Infinite Abelian subgroup of infinite non Abelian group example












8












$begingroup$


My thought is that we may take GL(2,F) as the group and this is obviously infinite and non abelian since matrix multiplication does not commute. Then I thought that if we make $langle grangle$, for some $g$ in $mathrm{GL}(2,F)$, which will be cyclic and hence Abelian, for instance:
$ g=
bigg[
begin{matrix}
1&0\0&2
end{matrix}
bigg]
$. Then $g^n$ will be in the form $ g^n=
bigg[
begin{matrix}
1&0\0&2^n
end{matrix}
bigg]
$. This is obviously infinite since $g^n=e Leftrightarrow n = 0$.
Would this example work? Much thanks in advance!






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $A$ is an infinite abelian group and $H$ is a finite, non-abelian group then $Atimes H$ works. [Also, you could take $F=mathbb{Z}$ in your example to get something easy to work with, but which isn't a field :-) ]
    $endgroup$
    – user1729
    12 hours ago


















8












$begingroup$


My thought is that we may take GL(2,F) as the group and this is obviously infinite and non abelian since matrix multiplication does not commute. Then I thought that if we make $langle grangle$, for some $g$ in $mathrm{GL}(2,F)$, which will be cyclic and hence Abelian, for instance:
$ g=
bigg[
begin{matrix}
1&0\0&2
end{matrix}
bigg]
$. Then $g^n$ will be in the form $ g^n=
bigg[
begin{matrix}
1&0\0&2^n
end{matrix}
bigg]
$. This is obviously infinite since $g^n=e Leftrightarrow n = 0$.
Would this example work? Much thanks in advance!






abstract-algebra group-theory







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    If $A$ is an infinite abelian group and $H$ is a finite, non-abelian group then $Atimes H$ works. [Also, you could take $F=mathbb{Z}$ in your example to get something easy to work with, but which isn't a field :-) ]
    $endgroup$
    – user1729
    12 hours ago
















8












8








8





$begingroup$


My thought is that we may take GL(2,F) as the group and this is obviously infinite and non abelian since matrix multiplication does not commute. Then I thought that if we make $langle grangle$, for some $g$ in $mathrm{GL}(2,F)$, which will be cyclic and hence Abelian, for instance:
$ g=
bigg[
begin{matrix}
1&0\0&2
end{matrix}
bigg]
$. Then $g^n$ will be in the form $ g^n=
bigg[
begin{matrix}
1&0\0&2^n
end{matrix}
bigg]
$. This is obviously infinite since $g^n=e Leftrightarrow n = 0$.
Would this example work? Much thanks in advance!






abstract-algebra group-theory







share|cite|improve this question











$endgroup$




My thought is that we may take GL(2,F) as the group and this is obviously infinite and non abelian since matrix multiplication does not commute. Then I thought that if we make $langle grangle$, for some $g$ in $mathrm{GL}(2,F)$, which will be cyclic and hence Abelian, for instance:
$ g=
bigg[
begin{matrix}
1&0\0&2
end{matrix}
bigg]
$. Then $g^n$ will be in the form $ g^n=
bigg[
begin{matrix}
1&0\0&2^n
end{matrix}
bigg]
$. This is obviously infinite since $g^n=e Leftrightarrow n = 0$.
Would this example work? Much thanks in advance!






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 12 hours ago









user1729

17.6k64294




17.6k64294










asked 12 hours ago









JustWanderingJustWandering

562




562








  • 3




    $begingroup$
    If $A$ is an infinite abelian group and $H$ is a finite, non-abelian group then $Atimes H$ works. [Also, you could take $F=mathbb{Z}$ in your example to get something easy to work with, but which isn't a field :-) ]
    $endgroup$
    – user1729
    12 hours ago
















  • 3




    $begingroup$
    If $A$ is an infinite abelian group and $H$ is a finite, non-abelian group then $Atimes H$ works. [Also, you could take $F=mathbb{Z}$ in your example to get something easy to work with, but which isn't a field :-) ]
    $endgroup$
    – user1729
    12 hours ago










3




3




$begingroup$
If $A$ is an infinite abelian group and $H$ is a finite, non-abelian group then $Atimes H$ works. [Also, you could take $F=mathbb{Z}$ in your example to get something easy to work with, but which isn't a field :-) ]
$endgroup$
– user1729
12 hours ago






$begingroup$
If $A$ is an infinite abelian group and $H$ is a finite, non-abelian group then $Atimes H$ works. [Also, you could take $F=mathbb{Z}$ in your example to get something easy to work with, but which isn't a field :-) ]
$endgroup$
– user1729
12 hours ago












5 Answers
5






active

oldest

votes


















9












$begingroup$

The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.






share|cite|improve this answer









$endgroup$









  • 7




    $begingroup$
    It's funny because the groups aren't simple.
    $endgroup$
    – Servaes
    12 hours ago



















7












$begingroup$

This example works indeed, if $F$ is infinite and $2^nneq1$ in $F$ for all non-zero $ninBbb{Z}$. This is satisfied for obvious candidates for $F$ such as $Bbb{R}$, $Bbb{C}$ and $Bbb{Q}$, but fails for other candidates such as the finite fields $Bbb{F}_q$, but also infinite fields of positive characteristic such as $Bbb{F}_p(T)$.



Assuming $F$ is a field, the condition that $2^nneq1$ for all non-zero $ninBbb{Z}$ is equivalent to $operatorname{char}F=0$, from which it follows that $F$ is infinite. So your example works if and only if $operatorname{char}F=0$.






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!



    It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$






    share|cite|improve this answer











    $endgroup$





















      1












      $begingroup$

      A simple example: let $G = S(mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions ${(n,n+1) |, n in mathbb Z, n , rm{ even} }$. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.






      share|cite|improve this answer









      $endgroup$





















        -1












        $begingroup$

        Yes it works if you take $F$ to be an infinite field for example.



        Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.



        I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $mathrm{diag}(x,y)^{-1} = mathrm{diag}(x^{-1}, y^{-1})$.



        This subgroup is isomorphic to $F^times oplus F^times$, which is abelian and infinite if $F$ is.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          It does not work for infinite fields such as $Bbb{F}_p(T)$.
          $endgroup$
          – Servaes
          12 hours ago










        • $begingroup$
          ah obviously yes.
          $endgroup$
          – lush
          12 hours ago










        • $begingroup$
          Changed it @Servaes, I had forgotten that he asked for a particular example to work.
          $endgroup$
          – lush
          12 hours ago












        Your Answer





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        5 Answers
        5






        active

        oldest

        votes








        5 Answers
        5






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        9












        $begingroup$

        The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.






        share|cite|improve this answer









        $endgroup$









        • 7




          $begingroup$
          It's funny because the groups aren't simple.
          $endgroup$
          – Servaes
          12 hours ago
















        9












        $begingroup$

        The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.






        share|cite|improve this answer









        $endgroup$









        • 7




          $begingroup$
          It's funny because the groups aren't simple.
          $endgroup$
          – Servaes
          12 hours ago














        9












        9








        9





        $begingroup$

        The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.






        share|cite|improve this answer









        $endgroup$



        The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 12 hours ago









        lhflhf

        167k11172404




        167k11172404








        • 7




          $begingroup$
          It's funny because the groups aren't simple.
          $endgroup$
          – Servaes
          12 hours ago














        • 7




          $begingroup$
          It's funny because the groups aren't simple.
          $endgroup$
          – Servaes
          12 hours ago








        7




        7




        $begingroup$
        It's funny because the groups aren't simple.
        $endgroup$
        – Servaes
        12 hours ago




        $begingroup$
        It's funny because the groups aren't simple.
        $endgroup$
        – Servaes
        12 hours ago











        7












        $begingroup$

        This example works indeed, if $F$ is infinite and $2^nneq1$ in $F$ for all non-zero $ninBbb{Z}$. This is satisfied for obvious candidates for $F$ such as $Bbb{R}$, $Bbb{C}$ and $Bbb{Q}$, but fails for other candidates such as the finite fields $Bbb{F}_q$, but also infinite fields of positive characteristic such as $Bbb{F}_p(T)$.



        Assuming $F$ is a field, the condition that $2^nneq1$ for all non-zero $ninBbb{Z}$ is equivalent to $operatorname{char}F=0$, from which it follows that $F$ is infinite. So your example works if and only if $operatorname{char}F=0$.






        share|cite|improve this answer









        $endgroup$


















          7












          $begingroup$

          This example works indeed, if $F$ is infinite and $2^nneq1$ in $F$ for all non-zero $ninBbb{Z}$. This is satisfied for obvious candidates for $F$ such as $Bbb{R}$, $Bbb{C}$ and $Bbb{Q}$, but fails for other candidates such as the finite fields $Bbb{F}_q$, but also infinite fields of positive characteristic such as $Bbb{F}_p(T)$.



          Assuming $F$ is a field, the condition that $2^nneq1$ for all non-zero $ninBbb{Z}$ is equivalent to $operatorname{char}F=0$, from which it follows that $F$ is infinite. So your example works if and only if $operatorname{char}F=0$.






          share|cite|improve this answer









          $endgroup$
















            7












            7








            7





            $begingroup$

            This example works indeed, if $F$ is infinite and $2^nneq1$ in $F$ for all non-zero $ninBbb{Z}$. This is satisfied for obvious candidates for $F$ such as $Bbb{R}$, $Bbb{C}$ and $Bbb{Q}$, but fails for other candidates such as the finite fields $Bbb{F}_q$, but also infinite fields of positive characteristic such as $Bbb{F}_p(T)$.



            Assuming $F$ is a field, the condition that $2^nneq1$ for all non-zero $ninBbb{Z}$ is equivalent to $operatorname{char}F=0$, from which it follows that $F$ is infinite. So your example works if and only if $operatorname{char}F=0$.






            share|cite|improve this answer









            $endgroup$



            This example works indeed, if $F$ is infinite and $2^nneq1$ in $F$ for all non-zero $ninBbb{Z}$. This is satisfied for obvious candidates for $F$ such as $Bbb{R}$, $Bbb{C}$ and $Bbb{Q}$, but fails for other candidates such as the finite fields $Bbb{F}_q$, but also infinite fields of positive characteristic such as $Bbb{F}_p(T)$.



            Assuming $F$ is a field, the condition that $2^nneq1$ for all non-zero $ninBbb{Z}$ is equivalent to $operatorname{char}F=0$, from which it follows that $F$ is infinite. So your example works if and only if $operatorname{char}F=0$.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered 12 hours ago









            ServaesServaes

            29.8k342101




            29.8k342101























                4












                $begingroup$

                Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!



                It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$






                share|cite|improve this answer











                $endgroup$


















                  4












                  $begingroup$

                  Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!



                  It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$






                  share|cite|improve this answer











                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!



                    It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$






                    share|cite|improve this answer











                    $endgroup$



                    Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!



                    It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited 12 hours ago

























                    answered 12 hours ago









                    Cameron BuieCameron Buie

                    86.4k773161




                    86.4k773161























                        1












                        $begingroup$

                        A simple example: let $G = S(mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions ${(n,n+1) |, n in mathbb Z, n , rm{ even} }$. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.






                        share|cite|improve this answer









                        $endgroup$


















                          1












                          $begingroup$

                          A simple example: let $G = S(mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions ${(n,n+1) |, n in mathbb Z, n , rm{ even} }$. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.






                          share|cite|improve this answer









                          $endgroup$
















                            1












                            1








                            1





                            $begingroup$

                            A simple example: let $G = S(mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions ${(n,n+1) |, n in mathbb Z, n , rm{ even} }$. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.






                            share|cite|improve this answer









                            $endgroup$



                            A simple example: let $G = S(mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions ${(n,n+1) |, n in mathbb Z, n , rm{ even} }$. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered 9 hours ago









                            John ColemanJohn Coleman

                            3,99311224




                            3,99311224























                                -1












                                $begingroup$

                                Yes it works if you take $F$ to be an infinite field for example.



                                Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.



                                I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $mathrm{diag}(x,y)^{-1} = mathrm{diag}(x^{-1}, y^{-1})$.



                                This subgroup is isomorphic to $F^times oplus F^times$, which is abelian and infinite if $F$ is.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  It does not work for infinite fields such as $Bbb{F}_p(T)$.
                                  $endgroup$
                                  – Servaes
                                  12 hours ago










                                • $begingroup$
                                  ah obviously yes.
                                  $endgroup$
                                  – lush
                                  12 hours ago










                                • $begingroup$
                                  Changed it @Servaes, I had forgotten that he asked for a particular example to work.
                                  $endgroup$
                                  – lush
                                  12 hours ago
















                                -1












                                $begingroup$

                                Yes it works if you take $F$ to be an infinite field for example.



                                Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.



                                I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $mathrm{diag}(x,y)^{-1} = mathrm{diag}(x^{-1}, y^{-1})$.



                                This subgroup is isomorphic to $F^times oplus F^times$, which is abelian and infinite if $F$ is.






                                share|cite|improve this answer











                                $endgroup$













                                • $begingroup$
                                  It does not work for infinite fields such as $Bbb{F}_p(T)$.
                                  $endgroup$
                                  – Servaes
                                  12 hours ago










                                • $begingroup$
                                  ah obviously yes.
                                  $endgroup$
                                  – lush
                                  12 hours ago










                                • $begingroup$
                                  Changed it @Servaes, I had forgotten that he asked for a particular example to work.
                                  $endgroup$
                                  – lush
                                  12 hours ago














                                -1












                                -1








                                -1





                                $begingroup$

                                Yes it works if you take $F$ to be an infinite field for example.



                                Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.



                                I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $mathrm{diag}(x,y)^{-1} = mathrm{diag}(x^{-1}, y^{-1})$.



                                This subgroup is isomorphic to $F^times oplus F^times$, which is abelian and infinite if $F$ is.






                                share|cite|improve this answer











                                $endgroup$



                                Yes it works if you take $F$ to be an infinite field for example.



                                Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.



                                I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $mathrm{diag}(x,y)^{-1} = mathrm{diag}(x^{-1}, y^{-1})$.



                                This subgroup is isomorphic to $F^times oplus F^times$, which is abelian and infinite if $F$ is.







                                share|cite|improve this answer














                                share|cite|improve this answer



                                share|cite|improve this answer








                                edited 12 hours ago

























                                answered 12 hours ago









                                lushlush

                                732115




                                732115












                                • $begingroup$
                                  It does not work for infinite fields such as $Bbb{F}_p(T)$.
                                  $endgroup$
                                  – Servaes
                                  12 hours ago










                                • $begingroup$
                                  ah obviously yes.
                                  $endgroup$
                                  – lush
                                  12 hours ago










                                • $begingroup$
                                  Changed it @Servaes, I had forgotten that he asked for a particular example to work.
                                  $endgroup$
                                  – lush
                                  12 hours ago


















                                • $begingroup$
                                  It does not work for infinite fields such as $Bbb{F}_p(T)$.
                                  $endgroup$
                                  – Servaes
                                  12 hours ago










                                • $begingroup$
                                  ah obviously yes.
                                  $endgroup$
                                  – lush
                                  12 hours ago










                                • $begingroup$
                                  Changed it @Servaes, I had forgotten that he asked for a particular example to work.
                                  $endgroup$
                                  – lush
                                  12 hours ago
















                                $begingroup$
                                It does not work for infinite fields such as $Bbb{F}_p(T)$.
                                $endgroup$
                                – Servaes
                                12 hours ago




                                $begingroup$
                                It does not work for infinite fields such as $Bbb{F}_p(T)$.
                                $endgroup$
                                – Servaes
                                12 hours ago












                                $begingroup$
                                ah obviously yes.
                                $endgroup$
                                – lush
                                12 hours ago




                                $begingroup$
                                ah obviously yes.
                                $endgroup$
                                – lush
                                12 hours ago












                                $begingroup$
                                Changed it @Servaes, I had forgotten that he asked for a particular example to work.
                                $endgroup$
                                – lush
                                12 hours ago




                                $begingroup$
                                Changed it @Servaes, I had forgotten that he asked for a particular example to work.
                                $endgroup$
                                – lush
                                12 hours ago


















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                                香粉寮

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                                香粉寮 香粉寮 （英文： Heung Fan Liu ）位於香港新界沙田區大圍，其北面山坡為大圍村原居民用以安葬之地。 目录 1 命名 2 歷史 2.1 天體活動 3 相關參見 4 參考文獻 5 外部連結 命名 香粉寮中的「香粉」是指該地一帶所種植的香樹上的粉末，而「寮」則指以前當地居民在該地所搭建的寮屋。 歷史 以前，當地居民以出產香料為生，並在當地種植香木，靠旁邊的城門河水力磨製香粉。 天體活動 在1930年代，該地曾以舉辦天體活動而聞名。 在1932年，當時一名連伯氏（Lanepart）的拉脫維亞人成立香港裸體運動會（ Hong Kong Nudist Society ），在該處進行天體活動，引來香港傳媒廣泛報導 [1] 。當時華人社會對於裸體的觀點，與外籍人士大為有異，也出於好奇，探頭探腦令天體營不勝其擾，天體營最終搬離，遠走馬灣島中鑊底灣，直至1960年代，由於參與人數不足，於鑊底灣的天體活動亦告終止。 相關參見 美田邨 恆峰花園 翠景花園 香粉寮街 碧田街 香粉寮發展項目 參考文獻 ^ 拂袖生. <<初談香港早期天體運動>>(上). [ 2010-12-26 ] .   外部連結 爾東. 趣談新界街道. 明報出版社. 2006: 165. ISBN  9628918311.   这是一篇與香港地理相關的小作品。你可以通过编辑或修订扩充其内容。 查 论 编 This page is only for reference, If you need detailed information, please check here
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                                GameSpot

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                                body.skin-minerva .mw-parser-output table.infobox caption{text-align:center} GameSpot 戰地風雲：惡名昭彰2在Gamespot的評論 网站类型 新聞 持有者 CBS 创始人 Pete Deemer Vince Broady Jon Epstein 网站 http://www.gamespot.com/ 注册 Optional (free and paid) 推出时间 1996年5月1日 [1] GameSpot （中国大陆：游戏基地），於1996年5月由Pete Deemer和Vince Broady創立，是一個提供新聞、評論、預告片、下載及其他的相關資訊的電子遊戲網站。GameSpot被一間後來被CNET收購的企業ZDNet所收購。根據Alexa，GameSpot.com是200個網路擁擠最嚴重的網站之一。 除了由GameSpot員工創作的內容，網站還允許用戶寫評論、網誌、之後在網路論壇分享。一些在CNET旗下的GameFAQs分享。 2004年， GameSpot被Spike TV的觀眾選上「電子遊戲賞節目」贏得「最傑出遊戲網站。 [2] 其他的遊戲網站還有IGN、1UP.com、GameSpy是它最大的競爭對手。2008年，根據Compete.com的統計，「gamespot.com」吸引了最少6000萬人的點擊率。 [3] GameSpot的主頁鏈結了到最近新聞、評論、預告、和一些有關遊戲機的入口：Wii、任天堂DS、電腦遊戲、Xbox 360、PSP、PlayStation 2、PlayStation 3。它還有一列「最受歡迎遊戲名單」，還有給用戶快速獲得遊戲資訊的搜尋器。GameSpot 還包括一些小範圍的遊戲機：任天堂64、GameCube、Game Boy Color、Game Boy Advance、Xbox、PlayStation、SEGA Saturn、Dreamcast、Neo Geo Pocket Color、N-Gage、手機遊戲。 目录 1 歷史 1.1 國際歷史 1.2 著名的員工 2 評論和分...
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                                璽

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                                國家象徵 事物 國號 國旗 國徽 國歌 國旗歌 國慶日 國家格言 國家代表色 國籍標誌 國家代碼 國家航空公司 國技 國樂 國璽 國石 （ 日语 ： 国石 ） 國菜 人物或拟人形象 國家元首 国父 国家化身 非人生物 國獸 （ 英语 ： List of national animals ） 國鳥 國花 國樹 查 论 编 本條目 論述以部分區域為主 ，未必具有普世通用的觀點。 請 協助補充內容 以避免偏頗，或討論本文的問題。   「 国玺 」重定向至此。關於法國法兰西岛大区上塞纳省的一个市镇，詳見「 索城 」。 太平天國玉璽 璽 是帝王用以署名的印章，武则天以后则多以 宝 名之。「璽」字的由來是秦始皇以和氏璧（一說是用藍田玉）製造玉璽，璽為印記之意。古時不少國家的君主均有專用的印章為印記，其中又以秦始皇的 傳國璽 最为知名。現代也以「璽」稱呼國家使用的印章，稱為 國璽 。 目录 1 中國 2 日本 3 參見 4 参考文献 5 外部链接 中國 秦、兩漢、新、魏、西晉、前趙、東晉、宋、南齊、梁、陳、隋、唐、後梁及後唐各朝沿用秦始皇之傳國玉璽，上面刻有「受命於天，既壽永昌」，以和氏璧製造，其中在王莽篡漢時，曾向當時的太皇太后王政君索取傳國玉璽，王政君非常憤怒，便將玉璽取出摔砸於地，為此傳國璽還碎了一角，雖然日後王莽以黃金補上缺角，卻仍看的出摔碎的痕跡。傳國玉璽在後唐末帝李從珂自焚之時失蹤。 在宋朝以前，皇帝通常也同时沿用秦朝 乘舆七玺 的制度，即在仅用于礼仪场合的传国玺外，另作皇帝日常办公用玺六方（皇帝之玺、天子之玺、皇帝行玺、天子行玺、皇帝信玺、天子信玺），重大礼仪时与传国玺一道作为礼器同列于御座御舆之前，象征皇权至上。惟唐初失传国玺，唐太宗另作一方宝玺代替，后复得传国玺，则自作宝玺易名“皇帝神玺”，传国玺称“受命玺”，与皇帝天子六玺合为乘舆八玺。五季纷扰，玺制混乱，至宋，皇帝就只用三方宝玺，一方为礼仪用玺受命宝（傳國璽此時已失踪，該玺為自作），另两方为日常办公所用的“书诏之宝”与“天下合同之宝”。宋徽宗时虽复活乘舆八玺之制，但为了强调自己的...
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