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How would I go about showing that in $S_7$, the equation $x^2=(1234)$ has no solutions but the equation $x^3=(1234)$ has at least two solutions?

Martin Sleziak
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luke
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4 Answers4

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This answer is not a theoretical approach but it is base on using GAP. By using the following codes in GAP environment; you can easily find the desired elements in second equation:

S_7:=SymmetricGroup(7);

e:=Elements(S_7);;

for i in [1..Size(e)] do if e[i]^3=(1,2,3,4) then Print(e[i],"\n"); fi; od;

(1,4,3,2)

(1,4,3,2)(5,6,7)

(1,4,3,2)(5,7,6)

Community
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Mikasa
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    A bit more succinct is: for alpha in SymmetricGroup(7) do if(alpha^3=(1,2,3,4)) then Print(alpha,"\n"); fi; od; – Douglas S. Stones Mar 29 '13 at 12:53
  • @DouglasS.Stones: Thank you so mcuh. You did it so fast just by one line. Are you in GAP forum? Have you ever seen the questions there? – Mikasa Mar 29 '13 at 15:41
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You can show that the second equation has at least two solutions by finding examples.

You can show that there is no $x$ satisfying the first equation by considering what the order of $x$ would be, and what the possible orders of elements of $S_7$ are.

Jonas Meyer
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  • Is there a good way to go about finding solutions to the second equation? – luke Mar 29 '13 at 06:33
  • @zach: consider what the order of the element must be, and work from there. –  Mar 29 '13 at 06:43
  • @zach: Can you find a solution in $S_4$? Try. If you get that, try thinking about what you can do with ${5,6,7}$ to get another solution in $S_7$. – Jonas Meyer Mar 29 '13 at 07:28
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You could approach the second question by considering what the orbit under $x$ of the point $1$ (the values obtained from it by repeatedly applying $x$) should look like. You know that $x^3$ sends $1$ successively to $2,3,4$ and then back to $1$ (that is $(x^3)^4(1)=1$) so you know the size of the orbit must divide $12$ (while there is not enough room for it to be equal to $12$), and that it must be at least $4$. In fact you can easily argue that $4$ must divide the size of the orbit (look what $x$ and $x^2$ do when applied to the subset $\{1,2,3,4\}$), so the orbit must be equal to $\{1,2,3,4\}$. So restricted to those elements $x$ must be a $4$-cycle, and restricted to the remaining elements $x^3$ must be the identity. Now you can determine all solutions for $x$.

Marc van Leeuwen
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Hint: You could consider the cycle structure of the elements in $\text{S}_7$. Which of them can make its square a 4-cycle, which of them can make its cube a 4-cycle?

Easy
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