Question
Suppose $G$ is a finite group of order $n$ and $m$ is relatively prime to $n$. If $g \in G$ and $g^{m}=e$, prove that $g=e$.
Step 1
Now, let's consider the element $g^1 = g^{mx + ny} = (g^m)^x \cdot (g^n)^y$. We are given that $g^m = e$, and by Lagrange's theorem, we know that $g^n = e$ since the order of $G$ is $n$. Show more…
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ELEMENTARY PROPERTIES OF GROUPS
E
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