Jacobson's proof of $D_n = \langle x,y\mid x^n=y^2=1, (xy)^2 = 1\rangle$

Question

I still don't understand what free groups are all about (I do know how to get them but I don't understand what they truly are) so this might be a very basic question.

Jacobson claims that the dihedral group is generated by these relations: $x^n,y^2,xyxy$ in the free group generated by $x$ and $y$.

He says first that if $R$ denotes the rotation by $2\pi/n$ and $S$ denotes the reflection in the $x$-axis, then $R^n=S^2= 1$ and $(RS)^2 =1$. So far so good.

Next he says that, then, $D_n$ is a homomorphic image of $FG^{(2)}/K$, where $K$ is the normal subgroup generated by $x^n,y^2,xyxy$. Then he proceeds to prove that $FG^{(2)}/K$ has order $\le 2n$, and he claims that this shows $D_n \cong FG^{(2)}/K$.

I have no idea what he is doing. Can someone please explain?

Answer 1

I'll try to explain what (At least i think) happens: You have the elements $R$ and $S$ in the group $D_n$ and know that these elements satisfy the relations given in the brackets. Then we can look at the grouphomomorphism $$ \phi: FG^{(2)} \to D_n $$ defined by $x \mapsto R, y \mapsto S$. Since $D_n = <R,S>$ this homomorphism is still surjective. The subgroup $K$ that you mentioned is in the kernel of this homomorphism, so by the fundamental theorem of group homomorphisms we get a well defined surjective homomorphism $$ \phi': FG^{(2)}/K \to D_n$$ Now, since $|D_n|=2n$ and he shows that $|FG^{(2)}|\leq 2n$ we have a surjective map from an at most smaller set in another set, so it has to be bijective, therefore $\phi'$ is an isomorphism. Hope that helps.

Answer 2

Define a group homomorphism $$\varphi \colon FG^{(2)} \to D_{2n}, \quad \varphi(x)=R, \,\, \varphi(y) = S.$$ It is clearly surjective. Since $R^n=S^2 = (RS)^2=1$, it follows that $K \subseteq \ker \varphi$, so there is an induced, surjective homomorphism $$\bar{\varphi} \colon FG^{(2)}/K \to D_{2n}.$$ But $|FG^{(2)}/K| \leq 2n= |D_{2n}|$, so $\bar{\varphi}$ is actually an isomorphism.