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Given $G$ group and $a \in G$ , $a^n = e$, $\gcd(m,n) = 1$, prove that $\exists b \in G $ that $a = b^m$.

Well, I'm not sure how to approach this, any hints are welcomed.

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    Hint: there exist integers $r$ and $s$ such that $rm + sn = 1$. So $a^1 = \dots$ – 2017-01-08
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    More generally, if $a$ and $c$ commute, and $a^n = c^m$, then there exists $b$ such that $a = b^m$ and $c = b^n$. – 2017-01-08
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    Proof: with $r$ and $s$ as in Alex Macedo's comment, put $b = a^rc^s$. – 2017-01-08

4 Answers 4

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Since $a^n=e$, then $G$ is a cyclic group generated by a. If $(m,n) = 1$,then $(-m,n)=1$ and $a^{-m}$ is another generator of the group $G$. Denote $k := -m$.

For the proof, suppose by contradiction. Suppose that $a^k$ is not a generator of $G$. So, the order of the element $a^k$ is not $n$, i. e. there exists a natural number $n_1 < n$ such that $({a^k})^{n_1}=a^{kn_1}=e$. Since the order of the element $a$ is $n$, then $kn_1=nq$ and since $(n,k)=1$, then $n\mid n_1$, i. e. $n_1\ge n$. This is a contradiction and as a result $a^k$ is a generator of the group $G$: $a^k=a^{-m}=b$, i. e. $b^m=a$. Choose $b = a^{-m}$.

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    For answers, and in general, you should use LaTeX formatting! This is highly unreadable! – 2017-01-08
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Due to gcd($m,n$)=1, $\exists$ $c,d \in \mathbb{Z}$ $|$ $1=cm+dn$ $\Rightarrow$ $a^1=a^{cm+dn} \Rightarrow a^{cm+dn}=a^{cm}a^{dn}=(a^{c})^m (a^n)^{d}=(a^c)^m e^d=(a^c)^m$.

Now, define as $b:=a^c$, and that's all!!

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    **Note** $\ c = {\large \frac{1}m}\bmod n\ $ so it's $\ \left(a^{\Large \frac{1}m}\right)^m\! = a\ $ with exponents interpreted mod $n,\,$ see my answer. – 2017-01-08
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    I don't get it... Could you explain yourself? – 2017-01-08
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    The point is that the solution of $\ b^m = a\ $ is simply $\ b = a^{\large \frac{1}m},\, $ just like in $\,\Bbb R.$ The Bezout computations obfuscate this key idea. – 2017-01-08
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Since $gcd(m,n)=1\implies$ by Bezout's Lemma, $\exists x,y\in \mathbb{Z}$ such that $mx+ny=1$.

So $a=a^1=a^{mx+ny}=(a^x)^m.(a^n)^y=(a^x)^m$.

Choose $b=a^x$.

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    **Note** $\ x = {\large \frac{1}m}\bmod n\ $ so it's $\ \left(a^{\Large \frac{1}m}\right)^m\! = a\ $ with exponents interpreted mod $n,\,$ see my answer. – 2017-01-08
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    This is perfect @seeker. Thank you! – 2017-01-08
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Hint $\,\ \gcd(m,n) = 1\,\Rightarrow\,{\large \frac{1}m}\bmod n\,$ exists $\,\Rightarrow \left[\,\color{#c00}a^{\Large\color{#c00}{ \frac{1}m}\!\bmod n} \right]^{\large\underset{\LARGE\, m}{\phantom{1}}}\! = a\ $

Remark $ $ Thus the solution of $\ b^m = a\ $ is simply $\ b = \color{#c00}{a^{\large \frac{1}m}},\, $ just like in $\,\Bbb R$

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    We used $\, j\equiv k\pmod n\,\Rightarrow\, a^{\large j} = a^{\large k}\,$ by $\,a^n = 1,\,$ i.e. exponents on $\,a\,$ can be interpretd mod $n$ $\quad$ – 2017-01-08
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    Thanks, what if $o(a) = 218$, what will be the result of $o(a^{63})$? can I use what we proved here on this? – 2017-01-08
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    @Ilan $\ a^{\large 63k}=1\,\Rightarrow\, 218\mid 63k.\ $ What does this imply about $k$ given $\,\gcd(218,63) = 1?\ \ $ – 2017-01-08
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    Hmm, I am not sure sir. ): – 2017-01-08
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    @Ilan Apply [Euclid's Lemma.](http://math.stackexchange.com/a/1162403/242) – 2017-01-08
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    I understood nothing from what you wrote there. :( – 2017-01-10
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    Maybe $k$ needs to be $218$? – 2017-01-10
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    because then $a^{{(218)}63} =e^{63} = e$ – 2017-01-10
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    @Ilan yes, since $218,63$ are coprime, $\,218\mid 63k\,\Rightarrow\,218\mid k\,$ By Euclid's Lemma, or by FTA, or by gcd properties, etc. – 2017-01-10
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    I see. so $o(a^{63}) = o(a) = 218$ – 2017-01-10
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    @Ilan Exactly $\phantom{}$ – 2017-01-10