}$. Multiplying on the right by $a^{-1}$ gives $a^{-1} = a^{i-j-1}$. Since $i > j$ and $i, j \in \mathbb{N}$, we know that $i-j-1 \geq 0$. If $i-j-1 = 0$, let $k=1$. Otherwise, let $k = i -j-1$. In either case, $a^k = a^{-1}$.
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