Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

In this question on yahoo answers , the answer says ,

"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 ?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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 $|H|\geq 25$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.