Elements $I_n+pXinmathbb{Z}^{ntimes n}$ with finite order in $operatorname{SL}(n,mathbb Z)$
Given $p$ an odd prime, $Xinmathbb{Z}^{ntimes n}$ a matrix.
If $I_{n}+pXin operatorname{SL}(n,mathbb Z)$ has finite order, prove $X=0$.
My idea:
let $m=p^rh$, $(p,h)=1$, s.t. $(I+pX)^m=I$.
let $(I+pX)^{p^r}=I+pY$, $I=(I+pX)^m=(I+pY)^h=(I+hpY+p^2YZ)$. $Y(hI+pZ)=0$.
If we can prove $Y=0$, then the following would be quite easy.
So how to prove $hI+pZ$ is invertible? Thanks for your time and help!
abstract-algebra group-theory
asked Nov 24 at 13:06
Andrews
374317
add a comment |
Given $p$ an odd prime, $Xinmathbb{Z}^{ntimes n}$ a matrix.
If $I_{n}+pXin operatorname{SL}(n,mathbb Z)$ has finite order, prove $X=0$.
My idea:
let $m=p^rh$, $(p,h)=1$, s.t. $(I+pX)^m=I$.
let $(I+pX)^{p^r}=I+pY$, $I=(I+pX)^m=(I+pY)^h=(I+hpY+p^2YZ)$. $Y(hI+pZ)=0$.
If we can prove $Y=0$, then the following would be quite easy.
So how to prove $hI+pZ$ is invertible? Thanks for your time and help!
abstract-algebra group-theory
asked Nov 24 at 13:06
Andrews
374317
2
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
– DonAntonio
Nov 24 at 13:12
1
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
– DonAntonio
Nov 24 at 13:52
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
– DonAntonio
Nov 24 at 14:06
@DonAntonio Sorry, my thought was wrong and misleading.
– Andrews
Nov 24 at 14:12
add a comment |
Given $p$ an odd prime, $Xinmathbb{Z}^{ntimes n}$ a matrix.
If $I_{n}+pXin operatorname{SL}(n,mathbb Z)$ has finite order, prove $X=0$.
My idea:
let $m=p^rh$, $(p,h)=1$, s.t. $(I+pX)^m=I$.
let $(I+pX)^{p^r}=I+pY$, $I=(I+pX)^m=(I+pY)^h=(I+hpY+p^2YZ)$. $Y(hI+pZ)=0$.
If we can prove $Y=0$, then the following would be quite easy.
So how to prove $hI+pZ$ is invertible? Thanks for your time and help!
abstract-algebra group-theory
asked Nov 24 at 13:06
Andrews
374317
Given $p$ an odd prime, $Xinmathbb{Z}^{ntimes n}$ a matrix.
If $I_{n}+pXin operatorname{SL}(n,mathbb Z)$ has finite order, prove $X=0$.
My idea:
let $m=p^rh$, $(p,h)=1$, s.t. $(I+pX)^m=I$.
let $(I+pX)^{p^r}=I+pY$, $I=(I+pX)^m=(I+pY)^h=(I+hpY+p^2YZ)$. $Y(hI+pZ)=0$.
If we can prove $Y=0$, then the following would be quite easy.
So how to prove $hI+pZ$ is invertible? Thanks for your time and help!
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 24 at 13:06
Andrews
374317
asked Nov 24 at 13:06
Andrews
374317
edited Nov 26 at 19:26
asked Nov 24 at 13:06
Andrews
374317
asked Nov 24 at 13:06
Andrews
374317
asked Nov 24 at 13:06
Andrews
374317
374317
2
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
– DonAntonio
Nov 24 at 13:12
1
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
– DonAntonio
Nov 24 at 13:52
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
– DonAntonio
Nov 24 at 14:06
@DonAntonio Sorry, my thought was wrong and misleading.
– Andrews
Nov 24 at 14:12
add a comment |
2
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
– DonAntonio
Nov 24 at 13:12
1
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
– DonAntonio
Nov 24 at 13:52
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
– DonAntonio
Nov 24 at 14:06
@DonAntonio Sorry, my thought was wrong and misleading.
– Andrews
Nov 24 at 14:12
2
2
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
– DonAntonio
Nov 24 at 13:12
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
– DonAntonio
Nov 24 at 13:12
1
1
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
– DonAntonio
Nov 24 at 13:52
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
– DonAntonio
Nov 24 at 13:52
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
– DonAntonio
Nov 24 at 14:06
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
– DonAntonio
Nov 24 at 14:06
@DonAntonio Sorry, my thought was wrong and misleading.
– Andrews
Nov 24 at 14:12
@DonAntonio Sorry, my thought was wrong and misleading.
– Andrews
Nov 24 at 14:12
add a comment |
1 Answer
1
active
oldest
votes
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 at 14:16
Hagen von Eitzen
275k21268495
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%2f3011533%2felements-i-npx-in-mathbbzn-times-n-with-finite-order-in-operatornames%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 at 14:16
Hagen von Eitzen
275k21268495
add a comment |
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 at 14:16
Hagen von Eitzen
275k21268495
add a comment |
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 at 14:16
Hagen von Eitzen
275k21268495
Assume the claim is false. Let $m>1$ be minimal with the property that $Xne0$ with $1+pX$ having order $m$ exists.
If $m=ab$ is composite with $a,b>1$, note that
$$(I+pX)^a=I+sum_{j=1}^a{achoose j}p^{j}X^j =I+pX'$$
with $X':=sum_{j=1}^a{achoose j}p^{j-1}X^jinBbb Z^{ntimes n}$. Then from $(I+pX')^b=(I+pX)^m=I$ and the minimality of $m$, we conclude that $X'=0$. But then $(1+pX)^a=I$, also contradicting minimality of $m$.
We conclude that the minimla $m$ cannot be composite. As ist certainly cannot be $=1$ either, we conclude that $m$ is prime.
If $(I+pX)^m=I$ with $Xne 0$, let $rge 1$ be maximal such that we can write $X=p^{r-1}Y$ with $YinBbb Z^{ntimes n}$.
Then
$$tag1 I=(I+pX)^m=(I+p^rY)^m=I+mp^rY+{mchoose 2}p^{2r}Y^2+p^{3r}cdot (ldots)$$
implies that $mp^rY$ must be a multiple of $p^{2r}$. This means that $m$ is a multiple of $p^r$, i.e., $m=p$ and $r=1$.
As $p$ is odd, the number $mchoose 2$ in $(1)$ is a multiple of $p$ so that
$$I=(I+pY)p=I+p^2Y+p^3cdot(ldots) $$
and so $Y$ is a multiple of $p$, contradiction.
answered Nov 24 at 14:16
Hagen von Eitzen
275k21268495
answered Nov 24 at 14:16
Hagen von Eitzen
275k21268495
answered Nov 24 at 14:16
Hagen von Eitzen
275k21268495
answered Nov 24 at 14:16
Hagen von Eitzen
275k21268495
275k21268495
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3011533%2felements-i-npx-in-mathbbzn-times-n-with-finite-order-in-operatornames%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
2
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
– DonAntonio
Nov 24 at 13:12
1
I really don't understand what you added about the even order. I also don't understand what that product there has anything to do...
– DonAntonio
Nov 24 at 13:52
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
– DonAntonio
Nov 24 at 14:06
@DonAntonio Sorry, my thought was wrong and misleading.
– Andrews
Nov 24 at 14:12