# ı need urgent help

• Dec 12th 2008, 01:36 AM
bogazichili
ı need urgent help
1. Let p be prime, and G be a finite group. If every element of G has order a power of p, then |G| = p^n for some n≥0. (Hint: Use Cauchy’s theorem.)

2. Tell as much as possible about the subgroups of a group of order 30 and of a group of order 40.

• Dec 12th 2008, 10:05 AM
ThePerfectHacker
Quote:

Originally Posted by bogazichili
1. Let p be prime, and G be a finite group. If every element of G has order a power of p, then |G| = p^n for some n≥0. (Hint: Use Cauchy’s theorem.)

Assume that there is $q\not = p$ such that $q$ divides order of $|G|$. Then by Cauchy's theorem there exists and element of order $q$ which is not a power of $p$ - a contradiction. Thus, all prime divisors of $|G|$ must be equal to $p$ and so $|G| = p^n$.

Quote:

2. Tell as much as possible about the subgroups of a group of order 30 and of a group of order 40.
I am not sure exactly what you are looking for.
But for order $30$ it can be shown (I looked up a classification table) that these groups are: $\mathbb{Z}_{30}, D_{15}, D_5 \times \mathbb{Z}_3, D_3 \times \mathbb{Z}_5$.