The maximim number of elements in the Alternating group of degree 28 I know if $r$ is a prime number, then $(r-1)!$ is maximum number of elements of same order in the alternating group $A_{r}$. What is maximum number of elements of same order in the alternating group $A_{28}$? 
 A: I just consulted a table of the 1883 conjugacy classes of $A_{28}$ and counted to find the most numerous.
There are $(28!)/27$ elements of order 27 in $A_{28}$. This is largest number of elements of the same order. The second most numerous order is 24 with $7632609771546411536941056000$ elements, coming from 93 different conjugacy classes.
A single $k$-cycle has centralizer of type $C_k \times S_{n-k}$ inside $S_n$ so if $n-k$ is small, it has many conjugates.
We cannot use $k=28$ in $A_n$ however, for the simple reason that $A_n$ does not contain cycles of even length. The runner up is $k=27$.
So in general, $A_{2n}$ has at least $((2n)!)/(2n-1)$ elements of order $2n-1$ (namely the $(2n-1)$-cycles), which is a fair amount. $A_{16}$ and $A_{22}$ have extra elements of order $15$ and $21$ respectively.
By comparison, $A_{2n+1}$ has at least $((2n+1)!)/(2n+1) = (2n)!$ elements of order $2n+1$. This is similar to your claim for $r=2n+1$ prime.
For $1 \leq r \leq 50$ the most numerous order is $r$ if $r$ is odd, and $r-1$ if $r$ is even. One might ask in a separate question if this patterns holds indefinitely.
A: I would appreciate if you let me know what's the refrence for the first claim which says "if $r$  is a prime number, then $(r−1)!$  is maximum number of elements of same order in the alternating group $A r $ . "
