# Thread: Order of the Centre of a Subgroup

1. ## 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

2. 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.

3. 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?

4. 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?
I wanted to do that!!

Haha! I'm just kidding!

5. 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.