Prove that $g mapsto g^a$ is a permutation












1












$begingroup$


"Prove that $g mapsto g^a$ for any $a in mathbb{Z}$ that is prime to the group order is a Permutation on any finite group G."



Can somebody please help me?



My idea was to show that $g,h in G$ with $g^a=h^a$ implies $g=h$, but I struggle to show that.






abstract-algebra group-theory permutations







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
    $endgroup$
    – Ethan Bolker
    Dec 12 '18 at 13:23












  • $begingroup$
    You´re sure it´s not true? I have to show that this map is bijective.
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 13:30






  • 1




    $begingroup$
    Wait...missed one argument: a has to be prime to the group order!
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 13:33
















1












$begingroup$


"Prove that $g mapsto g^a$ for any $a in mathbb{Z}$ that is prime to the group order is a Permutation on any finite group G."



Can somebody please help me?



My idea was to show that $g,h in G$ with $g^a=h^a$ implies $g=h$, but I struggle to show that.






abstract-algebra group-theory permutations







share|cite|improve this question











$endgroup$








  • 4




    $begingroup$
    You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
    $endgroup$
    – Ethan Bolker
    Dec 12 '18 at 13:23












  • $begingroup$
    You´re sure it´s not true? I have to show that this map is bijective.
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 13:30






  • 1




    $begingroup$
    Wait...missed one argument: a has to be prime to the group order!
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 13:33














1












1








1





$begingroup$


"Prove that $g mapsto g^a$ for any $a in mathbb{Z}$ that is prime to the group order is a Permutation on any finite group G."



Can somebody please help me?



My idea was to show that $g,h in G$ with $g^a=h^a$ implies $g=h$, but I struggle to show that.






abstract-algebra group-theory permutations







share|cite|improve this question











$endgroup$




"Prove that $g mapsto g^a$ for any $a in mathbb{Z}$ that is prime to the group order is a Permutation on any finite group G."



Can somebody please help me?



My idea was to show that $g,h in G$ with $g^a=h^a$ implies $g=h$, but I struggle to show that.






abstract-algebra group-theory permutations




abstract-algebra group-theory permutations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 6:06









Martin Sleziak

44.7k10119272




44.7k10119272










asked Dec 12 '18 at 13:21









FabianSchneiderFabianSchneider

706




706








  • 4




    $begingroup$
    You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
    $endgroup$
    – Ethan Bolker
    Dec 12 '18 at 13:23












  • $begingroup$
    You´re sure it´s not true? I have to show that this map is bijective.
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 13:30






  • 1




    $begingroup$
    Wait...missed one argument: a has to be prime to the group order!
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 13:33














  • 4




    $begingroup$
    You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
    $endgroup$
    – Ethan Bolker
    Dec 12 '18 at 13:23












  • $begingroup$
    You´re sure it´s not true? I have to show that this map is bijective.
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 13:30






  • 1




    $begingroup$
    Wait...missed one argument: a has to be prime to the group order!
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 13:33








4




4




$begingroup$
You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
$endgroup$
– Ethan Bolker
Dec 12 '18 at 13:23






$begingroup$
You can't prove it because it's not true. What about squaring in the two element group? Are you sure you have all the hypotheses?
$endgroup$
– Ethan Bolker
Dec 12 '18 at 13:23














$begingroup$
You´re sure it´s not true? I have to show that this map is bijective.
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:30




$begingroup$
You´re sure it´s not true? I have to show that this map is bijective.
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:30




1




1




$begingroup$
Wait...missed one argument: a has to be prime to the group order!
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:33




$begingroup$
Wait...missed one argument: a has to be prime to the group order!
$endgroup$
– FabianSchneider
Dec 12 '18 at 13:33










1 Answer
1






active

oldest

votes


















1












$begingroup$

Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.



Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    THANK YOU very much!
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 20:33











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.



Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    THANK YOU very much!
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 20:33
















1












$begingroup$

Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.



Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    THANK YOU very much!
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 20:33














1












1








1





$begingroup$

Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.



Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.






share|cite|improve this answer









$endgroup$



Let $n$ be the order of $G$. SInce $(n,a)=1$, by Bezout's lemma there are $s,tin mathbb{Z}$ such that $sn+ta=1$.



Notice the map $hmapsto h^{t}$ is the inverse of the provided map in the problem. To see that notice that
begin{align*}
(g^a)^t&=g^{at}\
&=g^{at}g^{sn} quadquadtext{(since $g^n=1$)}\
&=g^{at+sn}\
&=g.
end{align*}
The other way around compostion give you similar result.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 12 '18 at 13:46









user9077user9077

1,239612




1,239612












  • $begingroup$
    THANK YOU very much!
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 20:33


















  • $begingroup$
    THANK YOU very much!
    $endgroup$
    – FabianSchneider
    Dec 12 '18 at 20:33
















$begingroup$
THANK YOU very much!
$endgroup$
– FabianSchneider
Dec 12 '18 at 20:33




$begingroup$
THANK YOU very much!
$endgroup$
– FabianSchneider
Dec 12 '18 at 20:33


















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