# On Group of order $30$ and $60$.

"with $t = 6$, then there are 6 * (5 - 1) = 24 elements of order $5$ "

my question is , how did " 6 * ( 5 - 1 ) " come from ?

Which theorem is used here ?

The same question for " with $s = 10$, there are at least 10 * (3-1) = 20 elements of order $3$"

Another question:

In dummit and foote , 3rd ed , in page $145$

in prop $21$ which proves that if $G$ is a group of order 60 and $G$ has more than one sylow 5-subgroup then G is simple

The proof says ,

suppose the oposite ,

let $H$ be a proper normal subgroup of $G$ if $5$ divides $|H|$ then $H$ contains a sylow 5-subgroup of $G$ and since $H$ is normal then it contains all the conjugates of this sylow 5-subgroup , so , in particular $|H|$ is bigger than or equal to $( 1+(6 . 4) ) = 25$

My question is , how did the calculation " $|H|$ is bigger than or equal to $( 1+(6 . 4) ) = 25$ " come from ?

-
Any two distinct groups of order $5$ must have trivial intersection. If there was a nonidentity element $x$ contained in both, then $x$ would necessarily have order $5$, and thus generate each group, so they would in fact be the same.
So if there are $6$ Sylow $5$-subgroups, each nonidentity element has order $5$. There are $4$ nonidentity elements in each, and since each has trivial intersection, there is no danger of overcounting. So there must be at least $6\cdot 4=24$ elements of order $5$.
For the question about Dummit and Foote, $H$ must contain all $6$ conjugates of the Sylow $5$-subgroup. Again, the conjugates have trivial intersection, and each has $4$ elements not found in the others for the same reasoning as above. Including the identity element, $H$ must have at least $25$ distinct elements, so $|H|\geq 25$.