Proof that equation is solvable in abelian group

Question

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

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 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

score 2 · Answer 1 · edited Nov 21 '20 at 09:19

2

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

edited Nov 21 '20 at 09:19

Avisek Sharma

173
14

answered Oct 22 '20 at 14:24

Asinomás

105,651

score 2 · Accepted Answer · answered Oct 22 '20 at 14:34

2

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.

answered Oct 22 '20 at 14:34

Avisek Sharma

173
14

Proof that equation is solvable in abelian group

2 Answers2