# Order of the Centre of a Subgroup

• Mar 1st 2010, 09:39 PM
Haven
Order of the Centre of a Subgroup
I've been racking my brain over this problem:

Let $\displaystyle G$ be a group of order $\displaystyle p^n$
Prove the Center of $\displaystyle G$ cannot have order $\displaystyle p^{n-1}$

Naturally, I assume for a contradiction that $\displaystyle |Z(G)| = p^{n-1}$
By Lagrange's Theorem, there are $\displaystyle \frac{|G|}{|Z(G)|} = \frac{p^n}{p^{n-1}} = p$ distinct left cosets of $\displaystyle Z(G)$

Let $\displaystyle a \in G$.
If $\displaystyle a \in Z(G)$ then $\displaystyle aZ(G) = Z(G)$

This is as far as I get and I get the contraditiction. Any help would be greatly appreciated
• Mar 1st 2010, 09:42 PM
Drexel28
Quote:

Originally Posted by Haven
I've been racking my brain over this problem:

Let $\displaystyle G$ be a group of order $\displaystyle p^n$
Prove the Center of $\displaystyle G$ cannot have order $\displaystyle p^{n-1}$

Naturally, I assume for a contradiction that $\displaystyle |Z(G)| = p^{n-1}$
By Lagrange's Theorem, there are $\displaystyle \frac{|G|}{|Z(G)|} = \frac{p^n}{p^{n-1}} = p$ distinct left cosets of $\displaystyle Z(G)$

Let $\displaystyle a \in G$.
If $\displaystyle a \in Z(G)$ then $\displaystyle aZ(G) = Z(G)$

This is as far as I get and I get the contraditiction. Any help would be greatly appreciated

What have you learned up to in class. That will dictate members' answers.
• Mar 1st 2010, 09:51 PM
aliceinwonderland
Quote:

Originally Posted by Haven
I've been racking my brain over this problem:

Let $\displaystyle G$ be a group of order $\displaystyle p^n$
Prove the Center of $\displaystyle G$ cannot have order $\displaystyle p^{n-1}$

Naturally, I assume for a contradiction that $\displaystyle |Z(G)| = p^{n-1}$
By Lagrange's Theorem, there are $\displaystyle \frac{|G|}{|Z(G)|} = \frac{p^n}{p^{n-1}} = p$ distinct left cosets of $\displaystyle Z(G)$

Let $\displaystyle a \in G$.
If $\displaystyle a \in Z(G)$ then $\displaystyle aZ(G) = Z(G)$

This is as far as I get and I get the contraditiction. Any help would be greatly appreciated

Hint.

|G/Z(G)| = p (a prime number) ----> G/Z(G) is cyclic ------> G is abelian------>G/Z(G)=1. Contradiction?
• Mar 1st 2010, 09:53 PM
Drexel28
Quote:

Originally Posted by aliceinwonderland
Hint.

|G/Z(G)| = p (a prime number) ----> G/Z(G) is cyclic ------> G is abelian------>G/Z(G)=1. Contradiction?

(Angry) I wanted to do that!!

Haha! I'm just kidding! (Giggle)
• Mar 1st 2010, 09:56 PM
Haven
Quote:

Originally Posted by Drexel28
What have you learned up to in class. That will dictate members' answers.

I have not learned Quotient Groups yet, we're only allowed to use left cosets and Lagrange's Theorem.