The group of rigid motions of the cube is isomorphic to $S_4$.

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I want to solve the following exercise from Dummit & Foote. My attempt is down below. Is it correct? Thanks!




Show that the group of rigid motions of a cube is isomorphic to $S_4$.




My attempt:
Let us denote the vertices of the cube so that $1,2,3,4,1$ trace a square and $5,6,7,8$ are the vertices opposite to $1,2,3,4$. Let us also denote the pairs of opposite vertices $d_1,d_2,d_3,d_4$, where vertex $i$ is in $d_i$. To each rigid motion of the cube we associate a permutation of the set $A= d_i _i=1^4$. Denote this association by $varphi:G to S_4$, where $G$ is the group of those rigid motions, and we identified $S_A$ with $S_4$. By definition of function composition we can tell that $varphi$ is a group homomorphism.



We prove that $varphi$ is injective, using the trivial kernel characterisation:



Suppose $varphi(g)=1$ fixes all of the the pairs of opposite edges (that is we have $g(i) in i,i+4 $ for all $i$, where the numbers are reduced mod 8). Suppose $g$ sends vertex $1$ to its opposite $5$. Then the vertices $2,4,7$ adjacent to $1$ must be mapped to their opposite vertices as well. This is because out of the two seemingly possible options for their images, only one (the opposite vertex) is adjacent to $g(1)=5$. This completely determines $g$ to be the negation map which is not included in our group. The contradiction shows that we must have $g(1)=1$, and from that we can find similarly that $g$ is the identity mapping. Since $ker varphi$ is trivial $varphi$ is injective.



In order to show that it is surjective, observe that $S_4$ is generated by $(1 ; 2),(1 ; 2 ; 3 ; 4) $ (this is true because products of these two elements allow us to sort the numbers $1,2,3,4$ in any way we like). We now find elements in $G$ with images under $varphi$ being those generators. Observe that if $s$ is a $90^circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, followed by a rotation by $120^circ$ around the line through $2,6$ (so that $1$ is mapped to $3$), we have $varphi(s)=(1 ; 2)$ .Observe also that if $t$ is $90^circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, we have $varphi(t)=(1 ; 2 ; 3 ; 4)$. Now if $sigma in S_4$ is any permutation, we express in as a product involving $(1 ; 2),(1 ; 2 ; 3 ;4)$, and the corresponding product involving $s,t$ is mapped to $sigma$ by $varphi$. This proves $varphi$ is surjective. We conclude that $varphi$ is an isomorphism, so $G cong S_4$.










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  • It is hard to follow your surjectivity argument without a diagram indicating labeling the vertices. If this is a homework assignment, I would suggest including one in your write up that you turn in. It sounds like your $s$ sends $4$ to $3$ (the first transformation sends $4$ to $1$ and the second transformation sends $1$ to $3$), so it does not appear to be what you want in order for $phi(s) = (1, 2)$. There's an easier proof for surjectivity (see answer below).
    – Michael Joyce
    Apr 13 '14 at 7:58










  • @MichaelJoyce I'm sorry, it should have been $1$ mapped to $8$ in $s$.
    – user1337
    Apr 13 '14 at 8:59














up vote
4
down vote

favorite
2












I want to solve the following exercise from Dummit & Foote. My attempt is down below. Is it correct? Thanks!




Show that the group of rigid motions of a cube is isomorphic to $S_4$.




My attempt:
Let us denote the vertices of the cube so that $1,2,3,4,1$ trace a square and $5,6,7,8$ are the vertices opposite to $1,2,3,4$. Let us also denote the pairs of opposite vertices $d_1,d_2,d_3,d_4$, where vertex $i$ is in $d_i$. To each rigid motion of the cube we associate a permutation of the set $A= d_i _i=1^4$. Denote this association by $varphi:G to S_4$, where $G$ is the group of those rigid motions, and we identified $S_A$ with $S_4$. By definition of function composition we can tell that $varphi$ is a group homomorphism.



We prove that $varphi$ is injective, using the trivial kernel characterisation:



Suppose $varphi(g)=1$ fixes all of the the pairs of opposite edges (that is we have $g(i) in i,i+4 $ for all $i$, where the numbers are reduced mod 8). Suppose $g$ sends vertex $1$ to its opposite $5$. Then the vertices $2,4,7$ adjacent to $1$ must be mapped to their opposite vertices as well. This is because out of the two seemingly possible options for their images, only one (the opposite vertex) is adjacent to $g(1)=5$. This completely determines $g$ to be the negation map which is not included in our group. The contradiction shows that we must have $g(1)=1$, and from that we can find similarly that $g$ is the identity mapping. Since $ker varphi$ is trivial $varphi$ is injective.



In order to show that it is surjective, observe that $S_4$ is generated by $(1 ; 2),(1 ; 2 ; 3 ; 4) $ (this is true because products of these two elements allow us to sort the numbers $1,2,3,4$ in any way we like). We now find elements in $G$ with images under $varphi$ being those generators. Observe that if $s$ is a $90^circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, followed by a rotation by $120^circ$ around the line through $2,6$ (so that $1$ is mapped to $3$), we have $varphi(s)=(1 ; 2)$ .Observe also that if $t$ is $90^circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, we have $varphi(t)=(1 ; 2 ; 3 ; 4)$. Now if $sigma in S_4$ is any permutation, we express in as a product involving $(1 ; 2),(1 ; 2 ; 3 ;4)$, and the corresponding product involving $s,t$ is mapped to $sigma$ by $varphi$. This proves $varphi$ is surjective. We conclude that $varphi$ is an isomorphism, so $G cong S_4$.










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  • It is hard to follow your surjectivity argument without a diagram indicating labeling the vertices. If this is a homework assignment, I would suggest including one in your write up that you turn in. It sounds like your $s$ sends $4$ to $3$ (the first transformation sends $4$ to $1$ and the second transformation sends $1$ to $3$), so it does not appear to be what you want in order for $phi(s) = (1, 2)$. There's an easier proof for surjectivity (see answer below).
    – Michael Joyce
    Apr 13 '14 at 7:58










  • @MichaelJoyce I'm sorry, it should have been $1$ mapped to $8$ in $s$.
    – user1337
    Apr 13 '14 at 8:59












up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
2






2





I want to solve the following exercise from Dummit & Foote. My attempt is down below. Is it correct? Thanks!




Show that the group of rigid motions of a cube is isomorphic to $S_4$.




My attempt:
Let us denote the vertices of the cube so that $1,2,3,4,1$ trace a square and $5,6,7,8$ are the vertices opposite to $1,2,3,4$. Let us also denote the pairs of opposite vertices $d_1,d_2,d_3,d_4$, where vertex $i$ is in $d_i$. To each rigid motion of the cube we associate a permutation of the set $A= d_i _i=1^4$. Denote this association by $varphi:G to S_4$, where $G$ is the group of those rigid motions, and we identified $S_A$ with $S_4$. By definition of function composition we can tell that $varphi$ is a group homomorphism.



We prove that $varphi$ is injective, using the trivial kernel characterisation:



Suppose $varphi(g)=1$ fixes all of the the pairs of opposite edges (that is we have $g(i) in i,i+4 $ for all $i$, where the numbers are reduced mod 8). Suppose $g$ sends vertex $1$ to its opposite $5$. Then the vertices $2,4,7$ adjacent to $1$ must be mapped to their opposite vertices as well. This is because out of the two seemingly possible options for their images, only one (the opposite vertex) is adjacent to $g(1)=5$. This completely determines $g$ to be the negation map which is not included in our group. The contradiction shows that we must have $g(1)=1$, and from that we can find similarly that $g$ is the identity mapping. Since $ker varphi$ is trivial $varphi$ is injective.



In order to show that it is surjective, observe that $S_4$ is generated by $(1 ; 2),(1 ; 2 ; 3 ; 4) $ (this is true because products of these two elements allow us to sort the numbers $1,2,3,4$ in any way we like). We now find elements in $G$ with images under $varphi$ being those generators. Observe that if $s$ is a $90^circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, followed by a rotation by $120^circ$ around the line through $2,6$ (so that $1$ is mapped to $3$), we have $varphi(s)=(1 ; 2)$ .Observe also that if $t$ is $90^circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, we have $varphi(t)=(1 ; 2 ; 3 ; 4)$. Now if $sigma in S_4$ is any permutation, we express in as a product involving $(1 ; 2),(1 ; 2 ; 3 ;4)$, and the corresponding product involving $s,t$ is mapped to $sigma$ by $varphi$. This proves $varphi$ is surjective. We conclude that $varphi$ is an isomorphism, so $G cong S_4$.










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I want to solve the following exercise from Dummit & Foote. My attempt is down below. Is it correct? Thanks!




Show that the group of rigid motions of a cube is isomorphic to $S_4$.




My attempt:
Let us denote the vertices of the cube so that $1,2,3,4,1$ trace a square and $5,6,7,8$ are the vertices opposite to $1,2,3,4$. Let us also denote the pairs of opposite vertices $d_1,d_2,d_3,d_4$, where vertex $i$ is in $d_i$. To each rigid motion of the cube we associate a permutation of the set $A= d_i _i=1^4$. Denote this association by $varphi:G to S_4$, where $G$ is the group of those rigid motions, and we identified $S_A$ with $S_4$. By definition of function composition we can tell that $varphi$ is a group homomorphism.



We prove that $varphi$ is injective, using the trivial kernel characterisation:



Suppose $varphi(g)=1$ fixes all of the the pairs of opposite edges (that is we have $g(i) in i,i+4 $ for all $i$, where the numbers are reduced mod 8). Suppose $g$ sends vertex $1$ to its opposite $5$. Then the vertices $2,4,7$ adjacent to $1$ must be mapped to their opposite vertices as well. This is because out of the two seemingly possible options for their images, only one (the opposite vertex) is adjacent to $g(1)=5$. This completely determines $g$ to be the negation map which is not included in our group. The contradiction shows that we must have $g(1)=1$, and from that we can find similarly that $g$ is the identity mapping. Since $ker varphi$ is trivial $varphi$ is injective.



In order to show that it is surjective, observe that $S_4$ is generated by $(1 ; 2),(1 ; 2 ; 3 ; 4) $ (this is true because products of these two elements allow us to sort the numbers $1,2,3,4$ in any way we like). We now find elements in $G$ with images under $varphi$ being those generators. Observe that if $s$ is a $90^circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, followed by a rotation by $120^circ$ around the line through $2,6$ (so that $1$ is mapped to $3$), we have $varphi(s)=(1 ; 2)$ .Observe also that if $t$ is $90^circ$ rotation around the axis through the centres of the squares $1,2,3,4$ and $5,6,7,8$, such that $1$ is mapped to $2$, we have $varphi(t)=(1 ; 2 ; 3 ; 4)$. Now if $sigma in S_4$ is any permutation, we express in as a product involving $(1 ; 2),(1 ; 2 ; 3 ;4)$, and the corresponding product involving $s,t$ is mapped to $sigma$ by $varphi$. This proves $varphi$ is surjective. We conclude that $varphi$ is an isomorphism, so $G cong S_4$.







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edited Feb 15 '15 at 9:56









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  • It is hard to follow your surjectivity argument without a diagram indicating labeling the vertices. If this is a homework assignment, I would suggest including one in your write up that you turn in. It sounds like your $s$ sends $4$ to $3$ (the first transformation sends $4$ to $1$ and the second transformation sends $1$ to $3$), so it does not appear to be what you want in order for $phi(s) = (1, 2)$. There's an easier proof for surjectivity (see answer below).
    – Michael Joyce
    Apr 13 '14 at 7:58










  • @MichaelJoyce I'm sorry, it should have been $1$ mapped to $8$ in $s$.
    – user1337
    Apr 13 '14 at 8:59
















  • It is hard to follow your surjectivity argument without a diagram indicating labeling the vertices. If this is a homework assignment, I would suggest including one in your write up that you turn in. It sounds like your $s$ sends $4$ to $3$ (the first transformation sends $4$ to $1$ and the second transformation sends $1$ to $3$), so it does not appear to be what you want in order for $phi(s) = (1, 2)$. There's an easier proof for surjectivity (see answer below).
    – Michael Joyce
    Apr 13 '14 at 7:58










  • @MichaelJoyce I'm sorry, it should have been $1$ mapped to $8$ in $s$.
    – user1337
    Apr 13 '14 at 8:59















It is hard to follow your surjectivity argument without a diagram indicating labeling the vertices. If this is a homework assignment, I would suggest including one in your write up that you turn in. It sounds like your $s$ sends $4$ to $3$ (the first transformation sends $4$ to $1$ and the second transformation sends $1$ to $3$), so it does not appear to be what you want in order for $phi(s) = (1, 2)$. There's an easier proof for surjectivity (see answer below).
– Michael Joyce
Apr 13 '14 at 7:58




It is hard to follow your surjectivity argument without a diagram indicating labeling the vertices. If this is a homework assignment, I would suggest including one in your write up that you turn in. It sounds like your $s$ sends $4$ to $3$ (the first transformation sends $4$ to $1$ and the second transformation sends $1$ to $3$), so it does not appear to be what you want in order for $phi(s) = (1, 2)$. There's an easier proof for surjectivity (see answer below).
– Michael Joyce
Apr 13 '14 at 7:58












@MichaelJoyce I'm sorry, it should have been $1$ mapped to $8$ in $s$.
– user1337
Apr 13 '14 at 8:59




@MichaelJoyce I'm sorry, it should have been $1$ mapped to $8$ in $s$.
– user1337
Apr 13 '14 at 8:59










2 Answers
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The standard way of proving that $phi$ is surjective is the following. First, show that $G$ has $24$ elements. Then, since you've proven that $phi : G rightarrow S_4$ is an injective function between two sets of the same cardinality, it follows that $phi$ must be surjective. (For if not, the image of $phi$ would have $< 24$ elements in it, so by the pigeonhole principle, there would be some $pi in S_4$ with at least two distinct elements mapping to it via $phi$, contradicting the injectivity of $phi$.)



To show that $G$ has $24$ elements, use the orbit-stabilizer theorem. For example, $G$ acts on the set of $6$ faces of the cube, and the stabilizer of a face is a cyclic group of order $4$ generated by a $90^circ$ degree rotation. Thus, $G$ has $6 cdot 4 = 24$ elements. You could also use the action of $G$ on vertices ($8 cdot 3$), edges ($12 cdot 2)$ or diagonals ($4 cdot 6$) instead of faces.






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  • Thank you! This is great because I already have that $|G|=24$ from a previous exercise.
    – user1337
    Apr 13 '14 at 9:00

















up vote
2
down vote



+200










Surjective: The group of rigid motions of a cube contains $24$ elements, same as $S_4$. Proof - A cube has $6$ sides. If a particular side is facing upward, then there are four possible rotations of the cube that will preserve the upward-facing side. Hence, the order of the group is $6times 4 = 24$.



Injective: A cube has $4$ diagonals. For one of the diagonals's head and tale attach $1$, for another $2$ and so on. We choose tale and head of a diagonal same, since there is no way to have rigid motion for a cube to change head and tale and still the diagonal remains in same orientation. For first diagonal you can attach any one of $1$, $2$, $3$, or $4$, as you could choose for first place in $S_4$. For the second diagonal it remains $3$ numbers to choose to attach as same for the second place in $S_4$. And till the last one, and we are done.



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    The standard way of proving that $phi$ is surjective is the following. First, show that $G$ has $24$ elements. Then, since you've proven that $phi : G rightarrow S_4$ is an injective function between two sets of the same cardinality, it follows that $phi$ must be surjective. (For if not, the image of $phi$ would have $< 24$ elements in it, so by the pigeonhole principle, there would be some $pi in S_4$ with at least two distinct elements mapping to it via $phi$, contradicting the injectivity of $phi$.)



    To show that $G$ has $24$ elements, use the orbit-stabilizer theorem. For example, $G$ acts on the set of $6$ faces of the cube, and the stabilizer of a face is a cyclic group of order $4$ generated by a $90^circ$ degree rotation. Thus, $G$ has $6 cdot 4 = 24$ elements. You could also use the action of $G$ on vertices ($8 cdot 3$), edges ($12 cdot 2)$ or diagonals ($4 cdot 6$) instead of faces.






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    • Thank you! This is great because I already have that $|G|=24$ from a previous exercise.
      – user1337
      Apr 13 '14 at 9:00














    up vote
    5
    down vote



    accepted










    The standard way of proving that $phi$ is surjective is the following. First, show that $G$ has $24$ elements. Then, since you've proven that $phi : G rightarrow S_4$ is an injective function between two sets of the same cardinality, it follows that $phi$ must be surjective. (For if not, the image of $phi$ would have $< 24$ elements in it, so by the pigeonhole principle, there would be some $pi in S_4$ with at least two distinct elements mapping to it via $phi$, contradicting the injectivity of $phi$.)



    To show that $G$ has $24$ elements, use the orbit-stabilizer theorem. For example, $G$ acts on the set of $6$ faces of the cube, and the stabilizer of a face is a cyclic group of order $4$ generated by a $90^circ$ degree rotation. Thus, $G$ has $6 cdot 4 = 24$ elements. You could also use the action of $G$ on vertices ($8 cdot 3$), edges ($12 cdot 2)$ or diagonals ($4 cdot 6$) instead of faces.






    share|cite|improve this answer




















    • Thank you! This is great because I already have that $|G|=24$ from a previous exercise.
      – user1337
      Apr 13 '14 at 9:00












    up vote
    5
    down vote



    accepted







    up vote
    5
    down vote



    accepted






    The standard way of proving that $phi$ is surjective is the following. First, show that $G$ has $24$ elements. Then, since you've proven that $phi : G rightarrow S_4$ is an injective function between two sets of the same cardinality, it follows that $phi$ must be surjective. (For if not, the image of $phi$ would have $< 24$ elements in it, so by the pigeonhole principle, there would be some $pi in S_4$ with at least two distinct elements mapping to it via $phi$, contradicting the injectivity of $phi$.)



    To show that $G$ has $24$ elements, use the orbit-stabilizer theorem. For example, $G$ acts on the set of $6$ faces of the cube, and the stabilizer of a face is a cyclic group of order $4$ generated by a $90^circ$ degree rotation. Thus, $G$ has $6 cdot 4 = 24$ elements. You could also use the action of $G$ on vertices ($8 cdot 3$), edges ($12 cdot 2)$ or diagonals ($4 cdot 6$) instead of faces.






    share|cite|improve this answer












    The standard way of proving that $phi$ is surjective is the following. First, show that $G$ has $24$ elements. Then, since you've proven that $phi : G rightarrow S_4$ is an injective function between two sets of the same cardinality, it follows that $phi$ must be surjective. (For if not, the image of $phi$ would have $< 24$ elements in it, so by the pigeonhole principle, there would be some $pi in S_4$ with at least two distinct elements mapping to it via $phi$, contradicting the injectivity of $phi$.)



    To show that $G$ has $24$ elements, use the orbit-stabilizer theorem. For example, $G$ acts on the set of $6$ faces of the cube, and the stabilizer of a face is a cyclic group of order $4$ generated by a $90^circ$ degree rotation. Thus, $G$ has $6 cdot 4 = 24$ elements. You could also use the action of $G$ on vertices ($8 cdot 3$), edges ($12 cdot 2)$ or diagonals ($4 cdot 6$) instead of faces.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Apr 13 '14 at 8:05









    Michael Joyce

    12.2k21738




    12.2k21738











    • Thank you! This is great because I already have that $|G|=24$ from a previous exercise.
      – user1337
      Apr 13 '14 at 9:00
















    • Thank you! This is great because I already have that $|G|=24$ from a previous exercise.
      – user1337
      Apr 13 '14 at 9:00















    Thank you! This is great because I already have that $|G|=24$ from a previous exercise.
    – user1337
    Apr 13 '14 at 9:00




    Thank you! This is great because I already have that $|G|=24$ from a previous exercise.
    – user1337
    Apr 13 '14 at 9:00










    up vote
    2
    down vote



    +200










    Surjective: The group of rigid motions of a cube contains $24$ elements, same as $S_4$. Proof - A cube has $6$ sides. If a particular side is facing upward, then there are four possible rotations of the cube that will preserve the upward-facing side. Hence, the order of the group is $6times 4 = 24$.



    Injective: A cube has $4$ diagonals. For one of the diagonals's head and tale attach $1$, for another $2$ and so on. We choose tale and head of a diagonal same, since there is no way to have rigid motion for a cube to change head and tale and still the diagonal remains in same orientation. For first diagonal you can attach any one of $1$, $2$, $3$, or $4$, as you could choose for first place in $S_4$. For the second diagonal it remains $3$ numbers to choose to attach as same for the second place in $S_4$. And till the last one, and we are done.



    pic






    share|cite|improve this answer


























      up vote
      2
      down vote



      +200










      Surjective: The group of rigid motions of a cube contains $24$ elements, same as $S_4$. Proof - A cube has $6$ sides. If a particular side is facing upward, then there are four possible rotations of the cube that will preserve the upward-facing side. Hence, the order of the group is $6times 4 = 24$.



      Injective: A cube has $4$ diagonals. For one of the diagonals's head and tale attach $1$, for another $2$ and so on. We choose tale and head of a diagonal same, since there is no way to have rigid motion for a cube to change head and tale and still the diagonal remains in same orientation. For first diagonal you can attach any one of $1$, $2$, $3$, or $4$, as you could choose for first place in $S_4$. For the second diagonal it remains $3$ numbers to choose to attach as same for the second place in $S_4$. And till the last one, and we are done.



      pic






      share|cite|improve this answer
























        up vote
        2
        down vote



        +200







        up vote
        2
        down vote



        +200




        +200




        Surjective: The group of rigid motions of a cube contains $24$ elements, same as $S_4$. Proof - A cube has $6$ sides. If a particular side is facing upward, then there are four possible rotations of the cube that will preserve the upward-facing side. Hence, the order of the group is $6times 4 = 24$.



        Injective: A cube has $4$ diagonals. For one of the diagonals's head and tale attach $1$, for another $2$ and so on. We choose tale and head of a diagonal same, since there is no way to have rigid motion for a cube to change head and tale and still the diagonal remains in same orientation. For first diagonal you can attach any one of $1$, $2$, $3$, or $4$, as you could choose for first place in $S_4$. For the second diagonal it remains $3$ numbers to choose to attach as same for the second place in $S_4$. And till the last one, and we are done.



        pic






        share|cite|improve this answer














        Surjective: The group of rigid motions of a cube contains $24$ elements, same as $S_4$. Proof - A cube has $6$ sides. If a particular side is facing upward, then there are four possible rotations of the cube that will preserve the upward-facing side. Hence, the order of the group is $6times 4 = 24$.



        Injective: A cube has $4$ diagonals. For one of the diagonals's head and tale attach $1$, for another $2$ and so on. We choose tale and head of a diagonal same, since there is no way to have rigid motion for a cube to change head and tale and still the diagonal remains in same orientation. For first diagonal you can attach any one of $1$, $2$, $3$, or $4$, as you could choose for first place in $S_4$. For the second diagonal it remains $3$ numbers to choose to attach as same for the second place in $S_4$. And till the last one, and we are done.



        pic







        share|cite|improve this answer














        share|cite|improve this answer



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        edited Feb 15 '15 at 3:23

























        answered Feb 15 '15 at 2:40









        L.G.

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