Let a be the unique element of any order in a group G.Is a in the centre of G?
$begingroup$
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
asked Dec 20 '18 at 5:21
Debprasad KunduDebprasad Kundu
112
$endgroup$
add a comment |
$begingroup$
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
asked Dec 20 '18 at 5:21
Debprasad KunduDebprasad Kundu
112
$endgroup$
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
add a comment |
$begingroup$
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
asked Dec 20 '18 at 5:21
Debprasad KunduDebprasad Kundu
112
$endgroup$
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
asked Dec 20 '18 at 5:21
Debprasad KunduDebprasad Kundu
112
asked Dec 20 '18 at 5:21
Debprasad KunduDebprasad Kundu
112
asked Dec 20 '18 at 5:21
Debprasad KunduDebprasad Kundu
112
asked Dec 20 '18 at 5:21
Debprasad KunduDebprasad Kundu
112
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
answered Dec 20 '18 at 5:30
zipirovichzipirovich
11.3k11731
$endgroup$
$begingroup$
Thanx a lot.I just missed that
$endgroup$
– Debprasad Kundu
Dec 20 '18 at 5:39
add a comment |
$begingroup$
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.
answered Dec 20 '18 at 6:06
Paul CottalordaPaul Cottalorda
3765
$endgroup$
$begingroup$
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
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047187%2flet-a-be-the-unique-element-of-any-order-in-a-group-g-is-a-in-the-centre-of-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
answered Dec 20 '18 at 5:30
zipirovichzipirovich
11.3k11731
$endgroup$
$begingroup$
Thanx a lot.I just missed that
$endgroup$
– Debprasad Kundu
Dec 20 '18 at 5:39
add a comment |
$begingroup$
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.
answered Dec 20 '18 at 5:30
zipirovichzipirovich
11.3k11731
$endgroup$
$begingroup$
Thanx a lot.I just missed that
$endgroup$
– Debprasad Kundu
Dec 20 '18 at 5:39
add a comment |
$begingroup$
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.
answered Dec 20 '18 at 5:30
zipirovichzipirovich
11.3k11731
$endgroup$
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.
answered Dec 20 '18 at 5:30
zipirovichzipirovich
11.3k11731
answered Dec 20 '18 at 5:30
zipirovichzipirovich
11.3k11731
answered Dec 20 '18 at 5:30
zipirovichzipirovich
11.3k11731
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 20 '18 at 6:06
Paul CottalordaPaul Cottalorda
3765
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 20 '18 at 6:06
Paul CottalordaPaul Cottalorda
3765
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Dec 20 '18 at 6:06
Paul CottalordaPaul Cottalorda
3765
$endgroup$
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.
answered Dec 20 '18 at 6:06
Paul CottalordaPaul Cottalorda
3765
edited Dec 21 '18 at 14:07
answered Dec 20 '18 at 6:06
Paul CottalordaPaul Cottalorda
3765
answered Dec 20 '18 at 6:06
Paul CottalordaPaul Cottalorda
3765
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
add a comment |
$begingroup$
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
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3047187%2flet-a-be-the-unique-element-of-any-order-in-a-group-g-is-a-in-the-centre-of-g%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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