Prove alternating group has half of the size of symmetric group Prove: For $n&gt;1$, $A_n$ has order $\frac{n!}{2}$.

The textbook writes the proof as follows:


  For each odd permutation $\alpha$, the permutation $(12)\alpha$ is even and, by the cancellation property in groups, $(12)\alpha \neq (12)\beta$,when $\alpha \neq \beta$. Thus, there are at least as many even permutations as there are odd ones.$\dots$


However, I don't understand why "$(12)\alpha \neq (12)\beta$ when $\alpha \neq \beta$" implies that there are at least as many even permutations as there are odd ones.
$\alpha$ and $\beta$ are odd permutations. Thus the statement says that for every pair of distinct odd permutations $\{ \alpha, \beta \}$, you can find an associated distinct pair of even permutations, namely $\{(12) \alpha, (12) \beta\}$. Therefore, the number of even permutations is greater than or equal to the number of odd permutations.
$\alpha$ and $\beta$ are odd permutations. Thus the statement says that for every pair of distinct odd permutations $\{ \alpha, \beta \}$, you can find an associated distinct pair of even permutations, namely $\{(12) \alpha, (12) \beta\}$. Therefore, the number of even permutations is greater than or equal to the number of odd permutations.