Prove that the order of Z(G) is either <e> or G

• February 9th 2010, 08:01 PM
crushingyen
Prove that the order of Z(G) is either <e> or G
Let $G$ be a group of order $pq$, with $p$ and $q$ (not necessarily distinct) primes. Prove that the center $Z(G) = $ or $Z(G) = G$.

I started off by doing a proof by contradiction.

Suppose that $Z(G) \neq $, $Z(G) \neq G$

Then $|Z(G)|$ divides either $p$ or $q$, which implies that $|Z(G)| = p$ or $|Z(G)| = q$

We know that $Z(G)$ is a normal subgroup.

Consider the quotient group $G/Z(G)$.

Using Larange's Theorem, we know that $|G|/|Z(G)|$ (*).

I broke it into two cases: 1: $|Z(G)| = p$, 2: $|Z(G)| = q$.

If under case 1, (*) evaluates to $p$. Otherwise, in case 2, (*) evaluates to $q$.

This is as far as I've gotten and I'm pretty stumped as how else to follow up after.
• February 9th 2010, 09:35 PM
Drexel28
Quote:

Originally Posted by crushingyen
Let $G$ be a group of order $pq$, with $p$ and $q$ (not necessarily distinct) primes. Prove that the center $Z(G) = $ or $Z(G) = G$.

I started off by doing a proof by contradiction.

Suppose that $Z(G) \neq $, $Z(G) \neq G$

Then $|Z(G)|$ divides either $p$ or $q$, which implies that $|Z(G)| = p$ or $|Z(G)| = q$

We know that $Z(G)$ is a normal subgroup.

Consider the quotient group $G/Z(G)$.

Using Larange's Theorem, we know that $|G|/|Z(G)|$ (*).

I broke it into two cases: 1: $|Z(G)| = p$, 2: $|Z(G)| = q$.

If under case 1, (*) evaluates to $p$. Otherwise, in case 2, (*) evaluates to $q$.

This is as far as I've gotten and I'm pretty stumped as how else to follow up after.

Lemma: If $G/\mathcal{Z}(G)$ is cyclic then, we have that $G$ is abelian.

Proof: Let $x,y\in G$, then clearly $x\mathcal{Z}(G),y\mathcal{Z}(G)\in G/\mathcal{Z}(G)$. But, since $G/\mathcal{Z}(G)$ is cyclic we have that

$x\mathcal{Z}(G)=g^m\mathcal{Z}(G)$ and $y\mathcal{Z}(G)=g^n\mathcal{Z}(G)$ for some $g\in G$. Thus, we have that $x=g^m\text{ } z_1,y=g^n\text{ }z_2$ for some $z_1,z_2\in\mathcal{Z}(G)$ (this is because cosets form a partition). And so

$xy=g^m\text{ }z_1\text{ }g^n\text{ }z_2=z_2 \text{ }g^{m+n}\text{ }z_1=z_2\text{ }g^{n+m}\text{ }z_1=z_2\text{ }g^n\text{ }g^m\text{ }z_1$ $=g^n\text{ }z_2\text{ }g^m\text{ }z_1=yx$ (since $z_1,z_2\in\mathcal{Z}(G)$). The conclusion follows. $\blacksquare$

Now since $\mathcal{Z}(G)\lhd G$ we have, by Lagrange's theorem that $\left|\mathcal{Z}\left(G\right)\right|\mid\left|G\ right|=pq\implies \left|\mathcal{Z}\left(G\right)\right|=1,p,q,pq$. If it's the first or the last we're done. So, assume that it could be either $p$ or $q$. Then

$G/\mathcal{Z}(G)=p,q$ and since all prime ordered groups are cyclic it follows from the lemma that $G$ is abelian and so $\mathcal{Z}(G)=G$ which contradicts the assumption that $\left|\mathcal{Z}(G)\right|=p,q$.

The conclusion follows.