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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 $a^k=a^{-1}$.

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

Consider the sequence $a, a^2, a^3, ...$. Because, $G$ is finite, eventually, there will occur some element $a^k$ such that $k$ is an integer greater than the number of elements in $G$. Since $a^l \in G$ for all $l \in \mathbb{Z}$, there must have been a repeated element in the sequence $a, a^2, ...$. Consider these two repeated elements, and call them $a^i$ and $a^j$, such that $i > j$. Since we know $a^i = a^j$, we compute $$a^j = a^i = \underbrace{aaa...aaa}_{i \text{ times}} = (\underbrace{aaa...aaa}_{i-j \text{ times}})(\underbrace{aaa...aaa}_{j \text{ times}}) = a^{i-j}a^j.$$ Multiplying by the inverse of $a^j$ on the right side, we find that $a^j(a^j)^{-1} = a^{i-j}a^j(a^j)^{-1}$. By the property of associativity, we can rewrite this as $a^j(a^j)^{-1} = a^{i-j}(a^j(a^j)^{-1})$/ By the definition of the identity, it follows that $e = a^{i-j}e$. Using the definition of the identity again and rearranging the equation, we find that $a^{i-j} = e$. We then compute $$ e = a^{i-j} = aa^{i-j - 1}.$$

Thus,we know that $a^{j - i - 1}$ is the inverse of $a$. Because $i > j$, we know that$i - j - 1 \geq 0$. We must now consider the case that $i - j -1 \ne 0$. Assume, by way of contradiction, that $i - j - 1 = 0$. Because we know that $e=aa^{i-j-1}$, we know that $e=aa^0 = ae = a$. This implies that $a = e$. Since $e^p = e^{-1}$ for all positive integer values of $p$, we know that there exists another positive integer $k$ such that $a^k = a^{-1}$ in the case that $i - j - 1 = 0$

Thus, we know that there exists some positive integer $k$ such that $a^k = a^{-1}$.

From Ben: How do you know this integer $k$ is not 0?
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