When are two permutations $\alpha^2=\beta^2$ equal? Hi so i am trying to see for what $\alpha,\beta$ this holds true? I am operating on $S_5$. I know that if $\alpha=\epsilon$ then $\beta$ is any element os $S_5$ with the order 2 (there are 40 of these). What are the other possibilities?
I am doing this in order to show that for $S_5$ ever even permutation can be represented as a square of an element of $S_5$
 A: Here's a proof using your proposed strategy . . .

Clearly, any square in $S_5$ is an even permutation.

Thus, to prove that the set of square in $S_5$ is the same as the set of even permutations in $S_5$, it suffices to show that there are exactly $60$ distinct squares in $S_5$.

Any element of $S_5$ has order between $1$ and $6$ inclusive.

First let's consider what happens when we square an element of a given order.


*

*If an element has order at most $2$, then its square has order $1$.$\\[4pt]$

*If an element has order $3$, then it's a $3$-cycle, and the square of a $3$-cycle is a $3$-cycle, which has order $3$.$\\[4pt]$

*If an element has order $4$, then it's a $4$-cycle, $(a\;b\;c\;d)$ say, and its square is $(a\;c)(b\;d)$, which has order $2$.$\\[4pt]$

*If an element has order $5$, then it's a $5$-cycle, and the square of a $5$-cycle is a $5$-cycle, which has order $5$.$\\[4pt]$

*If an element has order $6$, then it has the form $(a\;b)(c\;d\;e)$, the product of a $2$-cycle $(a\;b)$ and a disjoint $3$-cycle $(c\;d\;e)$, and its square is the $3$-cycle $(e\;d\;c)$, which has order $3$.


From the above analysis, we see that there are no squares of order $4$ or $6$.

Next, let's count the squares . . .

Any element of odd order in $S_5$ is a square in its own cyclic group, so the elements of orders $1,3,5$ are all squares.


*

*There is only one element of order $1$.$\\[5pt]$

*An element of order $3$ is a $3$-cycle $(a\;b\;c)$. There are ${\large{\binom{5}{3}}}\!\cdot \!2 = 20$ such elements.
${\phantom{\Large{|}}}\!\!\!\!\!\!$Explanation: To construct $(a\;b\;c)$,
${\phantom{\LARGE{|}}}\;{\small{\bullet}}\;$Choose $\{a,b,c\}{:}\;{\large{\binom{5}{3}}}\;$choices.
${\phantom{\LARGE{|}}}\;{\small{\bullet}}\;$Choose which of $a,b,c$ comes after the smallest of $a,b,c$ in the cycle$\colon\,2\;$choices.$\\[5pt]$

*An element of order $5$ is a $5$-cycle $(a\;b\;c\;d\;e)$. There are $4! = 24$ such elements.
${\phantom{\Large{|}}}\!\!\!\!\!\!$Explanation: To construct $(a\;b\;c\;d\;e)$,
${\phantom{\LARGE{|}}}\;{\small{\bullet}}\;$Without loss of generality, assume $a = 1$.
${\phantom{\LARGE{|}}}\;{\small{\bullet}}\;$Choose an arbitrary order for $b,c,d,e\colon\,4!\;$choices.


Thus, we have $1 + 20 + 24 = 45$ elements of odd order, all of which are squares.

Next, we count the squares of order $2$ . . .


*

*Since a transposition is an odd permutation, it can't be a square.$\\[5pt]$

*Hence, if an element of order $2$ is a square, it must have the form $(a\;b)(c\;d)$, a product of two disjoint $2$-cycles. Moreover, every product $(a\;b)(c\;d)$ of two disjoint $2$-cycles is a square, since identically, $(a\;b)(c\;d) = (a\;c\;b\;d)^2$. There are ${\large{\binom{5}{4}}}\!\cdot \!3 = 15$ such elements.
${\phantom{\Large{|}}}\!\!\!\!\!\!$Explanation: To construct $(a\;b)(c\;d)$,
${\phantom{\LARGE{|}}}\;{\small{\bullet}}\;$Choose $\{a,b,c,d\}{:}\;{\large{\binom{5}{4}}}\;$choices.
${\phantom{\LARGE{|}}}\;{\small{\bullet}}\;$Choose which of $a,b,c,d$ transposes with the smallest of $a,b,c,d\colon\,3\;$choices.


Thus, there are $15$ squares of order $2$.

Since there are no squares of order $4$ or $6$, we've counted all the squares.

Thus, the number of squares in $S_5$ is $45 + 15 = 60$.

Therefore the set of square in $S_5$ is the same as the set of even permutations in $S_5$.
A: Write your even permutation as the product of disjoint cycles.
From the standard tricks
$$(a_1 \ a_2 \ \cdots \ a_n) = (a_{n-1} \ a_n)  \cdots (a_2 \ a_n) (a_1 \ a_n) = (a_1 \ a_2) (a_2 \ a_3) \cdots (a_{n-1} \ a_n)$$
you can write each even cycle as the product of an odd number of transpositions, and vice versa.
Since your permutation is even, you must end up with an even number of transpositions.
Thus, in your product-of-disjoint-cycles representation, you must have
$$\sum_i (1 + \text{length of cycle $i$}) \qquad \qquad\text{ is even}. \tag{1}$$
We cannot have a cycle of length 4 in our product-of-disjoint cycles representation because then our permutation would simply be this one cycle (we are in $S_5$ and only have $5$ objects to permute, so the fifth object must be fixed), but this cycle is an odd permutation. Thus the only even-length cycles we can have in our product-of-disjoint-cycles representation are cycles of length $2$. Condition (1) above then requires there to be an even number of cycles of length $2$.
Then you can use the characterization of square permutations here to conclude.

Equivalent alternate approach:
Let us characterize squares in $S_5$. If $(a_1 \ a_2 \cdots \ a_n)$ is a cycle, then
$$(a_1 \ a_2 \cdots \ a_n)^2 = \begin{cases}
\text{a cycle of length $n$} & \text{if $n$ is odd}
\\
(a_1 \ a_3 \ \cdots \ a_{n-1})(a_2 \ a_4 \cdots a_{n}) & \text{if $n$ is even}
\end{cases}$$
Thus if you write a permutation $\alpha$ as the product of disjoint cycles and square it, each odd cycle becomes another cycle of the same length, while even cycles split into two cycles of half length. This becomes the product-of-disjoint-cycle representation of $\alpha^2$. It is clear that any $\alpha^2$ will be the product of disjoint cycles of length $3$ or $5$ or $2$, with an even number of cycles of length $2$; it is not hard to show the converse holds as well (by "reversing" the computation above of squaring a cycle). Having characterized the set of squares in $S_5$, it remains to check that any even permutation must take this form. This boils down to showing that an even permutation cannot contain a cycle of length $4$ in its disjoint cycle representation (else, as we have mentioned above, it would simply be this cycle of length $4$, which is an odd permutation).
