Regarding this theorem, which is in Kassel pg 126, I have two questions. I have typed in the relevant material for reference.
1) How does $$U_1'\cong U[K]/(K^2-1)$$ imply $$U\cong U_1'/(K-1)$$? Kassel seems to imply that it suffices to prove the first isomorphism.
2) How do we conclude that "We then get an isomorphism from $U_1'$ to $U[K]/(K^2-1)$ by sending $E$ to $XK$, $F$ to $Y$, $K$ to $K$, and $L$ to $HK$."
Sincere thanks for any help!
To prove: If $q=1$, we have $$U_1'\cong U[K]/(K^2-1)$$ and $$U\cong U_1'/(K-1)$$
Kassel's Proof (paraphrased): We first prove the first isomorphism. Now, $U_1'$ is generated by $E$,$F$,$K$,$K^{-1}$,$L$ and Relations (\ref{or20prop.2.1}--\ref{or20prop.2.4}) in which $q$ has been replaced by $1$, namely
$KK^{-1}=K^{-1}K=1$ (2.5)
$KEK^{-1}=E,\ \ \ KFK^{-1}=F$ (2.6)
$[E,F]=L,\ \ \ K-K^{-1}=0$ (2.7)
$[L,E]=(EK+K^{-1}E$$[L,F]=-(FK+K^{-1}F)$ (2.8)
Relation 2.5 implies that $K$ commutes with $K^{-1}$. Relation 2.6 can be rewritten as $KE=EK$, $KF=FK$, hence $K$ commutes with $E$, $F$. Thus, $K$ is central.
The second relation in (2.7) gives $K=K^{-1}$, hence $K^2=1$. Hence, we can rewrite the Relations (2.8) as
$[L,E]=(EK+K^{-1}E)$ $=K^2(EK+K^{-1}E)\\$ $=EK+KE\\$ $=2EK$
$[L,F]=-(FK+K^{-1}F)\\$ $=-K^2(FK+K^{-1}F)\\$ $=-(FK+KF)\\$ $=-2FK$
We then get an isomorphism from $U_1'$ to $U[K]/(K^2-1)$ by sending $E$ to $XK$, $F$ to $Y$, $K$ to $K$, and $L$ to $HK$.
Recall that $$ \small U=\frac{\mathbb{k} \{X,Y,H\}}{([X,Y]-H, [H,X]-2X, [H,Y]+2Y)}\\ \small { U_1'=\frac{\mathbb{k}\{E,F,L,K,K^{-1}\}}{\begin{pmatrix}KK^{-1}-1& K^{-1}K-1& KE-EK& KF-FK\\ [E,F]-L& K^2-1& [L,E]-EK-K^{-1}E& [L,F]+FK+K^{-1}F\end{pmatrix}} } $$ Now we proceed to construction of isomorphisms.
First isomorphism Note that $$ \small U[K]=\frac{\mathbb{k} \{X,Y,H,K\}}{\begin{pmatrix}[X,Y]-H& [H,X]-2X& [H,Y]+2Y\\XK-KX& YK-KY& HK-KY\end{pmatrix}} $$ $$ \small \frac{U[K]}{(K^2-1)}=\frac{\mathbb{k} \{X,Y,H,K\}}{\begin{pmatrix}[X,Y]-H& [H,X]-2X& [H,Y]+2Y\\XK-KX& YK-KY& HK-KH\\ &K^2-1&\end{pmatrix}} $$ Denote ideals $$ \small \mathcal{\mathcal{I}}=\begin{pmatrix}KK^{-1}-1& K^{-1}K-1& KE-EK& KF-FK\\ [E,F]-L& K^2-1& [L,E]-EK-K^{-1}E& [L,F]+FK+K^{-1}F\end{pmatrix}\subset\mathbb{k}\{E,F,L,K,K^{-1}\}\\ \small \mathcal{J}=\begin{pmatrix}[X,Y]-H& [H,X]-2X& [H,Y]+2Y\\XK-KX& YK-KY& HK-KH\\ &K^2-1&\end{pmatrix}\subset\mathbb{k}\{X,Y,H,K\} $$ Consider homomorphisms of algebras $$ \small \Phi:\mathbb{k}\{E,F,L,K,K^{-1}\}\to \frac{U[K]}{(K^2-1)}\\ \Psi:\mathbb{k}\{X,Y,H,K\}\to U_1' $$ defined by $$ \small \Phi(E)=XK+\mathcal{J},\;\Phi(F)=Y+\mathcal{J},\;\Phi(L)=HK+\mathcal{J},\;\Phi(K)=K+\mathcal{J},\;\Phi(K^{-1})=K+\mathcal{J}\\ \small \Psi(X)=EK^{-1}+\mathcal{\mathcal{I}},\;\Psi(Y)=F+\mathcal{\mathcal{I}},\;\Psi(H)=LK^{-1}+\mathcal{\mathcal{I}},\;\Psi(K)=K+\mathcal{\mathcal{I}} $$ We are going to show that $\Phi(\mathcal{\mathcal{I}})\subset 0+\mathcal{J}$ and $\Psi(\mathcal{J})\subset 0+\mathcal{\mathcal{I}}$. Since $$ \small { \begin{align} \Phi(KK^{-1}-1)&=KK-1+\mathcal{J}=(K^2-1)+\mathcal{J}=0+\mathcal{J}\\ \Phi(K^{-1}K-1)&=KK-1+\mathcal{J}=(K^2-1)+\mathcal{J}=0+\mathcal{J}\\ \Phi(KE-EK)&=KXK-XKK+\mathcal{J}=(KX-XK)K+\mathcal{J}=0+\mathcal{J}\\ \Phi(KF-FK)&=KY-YK+\mathcal{J}=0+\mathcal{J}\\ \Phi([E,F]-L)&=XKY-YXK-HK+\mathcal{J}=XYK-YXK-HK+\mathcal{J}\\ &=([X,Y]-H)K+\mathcal{J}=0+\mathcal{J}\\ \Phi(K^2-1)&=K^2-1+\mathcal{J}=0+\mathcal{J}\\ \Phi([L,E]-EK-K^{-1}E)&=HKXK-XKHK-XKK-KXK+\mathcal{J}\\ &=HXK^2-XHK^2-XK^2-XK^2+\mathcal{J}=([H,X]-2X)K^2+\mathcal{J}=0+\mathcal{J}\\ \Phi([L,F]+FK+K^{-1}F)&=HKY-YHK+YK+KY+\mathcal{J}\\ &=HYK-YHK+YK+YK+\mathcal{J}=([H,Y]+2Y)K+\mathcal{J}=0+\mathcal{J}\\ \end{align} } $$ then all relations forming $\mathcal{\mathcal{I}}$ are mapped to $0+\mathcal{J}$ under $\Phi$. Hence $\Phi(\mathcal{\mathcal{I}})\subset 0+\mathcal{J}$. Since $$ \small{ \begin{align} \Psi(XK-KX)&=EK^{-1}K-KEK^{-1}+\mathcal{\mathcal{I}}=E-E+\mathcal{\mathcal{I}}=0+\mathcal{\mathcal{I}}\\ \Psi (YK-KY)&=FK-KF+\mathcal{\mathcal{I}}=0+\mathcal{\mathcal{I}}\\ \Psi (HK-KH)&=LK^{-1}K-KLK^{-1}+\mathcal{\mathcal{I}}=L-L+\mathcal{I}=0+\mathcal{I}\\ \Psi(K^2-1)&=K^2-1+\mathcal{I}=0+\mathcal{I}\\ \Psi([X,Y]-H)&=EK^{-1}F-FEK^{-1}-LK^{-1}+\mathcal{I}\\ &=EFK-FEK-LK+\mathcal{I}=([E,F]-L)K+\mathcal{I}=0+\mathcal{I}\\ \Psi([H,X]-2X)&=LK^{-1}EK^{-1}-EK^{-1}LK^{-1}-2EK^{-1}+\mathcal{I}\\ &=LKEK-EKLK-2EK=LEKK-ELKK-EK-KE+\mathcal{I}\\ &=[E,L]K^2-EK-K^{-1}E+\mathcal{I}=[E,L]-EK-K^{-1}E=0+\mathcal{I}\\ \Psi([H,Y]+2Y)&=LK^{-1}F-FLK^{-1}+2F+\mathcal{I}\\ &=LKF-FLK+2FK^{-1}K+\mathcal{I}=([L,F]+2FK^{-1})K+\mathcal{I}\\ &=([L,F]+FK^{-1}+FK^{-1})K=0+\mathcal{J}\\ \end{align} } $$ then all relations forming $\mathcal{J}$ are mapped to $0+\mathcal{I}$ under $\Psi$. Hence $\Psi(\mathcal{J})\subset 0+\mathcal{I}$ Given this inclusions one can say that there exists well defined homomorphisms of algebras $$ \small{ \varphi:U_1'\to\frac{U[K]}{(K^2-1)}\qquad\psi:\frac{U[K]}{(K^2-1)}\to U_1' } $$ well defined by $$ \small { \varphi(E+\mathcal{I})=XK+\mathcal{J},\;\varphi(F+\mathcal{I})=Y+\mathcal{J},\;\varphi(L+\mathcal{I})=HK+\mathcal{J}\\ \varphi(K+\mathcal{I})=K+\mathcal{J},\;\varphi(K^{-1}+\mathcal{I})=K+\mathcal{J}\\ \psi(X+\mathcal{J})=EK^{-1}+\mathcal{I},\;\psi(Y+\mathcal{J})=F+\mathcal{I},\;\psi(H+\mathcal{J})=LK^{-1}+\mathcal{I},\;\psi(K+\mathcal{J})=K+\mathcal{I} } $$ Note that $$ \small{ (\varphi\circ\psi)(X+\mathcal{J})=X+\mathcal{J},\qquad (\varphi\circ\psi)(Y+\mathcal{J})=Y+\mathcal{J},\\ (\varphi\circ\psi)(H+\mathcal{J})=H+\mathcal{J},\qquad (\varphi\circ\psi)(K+\mathcal{J})=K+\mathcal{J}\\ (\psi\circ\varphi)(E+\mathcal{I})=E+\mathcal{I},\qquad (\psi\circ\varphi)(F+\mathcal{I})=F+\mathcal{I},\qquad (\psi\circ\varphi)(L+\mathcal{I})=L+\mathcal{I},\qquad (\psi\circ\varphi)(K+\mathcal{I})=K+\mathcal{I},\qquad (\psi\circ\varphi)(K^{-1}+\mathcal{I})=K^{-1}+\mathcal{I} } $$ Since $\varphi$ and $\psi$ are inverses of each on generators, then they are inverse homomorphisms to each other. Now we can say that the desired isomorphism is constructed.
Second isomorphism Note that $$ \small { \frac{U_1'}{(K-1)}=\frac{\mathbb{k}\{E,F,L,K,K^{-1}\}}{\begin{pmatrix}KK^{-1}-1& K^{-1}K-1& KE-EK& KF-FK\\ [E,F]-L& K^2-1& [L,E]-EK-K^{-1}E& [L,F]+FK+K^{-1}F\\ &K - 1&\end{pmatrix}} } $$ Denote ideals $$ \small \mathcal{I}=\begin{pmatrix}KK^{-1}-1& K^{-1}K-1& KE-EK& KF-FK\\ [E,F]-L& K^2-1& [L,E]-EK-K^{-1}E& [L,F]+FK+K^{-1}F\\ &K - 1&\end{pmatrix}\subset\mathbb{k}\{E,F,L,K,K^{-1}\}\\ \small \mathcal{J}=([X,Y]-H, [H,X]-2X, [H,Y]+2Y)\subset\mathbb{k}\{X,Y,H\} $$ Consider homomorphisms $$ \small \Phi:\mathbb{k}\{E,F,L,K,K^{-1}\}\to U\\ \Psi:\mathbb{k}\{X,Y,H\}\to \frac{U_1'}{(K-1)} $$ defined by $$ \small \Phi(E)=X+\mathcal{J}\;\Phi(F)=Y+\mathcal{J},\;\Phi(L)=H+\mathcal{J},\;\Phi(K)=1+\mathcal{J},\;\Phi(K^{-1})=1+\mathcal{J}\\ \small \Psi(X)=E+\mathcal{I},\;\Psi(Y)=F+\mathcal{I},\;\Psi(H)=L+\mathcal{I} $$ We are going to show that $\Phi(\mathcal{I})\subset 0+\mathcal{J}$ and $\Psi(\mathcal{J})\subset 0+\mathcal{I}$. Since $$ \small { \begin{align} \Phi(KK^{-1}-1)&=1\cdot 1-1+\mathcal{J}=0+\mathcal{J}\\ \Phi(K^{-1}K-1)&=1\cdot 1-1+\mathcal{J}=0+\mathcal{J}\\ \Phi(KE-EK)&=1X-X1+\mathcal{J}=0+\mathcal{J}\\ \Phi(KF-FK)&=1Y-Y1+\mathcal{J}=0+\mathcal{J}\\ \Phi([E,F]-L)&=XY-YX-H+\mathcal{J}=([X,Y]-H)+\mathcal{J}=0+\mathcal{J}\\ \Phi(K^2-1)&=1^2-1+\mathcal{J}=0+\mathcal{J}\\ \Phi([L,E]-EK-K^{-1}E)&=HX-XH-X1-1X+\mathcal{J}=([H,X]-2X)+\mathcal{J}=0+\mathcal{J}\\ \Phi([L,F]+FK+K^{-1}F)&=HY-YH+Y1+1Y+\mathcal{J}=([H,Y]+2Y)+\mathcal{J}=0+\mathcal{J}\\ \end{align} } $$ then all relations forming $\mathcal{I}$ are mapped to $0+\mathcal{J}$ under $\Phi$. Hence $\Phi(\mathcal{I})\subset 0+\mathcal{J}$. Since $$ \small{ \begin{align} \Psi([X,Y]-H)&=EF-FE-L+\mathcal{I}=([E,F]-L)+\mathcal{I}=0+\mathcal{I}\\ \Psi([H,X]-2X)&=LE-EL-2E+\mathcal{I}=[E,L]-E1-1E+\mathcal{I}\\ &=[E,L]-EK-K^{-1}E+\mathcal{I}=0+\mathcal{I}\\ \Psi([H,Y]+2Y)&=LF-FL+2F+\mathcal{I}=([L,F]+F1+1F)+\mathcal{I}\\ &=([L,F]+FK+K^{-1}F)+\mathcal{I}=0+\mathcal{J}\\ \end{align} } $$ then all relations forming $\mathcal{J}$ are mapped to $0+\mathcal{\mathcal{I}}$ under $\Psi$. Hence $\Psi(\mathcal{J})\subset 0+\mathcal{\mathcal{I}}$ Given this inclusions one can say that there exists well defined homomorphisms of algebras $$ \small{ \varphi:U\to\frac{U_1'}{(K-1)}\qquad\psi:\frac{U_1'}{(K-1)}\to U } $$ well defined by $$ \small { \varphi(E+\mathcal{I})=X+\mathcal{J}\;\varphi(F+\mathcal{I})=Y+\mathcal{J},\;\varphi(L+\mathcal{I})=H+\mathcal{J}\\ \varphi(K+\mathcal{I})=1+\mathcal{J},\;\varphi(K^{-1}+\mathcal{I})=1+\mathcal{J}\\ \psi(X+\mathcal{J})=E+\mathcal{I},\;\psi(Y+\mathcal{J})=F+\mathcal{I},\;\psi(H+\mathcal{J})=L+\mathcal{I} } $$ Note that $$ \small{ (\varphi\circ\psi)(X+\mathcal{J})=X+\mathcal{J}\qquad (\varphi\circ\psi)(Y+\mathcal{J})=Y+\mathcal{J}\qquad (\varphi\circ\psi)(H+\mathcal{J})=H+\mathcal{J}\\ (\psi\circ\varphi)(E+\mathcal{\mathcal{I}})=E+\mathcal{\mathcal{I}}\qquad (\psi\circ\varphi)(F+\mathcal{\mathcal{I}})=F+\mathcal{\mathcal{I}}\qquad (\psi\circ\varphi)(L+\mathcal{\mathcal{I}})=L+\mathcal{\mathcal{I}}\\ (\psi\circ\varphi)(K+\mathcal{\mathcal{I}})=K+\mathcal{\mathcal{I}}\qquad (\psi\circ\varphi)(K^{-1}+\mathcal{\mathcal{I}})=K^{-1}+\mathcal{\mathcal{I}}\\ } $$ Since $\varphi$ and $\psi$ are inverses of each other on generators, then they are inverse homomorphisms of each other. Now we can say that the desired isomorphism is constructed.