# Thread: ı need urgent help

1. ## 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.

( please help me... its emergency for me )

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

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 $\displaystyle 30$ it can be shown (I looked up a classification table) that these groups are: $\displaystyle \mathbb{Z}_{30}, D_{15}, D_5 \times \mathbb{Z}_3, D_3 \times \mathbb{Z}_5$.