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.
Given $G$ group and $a \in G$ , $a^n = e$, $\gcd(m,n) = 1$, prove $\exists b \in G $ that $a = b^m$.
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4Hint: there exist integers $r$ and $s$ such that $rm + sn = 1$. So $a^1 = \dots$ – Alex Macedo Jan 08 '17 at 17:47
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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 $c = b^n$. – Calum Gilhooley Jan 08 '17 at 18:03
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Proof: with $r$ and $s$ as in Alex Macedo's comment, put $b = a^rc^s$. – Calum Gilhooley Jan 08 '17 at 22:57
4 Answers
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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!!
Luis Martinez
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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. – Bill Dubuque Jan 08 '17 at 18:45
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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. – Bill Dubuque Jan 08 '17 at 18:53
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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 $b=a^x$.
seeker
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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. – Bill Dubuque Jan 08 '17 at 18:45
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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 $\,\Bbb R$
Bill Dubuque
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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 $n$ $\quad$ – Bill Dubuque Jan 08 '17 at 18:50
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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? – Ilan Aizelman WS Jan 08 '17 at 19:44
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@Ilan $\ a^{\large 63k}=1,\Rightarrow, 218\mid 63k.\ $ What does this imply about $k$ given $,\gcd(218,63) = 1?\ \ $ – Bill Dubuque Jan 08 '17 at 19:54
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1@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. – Bill Dubuque Jan 10 '17 at 15:22
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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 $b = a^{-m}$.
martini
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user404634
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For answers, and in general, you should use LaTeX formatting! This is highly unreadable! – The Count Jan 08 '17 at 18:17