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Let $G$ be an abelian group, $a \in G$ be an element of finite order, $(\text{ord} \, a, n) = 1$. Prove that the equation $x^n = a$ is solvable in the group $G$.

I tried to apply a corollary from Lagrange's theorem, but I am concerned that the group can be of infinite order, and I do not know what to do with it

Purple Piper
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    Please provide some details on what you understand, what you tried, and where you got stuck. This website is not for automatic homework solutions. – halrankard2 Oct 22 '20 at 14:22
  • See also [this question](https://math.stackexchange.com/questions/1064817/let-g-to-be-finite-abelian-group-of-order-og-let-n-to-be-prime-number-and). – Dietrich Burde Oct 22 '20 at 14:27
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    See the Theorem in the linked dupe (the proof works in any group, not only the case $U(\Bbb Z/n)$ done there) – Bill Dubuque Oct 22 '20 at 18:07

2 Answers2

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let $m= \text{ord}(a)$. and pick $k$ such that $k$ is congruent to the inverse of $n\pmod m$. We thus can write $kn$ as $dm+1$.

Notice that $(a^k)^n = a^{kn}=a^{dm+1} = a^{dm}\cdot a = e\cdot a = a.$

Avisek Sharma
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Asinomás
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If we put $m=\textrm{ord}(a)$ then $\textrm{gcd}(m,n)=1$. This gives there are integers $t$ and $s$ such that $tm+sn=1$. Now, $a= a^1=a^{tm+sn}=(a^m)^t\cdot (a^s)^n$
$=e_G^t\cdot (a^s)^n=(a^s)^n$.
Hence, $a^s$ is a solution.

Avisek Sharma
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