Sorry for the vagueness of the title, I couldn't think of a better way to put it. I just wanted to run a couple of simple questions past SE to check my reasoning is correct etc.

Find a permutation $\alpha \in S_7$ such that $\alpha^4 = (2143567)$

Since any permutation can be expressed as a product of disjoint cycles, $\alpha$ cannot be comprised of any disjoint transpositions as then $\alpha^4$ maps the elements of the transposition back to itself. $\alpha$ cannot have any disjoint $3$-cycles, $p_i$ as then $p_i^4 = p_i$. We can argue like this to see that $\alpha$ must be a $7$-cycle. So since $\alpha^4 = (2143567)$ in $\alpha$ there are three other elements between the consecutive elements in $\alpha^4$. So between elements $2$ and $1$ in $\alpha$ there are $3$ other elements, between $1 $ and $4$ in $\alpha$ there are $3$ other elements, there is only one such permutation and

$\alpha = (1362457)$

Clearly we could write $\alpha = (3624571)$, which satisfies the condition above, but this is the identical permutation. Therefore, $\alpha = (1362457)$ and is unique.

Find all permutation $\alpha \in S_7$ such that $\alpha^3 = (1234)$

Clearly, there is more than one such permutation as $\alpha$ could be a $4$-cycle with elements $1,2,3,4$ or $\alpha$ could be a $4$-cycle of elements $1,2,3,4$ and a $3$-cycle of elements $5,6,7$.

$\alpha$ cannot be a product of any disjoint transpositions as $(ij)^3 = (ij)$. Also $\alpha$ cannot have any disjoint $k$-cycles where $k\geq4$ as there would be the same length cycle in $\alpha^3$.

So one possible permutation is $\alpha_1 = (1432)$

There are no other possible permutations comprised of just one $4$-cycle for reasons discussed in the previous question. Now, there are $2$ possible ways to arrange the elements $5,6,7$ in a $3$-cycle, $(567)$ and $(576)$ so two further permutations are

$\alpha_2 = (1432)(567)$ and $\alpha_3 = (1432)(576)$

In total there are $3$ permutations that satisfy $\alpha^3 = (1234)$

If there are any errors with this, or if there is a better method to exhaust possibilities, I'd appreciate your input. Thanks.

  • $\begingroup$ For the first one, your assertion that if $\alpha$ has a $3$-cycle in its cycle decomposition, $\alpha^4=\alpha$ is wrong. However, it is still correct that $\alpha$ can't have a $3$-cycle in its cycle decomposition. Suppose it did. There are three possibilities: $\endgroup$ – Avi Steiner Mar 13 '13 at 17:57
  • $\begingroup$ @AviSteiner Thanks Avi. Is it because if $\alpha$ has a $3$-cycle then it must have either a $4$-cycle (which would be the identity in $\alpha^4$), two transpositions (which again would be the identity in $\alpha^4$) or a $3$-cycle, so one elements maps to itself which isn't the case. Also, I wrote it down 'incorrectly' I should've said "if $\alpha$ has a $3$-cycle, $p$, then $p^4 = p$, sorry for the confusion. $\endgroup$ – Noble. Mar 13 '13 at 18:02
  • $\begingroup$ Actually, you should have said that "if $\alpha$ has a $3$-cycle $p$, then $\alpha^4=p$." (Sorry about the gibberish of the original version of this comment. I have absolutely know idea how that happened.) $\endgroup$ – Avi Steiner Mar 13 '13 at 18:14
  • $\begingroup$ @AviSteiner Thanks, I was able to decipher the original message, and yes, I see how that's true. Thanks a lot. $\endgroup$ – Noble. Mar 13 '13 at 18:16

Hint: For the first permutation: $\alpha$ has order $7$, so $(\alpha^4)^2 = \alpha$.

  • $\begingroup$ Ah yes, that's a nice concise way of getting the result for $\alpha$, am I right in saying it is the only possible permutation? $\endgroup$ – Noble. Mar 13 '13 at 17:51
  • $\begingroup$ Yes, indeed, you are correct: you have the correct $\alpha$, and it is unique. $\endgroup$ – Namaste Mar 13 '13 at 17:52
  • $\begingroup$ Many thanks! Sometimes I wonder if I've completely gone off the track and inventing ways to solve problems (incorrectly). $\endgroup$ – Noble. Mar 13 '13 at 17:54
  • $\begingroup$ You thought well on the second problem, as well. Nice work, and thanks for showing your work!+1 $\endgroup$ – Namaste Mar 13 '13 at 17:55
  • $\begingroup$ Thanks a lot for checking. $\endgroup$ – Noble. Mar 13 '13 at 18:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.