Elements $I_n+pXinmathbb{Z}^{ntimes n}$ with finite order in $operatorname{SL}(n,mathbb Z)$












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 13:12






  • 1




    $begingroup$
    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...
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 13:52










  • $begingroup$
    Sorry, that really didn't help at all to my understanding. Perhaps someone else...
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 14:06










  • $begingroup$
    @DonAntonio Sorry, my thought was wrong and misleading.
    $endgroup$
    – Andrews
    Nov 24 '18 at 14:12
















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 13:12






  • 1




    $begingroup$
    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...
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 13:52










  • $begingroup$
    Sorry, that really didn't help at all to my understanding. Perhaps someone else...
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 14:06










  • $begingroup$
    @DonAntonio Sorry, my thought was wrong and misleading.
    $endgroup$
    – Andrews
    Nov 24 '18 at 14:12














1












1








1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 '18 at 19:26







Andrews

















asked Nov 24 '18 at 13:06









AndrewsAndrews

3831317




3831317








  • 2




    $begingroup$
    Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 13:12






  • 1




    $begingroup$
    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...
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 13:52










  • $begingroup$
    Sorry, that really didn't help at all to my understanding. Perhaps someone else...
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 14:06










  • $begingroup$
    @DonAntonio Sorry, my thought was wrong and misleading.
    $endgroup$
    – Andrews
    Nov 24 '18 at 14:12














  • 2




    $begingroup$
    Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 13:12






  • 1




    $begingroup$
    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...
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 13:52










  • $begingroup$
    Sorry, that really didn't help at all to my understanding. Perhaps someone else...
    $endgroup$
    – DonAntonio
    Nov 24 '18 at 14:06










  • $begingroup$
    @DonAntonio Sorry, my thought was wrong and misleading.
    $endgroup$
    – Andrews
    Nov 24 '18 at 14:12








2




2




$begingroup$
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
$endgroup$
– DonAntonio
Nov 24 '18 at 13:12




$begingroup$
Have you already done anything? Did you try some examples, or some cases of $;p;$ ? Anything at all?
$endgroup$
– DonAntonio
Nov 24 '18 at 13:12




1




1




$begingroup$
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...
$endgroup$
– DonAntonio
Nov 24 '18 at 13:52




$begingroup$
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...
$endgroup$
– DonAntonio
Nov 24 '18 at 13:52












$begingroup$
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
$endgroup$
– DonAntonio
Nov 24 '18 at 14:06




$begingroup$
Sorry, that really didn't help at all to my understanding. Perhaps someone else...
$endgroup$
– DonAntonio
Nov 24 '18 at 14:06












$begingroup$
@DonAntonio Sorry, my thought was wrong and misleading.
$endgroup$
– Andrews
Nov 24 '18 at 14:12




$begingroup$
@DonAntonio Sorry, my thought was wrong and misleading.
$endgroup$
– Andrews
Nov 24 '18 at 14:12










1 Answer
1






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7












$begingroup$

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.






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    active

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    7












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      7












      $begingroup$

      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.






      share|cite|improve this answer









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        7












        7








        7





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 '18 at 14:16









        Hagen von EitzenHagen von Eitzen

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