How orbits of $G$-sets and these characters are related












2












$begingroup$


I've been learning about induced representations recently and I've come across something which I'm very confused about;



For any $G$-set $X$, the number of orbits is equal to $(1_G, chi_{mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $chi_{mathbb{C}[X]}$ denotes the character of the permutation representation defined below;



$mathbb{C}[X]$ is the natural permutation representation with basis ${{e_x : x in X}}$ and the action of $g$ being $ge_{x} = e_{gx}$.



I have no idea why this is true. I know that $Ind_{H}^{G} mathbb{C} simeq mathbb{C}[G/H]$ and the explanation I've seen goes like this;



If $X$ is transitive, then $(1_G, chi_{mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, chi_{mathbb{C}[X]}) = (1_G, chi_{mathbb{C}[X_1] oplus cdots oplus mathbb{C}[X_n]})= (1_G, chi_{mathbb{C}[X_1]}) + ... + (1_G, chi_{mathbb{C}[X_n]}) = n$.



I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, chi_{mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?










share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I've been learning about induced representations recently and I've come across something which I'm very confused about;



    For any $G$-set $X$, the number of orbits is equal to $(1_G, chi_{mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $chi_{mathbb{C}[X]}$ denotes the character of the permutation representation defined below;



    $mathbb{C}[X]$ is the natural permutation representation with basis ${{e_x : x in X}}$ and the action of $g$ being $ge_{x} = e_{gx}$.



    I have no idea why this is true. I know that $Ind_{H}^{G} mathbb{C} simeq mathbb{C}[G/H]$ and the explanation I've seen goes like this;



    If $X$ is transitive, then $(1_G, chi_{mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, chi_{mathbb{C}[X]}) = (1_G, chi_{mathbb{C}[X_1] oplus cdots oplus mathbb{C}[X_n]})= (1_G, chi_{mathbb{C}[X_1]}) + ... + (1_G, chi_{mathbb{C}[X_n]}) = n$.



    I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, chi_{mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?










    share|cite|improve this question











    $endgroup$















      2












      2








      2


      1



      $begingroup$


      I've been learning about induced representations recently and I've come across something which I'm very confused about;



      For any $G$-set $X$, the number of orbits is equal to $(1_G, chi_{mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $chi_{mathbb{C}[X]}$ denotes the character of the permutation representation defined below;



      $mathbb{C}[X]$ is the natural permutation representation with basis ${{e_x : x in X}}$ and the action of $g$ being $ge_{x} = e_{gx}$.



      I have no idea why this is true. I know that $Ind_{H}^{G} mathbb{C} simeq mathbb{C}[G/H]$ and the explanation I've seen goes like this;



      If $X$ is transitive, then $(1_G, chi_{mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, chi_{mathbb{C}[X]}) = (1_G, chi_{mathbb{C}[X_1] oplus cdots oplus mathbb{C}[X_n]})= (1_G, chi_{mathbb{C}[X_1]}) + ... + (1_G, chi_{mathbb{C}[X_n]}) = n$.



      I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, chi_{mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?










      share|cite|improve this question











      $endgroup$




      I've been learning about induced representations recently and I've come across something which I'm very confused about;



      For any $G$-set $X$, the number of orbits is equal to $(1_G, chi_{mathbb{C}[X]})$ where $1_G$ denotes the character of the trivial representation and $chi_{mathbb{C}[X]}$ denotes the character of the permutation representation defined below;



      $mathbb{C}[X]$ is the natural permutation representation with basis ${{e_x : x in X}}$ and the action of $g$ being $ge_{x} = e_{gx}$.



      I have no idea why this is true. I know that $Ind_{H}^{G} mathbb{C} simeq mathbb{C}[G/H]$ and the explanation I've seen goes like this;



      If $X$ is transitive, then $(1_G, chi_{mathbb{C}[X]}) = 1$, which can be shown through Frobenius Reciprocity. If $X$ isn't transitive, then $X = bigcup_{i=1}^{n} X_i$ where the $X_i$ are transitive. Then, $(1_G, chi_{mathbb{C}[X]}) = (1_G, chi_{mathbb{C}[X_1] oplus cdots oplus mathbb{C}[X_n]})= (1_G, chi_{mathbb{C}[X_1]}) + ... + (1_G, chi_{mathbb{C}[X_n]}) = n$.



      I mean, I understand every line of the proof and can follow it, but why does the value $(1_G, chi_{mathbb{C}[X]})$ give you the number of orbits of a $G$-set? What does this value have to do with orbits of $G$-sets?







      abstract-algebra group-theory representation-theory group-actions group-rings






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 8 at 22:05









      Eric Wofsey

      182k12209337




      182k12209337










      asked Jan 8 at 19:16









      the manthe man

      711715




      711715






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



          Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.






          share|cite|improve this answer









          $endgroup$





















            2












            $begingroup$

            I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



            My solution follows from the orbit-stabilizer theorem.



            Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
            $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
            i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



            Then
            $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
            =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

            i.e. the inner product is the average number of elements of $X$ fixed.
            Rearranging the sums, we have
            $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
            Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
            $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
            but the index of the stabilizer equals the size of the orbit, so we have
            $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



            A note on notation



            For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



            Note



            Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).






            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%2f3066603%2fhow-orbits-of-g-sets-and-these-characters-are-related%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









              2












              $begingroup$

              The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



              Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



                Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



                  Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.






                  share|cite|improve this answer









                  $endgroup$



                  The inner product $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of intertwining maps from the trivial representation $mathbb{C}$ to $mathbb{C}[X]$. A linear map $T:mathbb{C}tomathbb{C}[X]$ is determined by $T(1)$, and is intertwining iff $T(1)$ is fixed by every element of $G$. So, $(1_G,chi_{mathbb{C}[X]})$ is just the dimension of the space of vectors in $mathbb{C}[X]$ which are fixed by every element of $G$.



                  Now, when is an element $sum c_xe_xinmathbb{C}[X]$ fixed by every element of $G$? Exactly when the coefficients $c_x$ are constant on each orbit (since the action of $G$ just permutes these coefficients transitively within each orbit). Letting $X_1,dots,X_n$ be the orbits, this means that the vectors $v_i=sum_{xin X_i} e_x$ are a basis for the space of vectors fixed by $G$, and in particular that space has dimension $n$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 8 at 22:01









                  Eric WofseyEric Wofsey

                  182k12209337




                  182k12209337























                      2












                      $begingroup$

                      I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



                      My solution follows from the orbit-stabilizer theorem.



                      Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
                      $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
                      i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



                      Then
                      $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
                      =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

                      i.e. the inner product is the average number of elements of $X$ fixed.
                      Rearranging the sums, we have
                      $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
                      Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
                      $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
                      but the index of the stabilizer equals the size of the orbit, so we have
                      $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



                      A note on notation



                      For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



                      Note



                      Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).






                      share|cite|improve this answer











                      $endgroup$


















                        2












                        $begingroup$

                        I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



                        My solution follows from the orbit-stabilizer theorem.



                        Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
                        $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
                        i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



                        Then
                        $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
                        =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

                        i.e. the inner product is the average number of elements of $X$ fixed.
                        Rearranging the sums, we have
                        $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
                        Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
                        $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
                        but the index of the stabilizer equals the size of the orbit, so we have
                        $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



                        A note on notation



                        For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



                        Note



                        Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).






                        share|cite|improve this answer











                        $endgroup$
















                          2












                          2








                          2





                          $begingroup$

                          I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



                          My solution follows from the orbit-stabilizer theorem.



                          Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
                          $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
                          i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



                          Then
                          $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
                          =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

                          i.e. the inner product is the average number of elements of $X$ fixed.
                          Rearranging the sums, we have
                          $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
                          Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
                          $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
                          but the index of the stabilizer equals the size of the orbit, so we have
                          $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



                          A note on notation



                          For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



                          Note



                          Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).






                          share|cite|improve this answer











                          $endgroup$



                          I really like Eric Wofsey's answer (+1), which was posted as I was writing this. However, this is a different solution, so I'll post it too.



                          My solution follows from the orbit-stabilizer theorem.



                          Let's take a closer look at what $newcommandCC{mathbb{C}}chi_{CC[X]}$ actually is.
                          $$chi_{CC[X]}(g) = newcommandtr{operatorname{tr}}tr(pi_g) = sum_{xin X} langle e_x,ge_xrangle = sum_{xin X} [gx=x], $$
                          i.e. $chi_{CC[X]}(g)$ counts the number of elements of $X$ fixed by $G$.



                          Then
                          $$(1_G,chi_{CC[X]}) = frac{1}{|G|}sum_{gin G} chi_{CC[X]}(g)
                          =frac{1}{|G|}sum_{gin G}sum_{xin X} [gx=x],$$

                          i.e. the inner product is the average number of elements of $X$ fixed.
                          Rearranging the sums, we have
                          $$(1_G,chi_{CC[X]})=frac{1}{|G|}sum_{xin X} sum_{gin G}[gx=x] = frac{1}{|G|} sum_{xin X} |operatorname{Stab}(x)|.$$
                          Now pull the cardinality of the group in, and recognize the term in the summation as the index of the stabilizer
                          $$(1_G,chi_{CC[X]})=sum_{xin X} frac{|newcommandStab{operatorname{Stab}}Stab(x)|}{|G|}=sum_{xin X}frac{1}{[G:Stab(x)]},$$
                          but the index of the stabilizer equals the size of the orbit, so we have
                          $$(1_G,chi_{CC[X]})=sum_{xin X} frac{1}{|Gx|} =sum_{text{orbits}} 1 =#{text{orbits}} $$



                          A note on notation



                          For a statement $P$, I use the notation $$[P]:=begin{cases} 1 & text{ $P$ is true} \ 0 & text{otherwise}.end{cases}$$



                          Note



                          Also this result is essentially just Burnside's lemma (modulo my comment on the character counting the number of fixed points).







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited Jan 8 at 22:19

























                          answered Jan 8 at 22:13









                          jgonjgon

                          13.5k22041




                          13.5k22041






























                              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%2f3066603%2fhow-orbits-of-g-sets-and-these-characters-are-related%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

                              Mario Kart Wii

                              The Binding of Isaac: Rebirth/Afterbirth

                              What does “Dominus providebit” mean?