Let a be the unique element of any order in a group G.Is a in the centre of G?












1












$begingroup$


I can prove that the result is true if o(a) is 2 by taking another element z=xax^(-1) and then by using uniqueness.But for in general case I cannot understand how can I proceed further to prove it.Please help me.Thank you in advance






abstract-algebra







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$endgroup$








  • 4




    $begingroup$
    In general, can you establish a relationship between the order of $a$ (any element in any group) and the order of $xax^{-1}$?
    $endgroup$
    – Greg Martin
    Dec 20 '18 at 5:29






  • 2




    $begingroup$
    $a^{-1}$ has the same order as $a$, so that $o(a)in{1,2}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 20 '18 at 6:51
















1












$begingroup$


I can prove that the result is true if o(a) is 2 by taking another element z=xax^(-1) and then by using uniqueness.But for in general case I cannot understand how can I proceed further to prove it.Please help me.Thank you in advance






abstract-algebra







share|cite|improve this question









$endgroup$








  • 4




    $begingroup$
    In general, can you establish a relationship between the order of $a$ (any element in any group) and the order of $xax^{-1}$?
    $endgroup$
    – Greg Martin
    Dec 20 '18 at 5:29






  • 2




    $begingroup$
    $a^{-1}$ has the same order as $a$, so that $o(a)in{1,2}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 20 '18 at 6:51














1












1








1





$begingroup$


I can prove that the result is true if o(a) is 2 by taking another element z=xax^(-1) and then by using uniqueness.But for in general case I cannot understand how can I proceed further to prove it.Please help me.Thank you in advance






abstract-algebra







share|cite|improve this question









$endgroup$




I can prove that the result is true if o(a) is 2 by taking another element z=xax^(-1) and then by using uniqueness.But for in general case I cannot understand how can I proceed further to prove it.Please help me.Thank you in advance






abstract-algebra




abstract-algebra






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share|cite|improve this question










asked Dec 20 '18 at 5:21









Debprasad KunduDebprasad Kundu

112




112








  • 4




    $begingroup$
    In general, can you establish a relationship between the order of $a$ (any element in any group) and the order of $xax^{-1}$?
    $endgroup$
    – Greg Martin
    Dec 20 '18 at 5:29






  • 2




    $begingroup$
    $a^{-1}$ has the same order as $a$, so that $o(a)in{1,2}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 20 '18 at 6:51














  • 4




    $begingroup$
    In general, can you establish a relationship between the order of $a$ (any element in any group) and the order of $xax^{-1}$?
    $endgroup$
    – Greg Martin
    Dec 20 '18 at 5:29






  • 2




    $begingroup$
    $a^{-1}$ has the same order as $a$, so that $o(a)in{1,2}$.
    $endgroup$
    – Lord Shark the Unknown
    Dec 20 '18 at 6:51








4




4




$begingroup$
In general, can you establish a relationship between the order of $a$ (any element in any group) and the order of $xax^{-1}$?
$endgroup$
– Greg Martin
Dec 20 '18 at 5:29




$begingroup$
In general, can you establish a relationship between the order of $a$ (any element in any group) and the order of $xax^{-1}$?
$endgroup$
– Greg Martin
Dec 20 '18 at 5:29




2




2




$begingroup$
$a^{-1}$ has the same order as $a$, so that $o(a)in{1,2}$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 6:51




$begingroup$
$a^{-1}$ has the same order as $a$, so that $o(a)in{1,2}$.
$endgroup$
– Lord Shark the Unknown
Dec 20 '18 at 6:51










2 Answers
2






active

oldest

votes


















1












$begingroup$

If you have a valid proof for the case when the order of $a$ is two, $o(a)=2$, then I'm pretty sure that you're done with the question — because the order of $a$ doesn't really matter here. The order of $z=xax^{-1}$ is equal to the order of $a$; so, just as you said, by uniqueness it implies that $xax^{-1}=a$, and you're done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanx a lot.I just missed that
    $endgroup$
    – Debprasad Kundu
    Dec 20 '18 at 5:39



















-1












$begingroup$

What you want to show is that $forall x in G, x.a.x^{-1} = a$. The thing you miss is probably that $(x.a.x^{-1})^k = x.a^k.x^{-1}$ for any integer $k$. If you write the product down, you will see that all the intermediate $x^{-1}.x$ terms cancel each other $(x.a.x^{-1})^n = (x.a.x^{-1})(x.a.x^{-1})ldots(x.a.x^{-1})$



For a more rigorous proof you can do it by induction.



From here you have $(x.a.x^{-1})^n = x.a^n.x^{-1} = x.x^{-1} =1$. This means that $x.a.x^{-1}$ has an order $m$ that divises $n$.



Now, we just need to show that $m=n$. For that, one can see that $1 = (x.a.x^{-1})^m = x.a^m.x^{-1}$ which implies $a^m=1$. By definition of the order of an element we can deduce $m=n$ since the order is the smallest positive integer that verifies this property.



We proved that $x.a.x^{-1}$ is of order $n$ which implies, by uniqueness of an element of such order, that $x.a.x^{-1} = a$ which is what we wanted.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm not the downvoter, but if you wrap your maths in dollar signs "$" instead of backticks it will be much easier to read
    $endgroup$
    – postmortes
    Dec 20 '18 at 9:29










  • $begingroup$
    Thank you very much for the tips, it is indeed much more pleasant to read
    $endgroup$
    – Paul Cottalorda
    Dec 21 '18 at 14:08












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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

If you have a valid proof for the case when the order of $a$ is two, $o(a)=2$, then I'm pretty sure that you're done with the question — because the order of $a$ doesn't really matter here. The order of $z=xax^{-1}$ is equal to the order of $a$; so, just as you said, by uniqueness it implies that $xax^{-1}=a$, and you're done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanx a lot.I just missed that
    $endgroup$
    – Debprasad Kundu
    Dec 20 '18 at 5:39
















1












$begingroup$

If you have a valid proof for the case when the order of $a$ is two, $o(a)=2$, then I'm pretty sure that you're done with the question — because the order of $a$ doesn't really matter here. The order of $z=xax^{-1}$ is equal to the order of $a$; so, just as you said, by uniqueness it implies that $xax^{-1}=a$, and you're done.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanx a lot.I just missed that
    $endgroup$
    – Debprasad Kundu
    Dec 20 '18 at 5:39














1












1








1





$begingroup$

If you have a valid proof for the case when the order of $a$ is two, $o(a)=2$, then I'm pretty sure that you're done with the question — because the order of $a$ doesn't really matter here. The order of $z=xax^{-1}$ is equal to the order of $a$; so, just as you said, by uniqueness it implies that $xax^{-1}=a$, and you're done.






share|cite|improve this answer









$endgroup$



If you have a valid proof for the case when the order of $a$ is two, $o(a)=2$, then I'm pretty sure that you're done with the question — because the order of $a$ doesn't really matter here. The order of $z=xax^{-1}$ is equal to the order of $a$; so, just as you said, by uniqueness it implies that $xax^{-1}=a$, and you're done.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 5:30









zipirovichzipirovich

11.3k11731




11.3k11731












  • $begingroup$
    Thanx a lot.I just missed that
    $endgroup$
    – Debprasad Kundu
    Dec 20 '18 at 5:39


















  • $begingroup$
    Thanx a lot.I just missed that
    $endgroup$
    – Debprasad Kundu
    Dec 20 '18 at 5:39
















$begingroup$
Thanx a lot.I just missed that
$endgroup$
– Debprasad Kundu
Dec 20 '18 at 5:39




$begingroup$
Thanx a lot.I just missed that
$endgroup$
– Debprasad Kundu
Dec 20 '18 at 5:39











-1












$begingroup$

What you want to show is that $forall x in G, x.a.x^{-1} = a$. The thing you miss is probably that $(x.a.x^{-1})^k = x.a^k.x^{-1}$ for any integer $k$. If you write the product down, you will see that all the intermediate $x^{-1}.x$ terms cancel each other $(x.a.x^{-1})^n = (x.a.x^{-1})(x.a.x^{-1})ldots(x.a.x^{-1})$



For a more rigorous proof you can do it by induction.



From here you have $(x.a.x^{-1})^n = x.a^n.x^{-1} = x.x^{-1} =1$. This means that $x.a.x^{-1}$ has an order $m$ that divises $n$.



Now, we just need to show that $m=n$. For that, one can see that $1 = (x.a.x^{-1})^m = x.a^m.x^{-1}$ which implies $a^m=1$. By definition of the order of an element we can deduce $m=n$ since the order is the smallest positive integer that verifies this property.



We proved that $x.a.x^{-1}$ is of order $n$ which implies, by uniqueness of an element of such order, that $x.a.x^{-1} = a$ which is what we wanted.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm not the downvoter, but if you wrap your maths in dollar signs "$" instead of backticks it will be much easier to read
    $endgroup$
    – postmortes
    Dec 20 '18 at 9:29










  • $begingroup$
    Thank you very much for the tips, it is indeed much more pleasant to read
    $endgroup$
    – Paul Cottalorda
    Dec 21 '18 at 14:08
















-1












$begingroup$

What you want to show is that $forall x in G, x.a.x^{-1} = a$. The thing you miss is probably that $(x.a.x^{-1})^k = x.a^k.x^{-1}$ for any integer $k$. If you write the product down, you will see that all the intermediate $x^{-1}.x$ terms cancel each other $(x.a.x^{-1})^n = (x.a.x^{-1})(x.a.x^{-1})ldots(x.a.x^{-1})$



For a more rigorous proof you can do it by induction.



From here you have $(x.a.x^{-1})^n = x.a^n.x^{-1} = x.x^{-1} =1$. This means that $x.a.x^{-1}$ has an order $m$ that divises $n$.



Now, we just need to show that $m=n$. For that, one can see that $1 = (x.a.x^{-1})^m = x.a^m.x^{-1}$ which implies $a^m=1$. By definition of the order of an element we can deduce $m=n$ since the order is the smallest positive integer that verifies this property.



We proved that $x.a.x^{-1}$ is of order $n$ which implies, by uniqueness of an element of such order, that $x.a.x^{-1} = a$ which is what we wanted.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I'm not the downvoter, but if you wrap your maths in dollar signs "$" instead of backticks it will be much easier to read
    $endgroup$
    – postmortes
    Dec 20 '18 at 9:29










  • $begingroup$
    Thank you very much for the tips, it is indeed much more pleasant to read
    $endgroup$
    – Paul Cottalorda
    Dec 21 '18 at 14:08














-1












-1








-1





$begingroup$

What you want to show is that $forall x in G, x.a.x^{-1} = a$. The thing you miss is probably that $(x.a.x^{-1})^k = x.a^k.x^{-1}$ for any integer $k$. If you write the product down, you will see that all the intermediate $x^{-1}.x$ terms cancel each other $(x.a.x^{-1})^n = (x.a.x^{-1})(x.a.x^{-1})ldots(x.a.x^{-1})$



For a more rigorous proof you can do it by induction.



From here you have $(x.a.x^{-1})^n = x.a^n.x^{-1} = x.x^{-1} =1$. This means that $x.a.x^{-1}$ has an order $m$ that divises $n$.



Now, we just need to show that $m=n$. For that, one can see that $1 = (x.a.x^{-1})^m = x.a^m.x^{-1}$ which implies $a^m=1$. By definition of the order of an element we can deduce $m=n$ since the order is the smallest positive integer that verifies this property.



We proved that $x.a.x^{-1}$ is of order $n$ which implies, by uniqueness of an element of such order, that $x.a.x^{-1} = a$ which is what we wanted.






share|cite|improve this answer











$endgroup$



What you want to show is that $forall x in G, x.a.x^{-1} = a$. The thing you miss is probably that $(x.a.x^{-1})^k = x.a^k.x^{-1}$ for any integer $k$. If you write the product down, you will see that all the intermediate $x^{-1}.x$ terms cancel each other $(x.a.x^{-1})^n = (x.a.x^{-1})(x.a.x^{-1})ldots(x.a.x^{-1})$



For a more rigorous proof you can do it by induction.



From here you have $(x.a.x^{-1})^n = x.a^n.x^{-1} = x.x^{-1} =1$. This means that $x.a.x^{-1}$ has an order $m$ that divises $n$.



Now, we just need to show that $m=n$. For that, one can see that $1 = (x.a.x^{-1})^m = x.a^m.x^{-1}$ which implies $a^m=1$. By definition of the order of an element we can deduce $m=n$ since the order is the smallest positive integer that verifies this property.



We proved that $x.a.x^{-1}$ is of order $n$ which implies, by uniqueness of an element of such order, that $x.a.x^{-1} = a$ which is what we wanted.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 21 '18 at 14:07

























answered Dec 20 '18 at 6:06









Paul CottalordaPaul Cottalorda

3765




3765












  • $begingroup$
    I'm not the downvoter, but if you wrap your maths in dollar signs "$" instead of backticks it will be much easier to read
    $endgroup$
    – postmortes
    Dec 20 '18 at 9:29










  • $begingroup$
    Thank you very much for the tips, it is indeed much more pleasant to read
    $endgroup$
    – Paul Cottalorda
    Dec 21 '18 at 14:08


















  • $begingroup$
    I'm not the downvoter, but if you wrap your maths in dollar signs "$" instead of backticks it will be much easier to read
    $endgroup$
    – postmortes
    Dec 20 '18 at 9:29










  • $begingroup$
    Thank you very much for the tips, it is indeed much more pleasant to read
    $endgroup$
    – Paul Cottalorda
    Dec 21 '18 at 14:08
















$begingroup$
I'm not the downvoter, but if you wrap your maths in dollar signs "$" instead of backticks it will be much easier to read
$endgroup$
– postmortes
Dec 20 '18 at 9:29




$begingroup$
I'm not the downvoter, but if you wrap your maths in dollar signs "$" instead of backticks it will be much easier to read
$endgroup$
– postmortes
Dec 20 '18 at 9:29












$begingroup$
Thank you very much for the tips, it is indeed much more pleasant to read
$endgroup$
– Paul Cottalorda
Dec 21 '18 at 14:08




$begingroup$
Thank you very much for the tips, it is indeed much more pleasant to read
$endgroup$
– Paul Cottalorda
Dec 21 '18 at 14:08


















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