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