## Applying Lagrange!: Groups of Prime Orders.

### June 29, 2010

Little post.  Because I love doing things that comments tell me to do, we’re going to use Lagrange to prove a neato theorem.  Now, normally, if I told you, “Hey, guy, I’ve got a group $G$ with $n$ elements.  What one is it?” you’d probably be unable to tell me!  Why?  Lots of different groups have the same order!  For example, if we’re talking about order 8, are we talkin’ $D_{8}$?  Are we talkin’ $Z_{8}$?  Are we talkin’ $Q_{8}$?  I just don’t know!

But!  If the order of the group is prime, Lagrange’s Theorem tells us that the group must be cyclic.  No way out of it!  But let’s prove it just to be sure.

Let’s introduce a little notation before we do this, just to make things easier.  Let’s denote the cyclic group generated by the element x as $\langle x\rangle$ with those little brace-things around it.  So we have that, by definition $\langle x\rangle = \{x^{n} | n\in {\mathbb Z}\}$.

Theorem (Classification of Groups of Prime Order):  If a group $G$ has prime order, then it is cyclic.

Proof. Okay, talk about slick proofs, I love this one: Cauchy’s theorem tells us there’s an element $x$ of order $p$, so consider the group $\langle x\rangle = \{1, x, x^2, x^3, \dots, x^{p-1}\}$.  This is a subgroup of the group $G$ and it has the same order.  Therefore, it must be the case that $\langle x\rangle = G$.  Then $G$ is cyclic.  Bam.  $\Box$

Alternate Proof. Alright, another proof of this.  Since $|G| = p > 1$ we have that there is a non-identity element in $G$.  Let this non-identity element be called $x$.  Then consider $\langle x\rangle$.  Certainly, we have that $|\langle x\rangle| > 1$ since it has at least the identity and the element $x$ in it, and by Lagrange, since it is a subgroup, it must divide the order of the group.  Since the order of the subgroup is greater than 1, and only 1 and $p$ divide the order of the group, it must be the case that $|\langle x\rangle| = p$, which means that $\langle x\rangle = G$$\Box$