Element of maximal order in alternating group $A_{10}$: how to prove this is it?












2












$begingroup$


I'm trying to solve this exercise here:




What's the maximum order of any element in $A_{10}$?




Here are my thoughts:



One way of doing this seems to be by writing down all possible elements in $S_{10}$ (I mean just per order, not actually every element) and then looking at those that are in $A_{10}$.



This seems a lot of work but for completeness I will post this here. So, (some) possible orders for elements in $S_{10}$ are:



$$
begin{matrix}
text{ disjoint cycles } & text{ order} \
(1)(1)(1)(1)(1)(1)(1)(1)(1)(1) & 1\
(1)(1)(1)(1)(1)(1)(1)(1)(2) & 2\
(1)(1)(1)(1)(1)(1)(1)(3) & 3\
(1)(1)(1)(1)(1)(1)(4) & 4\
(1)(1)(1)(1)(1)(5) & 5\
dots & \
(10) & 10 \
(7)(2)(1) & 14\
(5)(3)(2) & 30\
(5)(3)(1)(1) & 15 \
end{matrix}
$$



where $(k)$ denotes a cycle of length $k$.



So these are not all orders but I think I got the interesting ones. For example, I think $30$ is the highest possible order in $S_{10}$.



Also, I think $15$ is the highest order in $A_{10}$.



But my problem is I don't know how to prove it.




How can I prove that $15$ is the highest order in $A_{10}$?




And my other question is:




Is there a failsafe way to find elements of highest order? By that I mean one that makes sure I'm not missing any elements?




It seems kind of tricky: these are such large groups but they don't seem to contain any elements of large orders. Or do they? Maybe I'm missing something?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I don’t know if there’s an easy solution for general $A_n$ (or even $S_n$), but the function $g(n)$ that is the the largest order among the elements of $S_n$ is known as Landau’s function ( oeis.org/A000793 ), and the sequence of largest order elements in $A_n$ is also in OEIS here: oeis.org/A051593 .
    $endgroup$
    – Steve Kass
    Jan 5 '16 at 6:05
















2












$begingroup$


I'm trying to solve this exercise here:




What's the maximum order of any element in $A_{10}$?




Here are my thoughts:



One way of doing this seems to be by writing down all possible elements in $S_{10}$ (I mean just per order, not actually every element) and then looking at those that are in $A_{10}$.



This seems a lot of work but for completeness I will post this here. So, (some) possible orders for elements in $S_{10}$ are:



$$
begin{matrix}
text{ disjoint cycles } & text{ order} \
(1)(1)(1)(1)(1)(1)(1)(1)(1)(1) & 1\
(1)(1)(1)(1)(1)(1)(1)(1)(2) & 2\
(1)(1)(1)(1)(1)(1)(1)(3) & 3\
(1)(1)(1)(1)(1)(1)(4) & 4\
(1)(1)(1)(1)(1)(5) & 5\
dots & \
(10) & 10 \
(7)(2)(1) & 14\
(5)(3)(2) & 30\
(5)(3)(1)(1) & 15 \
end{matrix}
$$



where $(k)$ denotes a cycle of length $k$.



So these are not all orders but I think I got the interesting ones. For example, I think $30$ is the highest possible order in $S_{10}$.



Also, I think $15$ is the highest order in $A_{10}$.



But my problem is I don't know how to prove it.




How can I prove that $15$ is the highest order in $A_{10}$?




And my other question is:




Is there a failsafe way to find elements of highest order? By that I mean one that makes sure I'm not missing any elements?




It seems kind of tricky: these are such large groups but they don't seem to contain any elements of large orders. Or do they? Maybe I'm missing something?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I don’t know if there’s an easy solution for general $A_n$ (or even $S_n$), but the function $g(n)$ that is the the largest order among the elements of $S_n$ is known as Landau’s function ( oeis.org/A000793 ), and the sequence of largest order elements in $A_n$ is also in OEIS here: oeis.org/A051593 .
    $endgroup$
    – Steve Kass
    Jan 5 '16 at 6:05














2












2








2





$begingroup$


I'm trying to solve this exercise here:




What's the maximum order of any element in $A_{10}$?




Here are my thoughts:



One way of doing this seems to be by writing down all possible elements in $S_{10}$ (I mean just per order, not actually every element) and then looking at those that are in $A_{10}$.



This seems a lot of work but for completeness I will post this here. So, (some) possible orders for elements in $S_{10}$ are:



$$
begin{matrix}
text{ disjoint cycles } & text{ order} \
(1)(1)(1)(1)(1)(1)(1)(1)(1)(1) & 1\
(1)(1)(1)(1)(1)(1)(1)(1)(2) & 2\
(1)(1)(1)(1)(1)(1)(1)(3) & 3\
(1)(1)(1)(1)(1)(1)(4) & 4\
(1)(1)(1)(1)(1)(5) & 5\
dots & \
(10) & 10 \
(7)(2)(1) & 14\
(5)(3)(2) & 30\
(5)(3)(1)(1) & 15 \
end{matrix}
$$



where $(k)$ denotes a cycle of length $k$.



So these are not all orders but I think I got the interesting ones. For example, I think $30$ is the highest possible order in $S_{10}$.



Also, I think $15$ is the highest order in $A_{10}$.



But my problem is I don't know how to prove it.




How can I prove that $15$ is the highest order in $A_{10}$?




And my other question is:




Is there a failsafe way to find elements of highest order? By that I mean one that makes sure I'm not missing any elements?




It seems kind of tricky: these are such large groups but they don't seem to contain any elements of large orders. Or do they? Maybe I'm missing something?










share|cite|improve this question











$endgroup$




I'm trying to solve this exercise here:




What's the maximum order of any element in $A_{10}$?




Here are my thoughts:



One way of doing this seems to be by writing down all possible elements in $S_{10}$ (I mean just per order, not actually every element) and then looking at those that are in $A_{10}$.



This seems a lot of work but for completeness I will post this here. So, (some) possible orders for elements in $S_{10}$ are:



$$
begin{matrix}
text{ disjoint cycles } & text{ order} \
(1)(1)(1)(1)(1)(1)(1)(1)(1)(1) & 1\
(1)(1)(1)(1)(1)(1)(1)(1)(2) & 2\
(1)(1)(1)(1)(1)(1)(1)(3) & 3\
(1)(1)(1)(1)(1)(1)(4) & 4\
(1)(1)(1)(1)(1)(5) & 5\
dots & \
(10) & 10 \
(7)(2)(1) & 14\
(5)(3)(2) & 30\
(5)(3)(1)(1) & 15 \
end{matrix}
$$



where $(k)$ denotes a cycle of length $k$.



So these are not all orders but I think I got the interesting ones. For example, I think $30$ is the highest possible order in $S_{10}$.



Also, I think $15$ is the highest order in $A_{10}$.



But my problem is I don't know how to prove it.




How can I prove that $15$ is the highest order in $A_{10}$?




And my other question is:




Is there a failsafe way to find elements of highest order? By that I mean one that makes sure I'm not missing any elements?




It seems kind of tricky: these are such large groups but they don't seem to contain any elements of large orders. Or do they? Maybe I'm missing something?







group-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 '16 at 3:00







a student

















asked Jan 5 '16 at 5:57









a studenta student

1,958726




1,958726








  • 1




    $begingroup$
    I don’t know if there’s an easy solution for general $A_n$ (or even $S_n$), but the function $g(n)$ that is the the largest order among the elements of $S_n$ is known as Landau’s function ( oeis.org/A000793 ), and the sequence of largest order elements in $A_n$ is also in OEIS here: oeis.org/A051593 .
    $endgroup$
    – Steve Kass
    Jan 5 '16 at 6:05














  • 1




    $begingroup$
    I don’t know if there’s an easy solution for general $A_n$ (or even $S_n$), but the function $g(n)$ that is the the largest order among the elements of $S_n$ is known as Landau’s function ( oeis.org/A000793 ), and the sequence of largest order elements in $A_n$ is also in OEIS here: oeis.org/A051593 .
    $endgroup$
    – Steve Kass
    Jan 5 '16 at 6:05








1




1




$begingroup$
I don’t know if there’s an easy solution for general $A_n$ (or even $S_n$), but the function $g(n)$ that is the the largest order among the elements of $S_n$ is known as Landau’s function ( oeis.org/A000793 ), and the sequence of largest order elements in $A_n$ is also in OEIS here: oeis.org/A051593 .
$endgroup$
– Steve Kass
Jan 5 '16 at 6:05




$begingroup$
I don’t know if there’s an easy solution for general $A_n$ (or even $S_n$), but the function $g(n)$ that is the the largest order among the elements of $S_n$ is known as Landau’s function ( oeis.org/A000793 ), and the sequence of largest order elements in $A_n$ is also in OEIS here: oeis.org/A051593 .
$endgroup$
– Steve Kass
Jan 5 '16 at 6:05










2 Answers
2






active

oldest

votes


















1












$begingroup$

An element of $S_n$ of maximal order can always be chosen to have a cycle structure in which all the cycle lengths are prime powers for distinct primes. For $A_n$ there may possibly be an additional $2$-cycle to make the permutation even. This appears if and only if another cycle with length a power of $2$ is used. (Since cycles of odd length are even permutations and cycles of even length are odd permutations.)



We'll begin with the problem for $S_n$ and then worry about $A_n$.



The order of a permutation is the lcm of its cycle lengths. If you have a permutation that is not of the specified form, you can always bring it to that form, without lowering its order, through the following steps. If powers of the same prime $p$ appear in the lengths of two different cycles, you can obtain a shorter permutation (i.e. one that moves fewer elements) by dividing the length of the cycle with the lower power of $p$ by that power. Also, if you have a cycle of length $ab$, where $a, b > 1$ are relatively prime, you can replace them with cycles of length $a$ and $b$ and get a shorter permutation.



Thus the problem for $S_n$ is to choose prime powers whose sum is $leq n$ and whose product is as large as possible.



For $A_n$, it's almost the same, except that after you've performed the reductions above you may not have an even permutation. But any reductions that may have been performed will have freed up at least two elements that can then be used to make a $2$-cycle which will make the permutation even. The only apparent exception is if the only reduction to be performed is for a $6$-cycle ($a, b$ are $2, 3$), but in this case replacing a $6$-cycle with a $2$-cycle and a $3$-cycle doesn't actually change the permutation's parity.



So an element of $A_n$ of maximal order can always be chosen with relatively prime, prime-power cycle lengths and at most one additional $2$-cycle. In this case you either want to find the highest product possible of odd prime powers with sum $leq n$, or the highest product possible of prime powers including a power of $2$ whose sum is $leq n - 2$.



In the case of $A_{10}$, the possible prime powers are $2, 3, 4, 5, 7, 8, 9$.
Then the possibilities are:




  • $9$, order $9$.

  • $8 (+ 2)$, order $8$.

  • $7$, order $7$.

  • $7 + 3$, order $21$.

  • $5 + 3$, order $15$.

  • $5 + 2 (+2)$, order $10$.

  • $4 + 3 (+ 2)$, order $12$

  • $3 + 2 (+2)$, order $6$.

  • $3$, order $3$.

  • $2 (+2)$, order $2$.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So $3+7$ is not possible?
    $endgroup$
    – a student
    Jan 7 '16 at 2:00










  • $begingroup$
    @astudent Thanks for the correction. I've edited the answer.
    $endgroup$
    – David
    Jan 7 '16 at 2:53












  • $begingroup$
    Yes but this means that your method doesn't prevent missing numbers either. I am looking for a failsafe method.
    $endgroup$
    – a student
    Jan 7 '16 at 2:59










  • $begingroup$
    I think the only mistake was in the last step, where I tried to list all the sums of prime powers satisfying the criteria for $n = 10$. That part can be done systematically too. If you're more careful than I am (or program a computer to do it) then you won't make a mistake. I don't know what the method is on this site if you want an answer double-checked by other people. Maybe ask on Meta.
    $endgroup$
    – David
    Jan 7 '16 at 3:05





















0












$begingroup$

Answer : $21$



The permutations that are the products of a $7-cycle$ and a $3-cycle$, which are disjoint, have this order.
I usually list out all the elements of $S_{10}$ in terms of cycle lengths in an orderly fashion (for instance, keep the longest cycle leftmost). Then I scratch off the ones that don't belong in $A_{10}$. The rest is easy.






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    2 Answers
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    2 Answers
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    active

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    1












    $begingroup$

    An element of $S_n$ of maximal order can always be chosen to have a cycle structure in which all the cycle lengths are prime powers for distinct primes. For $A_n$ there may possibly be an additional $2$-cycle to make the permutation even. This appears if and only if another cycle with length a power of $2$ is used. (Since cycles of odd length are even permutations and cycles of even length are odd permutations.)



    We'll begin with the problem for $S_n$ and then worry about $A_n$.



    The order of a permutation is the lcm of its cycle lengths. If you have a permutation that is not of the specified form, you can always bring it to that form, without lowering its order, through the following steps. If powers of the same prime $p$ appear in the lengths of two different cycles, you can obtain a shorter permutation (i.e. one that moves fewer elements) by dividing the length of the cycle with the lower power of $p$ by that power. Also, if you have a cycle of length $ab$, where $a, b > 1$ are relatively prime, you can replace them with cycles of length $a$ and $b$ and get a shorter permutation.



    Thus the problem for $S_n$ is to choose prime powers whose sum is $leq n$ and whose product is as large as possible.



    For $A_n$, it's almost the same, except that after you've performed the reductions above you may not have an even permutation. But any reductions that may have been performed will have freed up at least two elements that can then be used to make a $2$-cycle which will make the permutation even. The only apparent exception is if the only reduction to be performed is for a $6$-cycle ($a, b$ are $2, 3$), but in this case replacing a $6$-cycle with a $2$-cycle and a $3$-cycle doesn't actually change the permutation's parity.



    So an element of $A_n$ of maximal order can always be chosen with relatively prime, prime-power cycle lengths and at most one additional $2$-cycle. In this case you either want to find the highest product possible of odd prime powers with sum $leq n$, or the highest product possible of prime powers including a power of $2$ whose sum is $leq n - 2$.



    In the case of $A_{10}$, the possible prime powers are $2, 3, 4, 5, 7, 8, 9$.
    Then the possibilities are:




    • $9$, order $9$.

    • $8 (+ 2)$, order $8$.

    • $7$, order $7$.

    • $7 + 3$, order $21$.

    • $5 + 3$, order $15$.

    • $5 + 2 (+2)$, order $10$.

    • $4 + 3 (+ 2)$, order $12$

    • $3 + 2 (+2)$, order $6$.

    • $3$, order $3$.

    • $2 (+2)$, order $2$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So $3+7$ is not possible?
      $endgroup$
      – a student
      Jan 7 '16 at 2:00










    • $begingroup$
      @astudent Thanks for the correction. I've edited the answer.
      $endgroup$
      – David
      Jan 7 '16 at 2:53












    • $begingroup$
      Yes but this means that your method doesn't prevent missing numbers either. I am looking for a failsafe method.
      $endgroup$
      – a student
      Jan 7 '16 at 2:59










    • $begingroup$
      I think the only mistake was in the last step, where I tried to list all the sums of prime powers satisfying the criteria for $n = 10$. That part can be done systematically too. If you're more careful than I am (or program a computer to do it) then you won't make a mistake. I don't know what the method is on this site if you want an answer double-checked by other people. Maybe ask on Meta.
      $endgroup$
      – David
      Jan 7 '16 at 3:05


















    1












    $begingroup$

    An element of $S_n$ of maximal order can always be chosen to have a cycle structure in which all the cycle lengths are prime powers for distinct primes. For $A_n$ there may possibly be an additional $2$-cycle to make the permutation even. This appears if and only if another cycle with length a power of $2$ is used. (Since cycles of odd length are even permutations and cycles of even length are odd permutations.)



    We'll begin with the problem for $S_n$ and then worry about $A_n$.



    The order of a permutation is the lcm of its cycle lengths. If you have a permutation that is not of the specified form, you can always bring it to that form, without lowering its order, through the following steps. If powers of the same prime $p$ appear in the lengths of two different cycles, you can obtain a shorter permutation (i.e. one that moves fewer elements) by dividing the length of the cycle with the lower power of $p$ by that power. Also, if you have a cycle of length $ab$, where $a, b > 1$ are relatively prime, you can replace them with cycles of length $a$ and $b$ and get a shorter permutation.



    Thus the problem for $S_n$ is to choose prime powers whose sum is $leq n$ and whose product is as large as possible.



    For $A_n$, it's almost the same, except that after you've performed the reductions above you may not have an even permutation. But any reductions that may have been performed will have freed up at least two elements that can then be used to make a $2$-cycle which will make the permutation even. The only apparent exception is if the only reduction to be performed is for a $6$-cycle ($a, b$ are $2, 3$), but in this case replacing a $6$-cycle with a $2$-cycle and a $3$-cycle doesn't actually change the permutation's parity.



    So an element of $A_n$ of maximal order can always be chosen with relatively prime, prime-power cycle lengths and at most one additional $2$-cycle. In this case you either want to find the highest product possible of odd prime powers with sum $leq n$, or the highest product possible of prime powers including a power of $2$ whose sum is $leq n - 2$.



    In the case of $A_{10}$, the possible prime powers are $2, 3, 4, 5, 7, 8, 9$.
    Then the possibilities are:




    • $9$, order $9$.

    • $8 (+ 2)$, order $8$.

    • $7$, order $7$.

    • $7 + 3$, order $21$.

    • $5 + 3$, order $15$.

    • $5 + 2 (+2)$, order $10$.

    • $4 + 3 (+ 2)$, order $12$

    • $3 + 2 (+2)$, order $6$.

    • $3$, order $3$.

    • $2 (+2)$, order $2$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      So $3+7$ is not possible?
      $endgroup$
      – a student
      Jan 7 '16 at 2:00










    • $begingroup$
      @astudent Thanks for the correction. I've edited the answer.
      $endgroup$
      – David
      Jan 7 '16 at 2:53












    • $begingroup$
      Yes but this means that your method doesn't prevent missing numbers either. I am looking for a failsafe method.
      $endgroup$
      – a student
      Jan 7 '16 at 2:59










    • $begingroup$
      I think the only mistake was in the last step, where I tried to list all the sums of prime powers satisfying the criteria for $n = 10$. That part can be done systematically too. If you're more careful than I am (or program a computer to do it) then you won't make a mistake. I don't know what the method is on this site if you want an answer double-checked by other people. Maybe ask on Meta.
      $endgroup$
      – David
      Jan 7 '16 at 3:05
















    1












    1








    1





    $begingroup$

    An element of $S_n$ of maximal order can always be chosen to have a cycle structure in which all the cycle lengths are prime powers for distinct primes. For $A_n$ there may possibly be an additional $2$-cycle to make the permutation even. This appears if and only if another cycle with length a power of $2$ is used. (Since cycles of odd length are even permutations and cycles of even length are odd permutations.)



    We'll begin with the problem for $S_n$ and then worry about $A_n$.



    The order of a permutation is the lcm of its cycle lengths. If you have a permutation that is not of the specified form, you can always bring it to that form, without lowering its order, through the following steps. If powers of the same prime $p$ appear in the lengths of two different cycles, you can obtain a shorter permutation (i.e. one that moves fewer elements) by dividing the length of the cycle with the lower power of $p$ by that power. Also, if you have a cycle of length $ab$, where $a, b > 1$ are relatively prime, you can replace them with cycles of length $a$ and $b$ and get a shorter permutation.



    Thus the problem for $S_n$ is to choose prime powers whose sum is $leq n$ and whose product is as large as possible.



    For $A_n$, it's almost the same, except that after you've performed the reductions above you may not have an even permutation. But any reductions that may have been performed will have freed up at least two elements that can then be used to make a $2$-cycle which will make the permutation even. The only apparent exception is if the only reduction to be performed is for a $6$-cycle ($a, b$ are $2, 3$), but in this case replacing a $6$-cycle with a $2$-cycle and a $3$-cycle doesn't actually change the permutation's parity.



    So an element of $A_n$ of maximal order can always be chosen with relatively prime, prime-power cycle lengths and at most one additional $2$-cycle. In this case you either want to find the highest product possible of odd prime powers with sum $leq n$, or the highest product possible of prime powers including a power of $2$ whose sum is $leq n - 2$.



    In the case of $A_{10}$, the possible prime powers are $2, 3, 4, 5, 7, 8, 9$.
    Then the possibilities are:




    • $9$, order $9$.

    • $8 (+ 2)$, order $8$.

    • $7$, order $7$.

    • $7 + 3$, order $21$.

    • $5 + 3$, order $15$.

    • $5 + 2 (+2)$, order $10$.

    • $4 + 3 (+ 2)$, order $12$

    • $3 + 2 (+2)$, order $6$.

    • $3$, order $3$.

    • $2 (+2)$, order $2$.






    share|cite|improve this answer











    $endgroup$



    An element of $S_n$ of maximal order can always be chosen to have a cycle structure in which all the cycle lengths are prime powers for distinct primes. For $A_n$ there may possibly be an additional $2$-cycle to make the permutation even. This appears if and only if another cycle with length a power of $2$ is used. (Since cycles of odd length are even permutations and cycles of even length are odd permutations.)



    We'll begin with the problem for $S_n$ and then worry about $A_n$.



    The order of a permutation is the lcm of its cycle lengths. If you have a permutation that is not of the specified form, you can always bring it to that form, without lowering its order, through the following steps. If powers of the same prime $p$ appear in the lengths of two different cycles, you can obtain a shorter permutation (i.e. one that moves fewer elements) by dividing the length of the cycle with the lower power of $p$ by that power. Also, if you have a cycle of length $ab$, where $a, b > 1$ are relatively prime, you can replace them with cycles of length $a$ and $b$ and get a shorter permutation.



    Thus the problem for $S_n$ is to choose prime powers whose sum is $leq n$ and whose product is as large as possible.



    For $A_n$, it's almost the same, except that after you've performed the reductions above you may not have an even permutation. But any reductions that may have been performed will have freed up at least two elements that can then be used to make a $2$-cycle which will make the permutation even. The only apparent exception is if the only reduction to be performed is for a $6$-cycle ($a, b$ are $2, 3$), but in this case replacing a $6$-cycle with a $2$-cycle and a $3$-cycle doesn't actually change the permutation's parity.



    So an element of $A_n$ of maximal order can always be chosen with relatively prime, prime-power cycle lengths and at most one additional $2$-cycle. In this case you either want to find the highest product possible of odd prime powers with sum $leq n$, or the highest product possible of prime powers including a power of $2$ whose sum is $leq n - 2$.



    In the case of $A_{10}$, the possible prime powers are $2, 3, 4, 5, 7, 8, 9$.
    Then the possibilities are:




    • $9$, order $9$.

    • $8 (+ 2)$, order $8$.

    • $7$, order $7$.

    • $7 + 3$, order $21$.

    • $5 + 3$, order $15$.

    • $5 + 2 (+2)$, order $10$.

    • $4 + 3 (+ 2)$, order $12$

    • $3 + 2 (+2)$, order $6$.

    • $3$, order $3$.

    • $2 (+2)$, order $2$.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 7 '16 at 2:53

























    answered Jan 5 '16 at 7:20









    DavidDavid

    5,233736




    5,233736












    • $begingroup$
      So $3+7$ is not possible?
      $endgroup$
      – a student
      Jan 7 '16 at 2:00










    • $begingroup$
      @astudent Thanks for the correction. I've edited the answer.
      $endgroup$
      – David
      Jan 7 '16 at 2:53












    • $begingroup$
      Yes but this means that your method doesn't prevent missing numbers either. I am looking for a failsafe method.
      $endgroup$
      – a student
      Jan 7 '16 at 2:59










    • $begingroup$
      I think the only mistake was in the last step, where I tried to list all the sums of prime powers satisfying the criteria for $n = 10$. That part can be done systematically too. If you're more careful than I am (or program a computer to do it) then you won't make a mistake. I don't know what the method is on this site if you want an answer double-checked by other people. Maybe ask on Meta.
      $endgroup$
      – David
      Jan 7 '16 at 3:05




















    • $begingroup$
      So $3+7$ is not possible?
      $endgroup$
      – a student
      Jan 7 '16 at 2:00










    • $begingroup$
      @astudent Thanks for the correction. I've edited the answer.
      $endgroup$
      – David
      Jan 7 '16 at 2:53












    • $begingroup$
      Yes but this means that your method doesn't prevent missing numbers either. I am looking for a failsafe method.
      $endgroup$
      – a student
      Jan 7 '16 at 2:59










    • $begingroup$
      I think the only mistake was in the last step, where I tried to list all the sums of prime powers satisfying the criteria for $n = 10$. That part can be done systematically too. If you're more careful than I am (or program a computer to do it) then you won't make a mistake. I don't know what the method is on this site if you want an answer double-checked by other people. Maybe ask on Meta.
      $endgroup$
      – David
      Jan 7 '16 at 3:05


















    $begingroup$
    So $3+7$ is not possible?
    $endgroup$
    – a student
    Jan 7 '16 at 2:00




    $begingroup$
    So $3+7$ is not possible?
    $endgroup$
    – a student
    Jan 7 '16 at 2:00












    $begingroup$
    @astudent Thanks for the correction. I've edited the answer.
    $endgroup$
    – David
    Jan 7 '16 at 2:53






    $begingroup$
    @astudent Thanks for the correction. I've edited the answer.
    $endgroup$
    – David
    Jan 7 '16 at 2:53














    $begingroup$
    Yes but this means that your method doesn't prevent missing numbers either. I am looking for a failsafe method.
    $endgroup$
    – a student
    Jan 7 '16 at 2:59




    $begingroup$
    Yes but this means that your method doesn't prevent missing numbers either. I am looking for a failsafe method.
    $endgroup$
    – a student
    Jan 7 '16 at 2:59












    $begingroup$
    I think the only mistake was in the last step, where I tried to list all the sums of prime powers satisfying the criteria for $n = 10$. That part can be done systematically too. If you're more careful than I am (or program a computer to do it) then you won't make a mistake. I don't know what the method is on this site if you want an answer double-checked by other people. Maybe ask on Meta.
    $endgroup$
    – David
    Jan 7 '16 at 3:05






    $begingroup$
    I think the only mistake was in the last step, where I tried to list all the sums of prime powers satisfying the criteria for $n = 10$. That part can be done systematically too. If you're more careful than I am (or program a computer to do it) then you won't make a mistake. I don't know what the method is on this site if you want an answer double-checked by other people. Maybe ask on Meta.
    $endgroup$
    – David
    Jan 7 '16 at 3:05













    0












    $begingroup$

    Answer : $21$



    The permutations that are the products of a $7-cycle$ and a $3-cycle$, which are disjoint, have this order.
    I usually list out all the elements of $S_{10}$ in terms of cycle lengths in an orderly fashion (for instance, keep the longest cycle leftmost). Then I scratch off the ones that don't belong in $A_{10}$. The rest is easy.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Answer : $21$



      The permutations that are the products of a $7-cycle$ and a $3-cycle$, which are disjoint, have this order.
      I usually list out all the elements of $S_{10}$ in terms of cycle lengths in an orderly fashion (for instance, keep the longest cycle leftmost). Then I scratch off the ones that don't belong in $A_{10}$. The rest is easy.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Answer : $21$



        The permutations that are the products of a $7-cycle$ and a $3-cycle$, which are disjoint, have this order.
        I usually list out all the elements of $S_{10}$ in terms of cycle lengths in an orderly fashion (for instance, keep the longest cycle leftmost). Then I scratch off the ones that don't belong in $A_{10}$. The rest is easy.






        share|cite|improve this answer









        $endgroup$



        Answer : $21$



        The permutations that are the products of a $7-cycle$ and a $3-cycle$, which are disjoint, have this order.
        I usually list out all the elements of $S_{10}$ in terms of cycle lengths in an orderly fashion (for instance, keep the longest cycle leftmost). Then I scratch off the ones that don't belong in $A_{10}$. The rest is easy.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 27 '17 at 14:34









        Hrit RoyHrit Roy

        884215




        884215






























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