# Alternating Group Proof

I am trying to show the following: let $\beta$ be an $r$-cycle. Show that $\beta \in A_n$ iff $r$ is odd. First I want to understand what the question is even stating as I am still a beginner in learning group-theory.

As I understand it, probably incorrectly, if $\beta$ is an r-cycle then $\beta = (a_1,a_2,a_3,...,a_r)$ for some $r \in \mathbb{Z}$.

$\bf{If-Direction}:$ if $r$ is odd then $\beta$ is the product of an even number of tranpositions $(a_1a_r)(a_1a_{r-1})(a_1a_{r-2})...(a_1a_2)$ which implies that $\beta \in A_n$.

$\bf{Only-If \ Direction:}$ if $\beta \in A_n$ then $\beta = (a_1a_r)(a_1a_{r-1})(a_1a_{r-2})...(a_1a_2)$ for some $r \in \mathbb{Z}$. Moreover, there must be an even number of these transpositions by the definition of $\beta \in A_n$. So $r$ must be odd since it is impossible for $r$ to be even and for there to be $r-1$ products of transpositions since $r-1$ was already shown to be even.

Is this a complete proof? Thank you in advance.

• looks good to me. – Quang Hoang Oct 1 '15 at 5:50

Let us say $r$ persons are seated in chairs around a round table. We want them to change their positions: every one should move counterclockwise to the next position. (this permutation is an $r$ cycle)
This can be achieved by making two persons at a time to exchange their positions. Call one person as the head. Let the head person move clockwise and exchange his position with the next person. Everytime this head person exchanges his position with the immediate next one clockwise (whoever it is). Once head person comes to the last position the process halts. Now note that desired $r$ cycle permutation is obtained. This involved exactly $r-1$ exchanges (transpositions). SO if $r$ is odd then any $r$ cycle would be an even permutation.