Proving a mapping is a group action












4












$begingroup$


Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.



My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.



For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.



Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?






group-theory finite-groups group-actions







share|cite|improve this question









$endgroup$

















    4












    $begingroup$


    Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.



    My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.



    For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.



    Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?






    group-theory finite-groups group-actions







    share|cite|improve this question









    $endgroup$















      4












      4








      4





      $begingroup$


      Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.



      My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.



      For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.



      Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?






      group-theory finite-groups group-actions







      share|cite|improve this question









      $endgroup$




      Let $G$ be a finite group, $S$ be the set of all subsets of $G$ of size $n$, and for $g in G$, $T in S$ define $g.T={gt: t in T}$.



      My course's notes says that this is a group action of $G$ on $S$, and "shows" it by first stating without proof that $g.T$ is also of size $n$, hence in $S$. To me it's not immediately obvious that this is true.



      For it to be true requires $gt_1 = gt_2 implies t_1=t_2$, i.e. two distinct elements of $T$ will always be mapped to distinct values by $g$. If $g$ maps two distinct elements of $T$ to the same value then the cardinality of $g.T$ will be less than $n$.



      Why is it impossible for $g$ to map two distinct values $t_1, t_2$ to the same value?






      group-theory finite-groups group-actions




      group-theory finite-groups group-actions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 7 hours ago









      cb7cb7

      1256




      1256






















          1 Answer
          1






          active

          oldest

          votes


















          5












          $begingroup$

          Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.






          share|cite|improve this answer









          $endgroup$













            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
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3135335%2fproving-a-mapping-is-a-group-action%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









            5












            $begingroup$

            Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.






            share|cite|improve this answer









            $endgroup$


















              5












              $begingroup$

              Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.






              share|cite|improve this answer









              $endgroup$
















                5












                5








                5





                $begingroup$

                Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.






                share|cite|improve this answer









                $endgroup$



                Suppose $gt_1=gt_2$. Multiply both sides by $g^{-1}$ from the left side and you will get $t_1=t_2$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 7 hours ago









                MarkMark

                9,579622




                9,579622






























                    draft saved

                    draft discarded




















































                    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.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3135335%2fproving-a-mapping-is-a-group-action%23new-answer', 'question_page');
                    }
                    );

                    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







                    -

                    Popular posts from this blog

                    How to make a Squid Proxy server?

                    Image
                    1 I want to set up squid proxy server for monitoring user activity in a network and Internet Web URL filtering for user. I also want to configure the user login, user log, user access group, and block unwanted web sites in my network. Please help me and guide me the full process to do this. server proxy squid share | improve this question edited Jul 31 '12 at 14:45 Jorge Castro 37.1k 107 422 617 asked Jul 31 '12 at 5:16 Goivind Tiwari Goivind Tiwari
                    Read more

                    Is this a new Fibonacci Identity?

                    Image
                    6 1 $begingroup$ I have found the following Fibonacci Identity (and proved it). If $F_n$ denotes the nth Fibonacci Number, we have the following identity begin{equation} F_{n-r+h}F_{n+k+g+1} - F_{n-r+g}F_{n+k+h+1} = (-1)^{n+r+h+1} F_{g-h}F_{k+r+1} end{equation} where $F_1 = F_2 = 1$ , $r leq n$ , $h leq g$ , and $n, g, k in mathbb{N}$ . It is not too hard to show that this identity subsumes Cassini's Identity, Catalan's Identity, Vajda's Idenity, and d'Ocagne's identity to name a few. I have done a pretty thorough literature review, and I have not found anything like this, but I am still wondering if anyone has seen this identity before? I found this by accident after noticing some patterns in some analysis work I was doing, so if this is already known I would be curious to see w
                    Read more

                    19世紀

                    Image
                    この記事は検証可能な参考文献や出典が全く示されていないか、不十分です。 出典を追加して記事の信頼性向上にご協力ください。 （ 2012年3月 ） 千年紀: 2千年紀 世紀: 18世紀 - 19世紀 - 20世紀 十年紀: 1800年代 1810年代 1820年代 1830年代 1840年代 1850年代 1860年代 1870年代 1880年代 1890年代 19世紀に君臨した大英帝国。 19世紀 （じゅうきゅうせいき）は、西暦1801年から西暦1900年までの100年間を指す世紀。 目次 1 19世紀の歴史 1.1 国民国家の成立 1.2 帝国主義の興隆 1.3 列強の植民地争奪戦 2 できごと 2.1 1800年代 2.2 1810年代 2.3 1820年代 2.4 1830年代 2.5 1840年代 2.6 1850年代 2.7 1860年代 2.8 1870年代 2.9 1880年代 2.10 1890年代 2.11 1900年代 3 文化 3.1 文学 3.2 音楽 3.3 思想 3.4 科学 3.5 技術 4 人物 4.1 ヨーロッパ 4.1.1 政治家・王族 4.1.1.1 フランス 4.1.1.2 オーストリア＝ハンガリー 4.1.1.3 ロシア 4.1.1.4 イギリス 4.1.1.5 ドイツ（プロイセンほかドイツ領邦を含む） 4.1.1.6 北欧 4.1.1.7 イタリア 4.1.1.8 スペイン 4.1.1.9 ベルギー 4.1.1.10 ギリシア 4.1.2 軍人 4.1.3 実業家 4.1.4 科学と技術 4.1.5 思想と哲学・人文諸学 4.1.6 宗教 4.1.7 文学 4.1.8 美術 4.1.9 音楽 4.1.10 社会事業家 4.1.11 探検家・旅行家 4.1.12 料
                    Read more