Every element in a finite (abelian) group $G$ is an $n$'th power if $\,\gcd(n,|G|)=1$

Question

How to prove that every element in a finite (abelian) group $G$ is an $n$'th power if $\,\gcd(n,|G|)=1$

Answer 1

Perhaps you don't know it by the name "Bezout Equality", but it is probably the easiest way to go here.

It says that if $\;x,y,\in\Bbb Z\;$ and if $\;\;$ g.c.d. $(x,y)=d\;$ , then there exist $\;m,n\in\Bbb Z\;$ with $\;mx+ny=d$ .

In our case, there exist $\;a,b\in\Bbb Z\;$ s.t. $\;a\mathcal O(G)+bn=1\;$ , and then for any $\;g\in G\;$ we get

$$g=g^1=g^{a\mathcal O(G)+bn}=\left(g^{\mathcal O(G)}\right)^a\left(g^b\right)^n=1 x^n=x^n\;,\;\;\text{with}\;\;x:=g^b$$

Answer 2

Here are my comments expanded into an answer which does not use Bezout's identity (I will in fact at the end use this to prove Bezout's identity purely using group theory).

Let's take a more general setup and prove something a bit stronger:
Let $G$ be any finite group and assume that $n$ is coprime to $|G|$. Then for any $x$ in $G$ there is a $y\in G$ such that $y^n = x$ and such that $y = x^m$ for some $m$.

Proof: Let $H = \langle x\rangle$ and consider the map from $H$ to itself given by $a\mapsto a^n$. This is easily seen to be a homomorphism since $H$ is abelian.

We also see that since $n$ is coprime to $|G|$ it is also coprime to $|H|$ and hence the only element $h\in H$ with $h^n = 1$ must be $1$ (by Lagrange). This means that the map above has trivial kernel, and hence that it is injective.

But since $H$ is finite, this also means that the map is surjective, and hence $x$ must be in the image, which precisely means that there is some $y\in H$ with $y^n = x$ (and since we have $y\in H$ we have $y = x^m$ for some $m$).

To prove Bezout's identity from the above, let two coprime numbers $k$ and $n$ be given. We need to find integers $a$ and $b$ such that $ak + bn = 1$. Note that by definition this means that we need to find an integer $b$ such that $bn\equiv 1\pmod{k}$.

To apply the above statement, let $G = \langle x\rangle$ be the cyclic group of order $k$. The above says that there is an integer $m$ such that $(x^m)^n = x$.

But $(x^m)^n = x^{mn}$, and if this equals $x$ then this means that $x^{mn - 1} = 1$, i.e. that $mn - 1$ is divisible by the order of $x$, which is $k$. By definition, this means that $mn - 1\equiv 0 \pmod{k}$ or in other words that $mn\equiv 1\pmod{k}$, so $m$ is the integer we were looking for (the $b$ above).

Answer 3

As in $U(\Bbb Z/m),\,$ it's easy to take $n$'th roots in a group if $\,n\,$ is coprime to the group order $\:\!|G|$.

Theorem $ $ Compute $\,n$'th root in group $\,G\,$ by raising to power $\frac{1}n\!\pmod{\!|G|}\,$ if $\,\gcd(n,|G|)\! =\! 1$

Let $\,\color{#0a0}{f=|G|}.\,$ By Bezout $\,\gcd(n,f)\!=\!1$ $\Rightarrow \exists\:\! n' \equiv \frac{1}n\equiv n^{-1}\!\pmod{\!f},\, $ so $\ \color{#c00}{nn' = 1 + fj}\,$ so $$ \bbox[8px,border:2px solid #96f]{{a^{\Large\color{c00} n} = b \iff a = b^{\:\!\Large n'},\ \ n' \equiv \frac{1}n\!\!\!\!\pmod{\!f}}}\qquad$$

$\begin{align}{\bf Proof}\ \ \ &b = a^{\large n}\,\Rightarrow\, b^{\large n'}\! = a^{\large \color{#c00}{nn'}}\! = a^{\large\color{#c00}{1+fj}} = a(\color{#0a0}{a^{\large f}})^{\large j} = a\\ &a = b^{\large n'}\!\Rightarrow\, a^{\large n} = b^{\large\color{#c00}{n'n}} = \,b^{\large\color{#c00}{1+fj}} = \,b(\color{#0a0}{b^{\large f}})^{\large j} = b \end{align}$

because, $ $ by Lagrange's Theorem, $\ \color{#0a0}{f=|G|\,\Rightarrow\, a^f = 1 = b^f}$

Answer 4

This is related to the OP. In fact, the converse of this statement is also true, i.e., if every element in a finite group $G$ is a $n$-th power, then $\gcd(n,|G|) =1$.

We prove this by contradiction. First we note that the map $G \to G: g \mapsto g^n$ is surjective by assumption. Since $G$ is finite, this map is also injective. Now we suppose that $\gcd(n,|G|)>1$, then there's a prime number $p$ dividing both $n$ and $|G|$. By Cauchy's theorem, there is an element $x\in G$ with order $p$. Since $x$ is a $n$-th power (by assumption), there is $y\in G$ such that $x = y^n$, and $|y^n|=p$, i.e., $y^{np}=(y^p)^n = 1$. Because the map $g\mapsto g^n$ is injective and $1^n = 1$, we know that $y^p=1$. This forces $|y|=p$. Therefore, $y^n = y^{p\cdot (n/p)} =1$. But then this contradicts the fact that $|y^n| = p$ and we proved the claim.