# Finding a permutation, and number of, from powers of the permutation

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.

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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: –  Avi Steiner Mar 13 '13 at 17:57
@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. –  Noble. Mar 13 '13 at 18:02
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.) –  Avi Steiner Mar 13 '13 at 18:14
@AviSteiner Thanks, I was able to decipher the original message, and yes, I see how that's true. Thanks a lot. –  Noble. Mar 13 '13 at 18:16

## 1 Answer

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

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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? –  Noble. Mar 13 '13 at 17:51
Yes, indeed, you are correct: you have the correct $\alpha$, and it is unique. –  amWhy Mar 13 '13 at 17:52
Many thanks! Sometimes I wonder if I've completely gone off the track and inventing ways to solve problems (incorrectly). –  Noble. Mar 13 '13 at 17:54
You thought well on the second problem, as well. Nice work, and thanks for showing your work!+1 –  amWhy Mar 13 '13 at 17:55
Thanks a lot for checking. –  Noble. Mar 13 '13 at 18:06