\input amstex
\input amssym.def
\nopagenumbers
\magnification=\magstep1
\centerline{\bf HW  \#2}
\bigskip
\centerline {\bf Math 506A}
\bigskip
\noindent 1. Let $R$ be any ring and $M$ an irreducible left $R$-module (so that $M$ has no nonzero proper $R$-submodule).  Show that the ring $\hom_R(M,M)$ of $R$-module homomorphisms from $M$ to itself is a division ring; that is, it satisfies all axioms of a field except for commutativity of multiplication.
\vskip .3in
\noindent 2. Recall that the dihedral group $D_8$ of order 8 is generated by two elements $x,y$ such that $x^4=y^2=1, yxy = yxy^{-1} = x^{-1}$.  This group is the group of symmetries of a square.  Use this fact to choose vertices of a suitable square and write down an irreducible two-dimensional real representation $\pi$ of $D_8$, computing the $2\times 2$ matrices $\pi(x)$ and $\pi(y)$ explicitly.
\vskip .3in
\noindent 3. Recall that the quaternion group $Q_8$, also of order 8 and generated by two elements $x,y$, this time such that $x^4 = 1, y^2 = x^2, yxy^{-1} = x^{-1}$, is {\sl not} isomorphic to any subgroup of the group $G=GL(2,\Bbb R)$ of $2\times 2$ invertible real matrices.  (Assume contrarily that $\pi$ is an isomorphism from $Q_8$ into a subgroup of $G$.  Show first that $\pi(x^2) = -I$, the negative of the identity matrix.  Replacing $\pi$ by a conjugate, you may assume that $\pi(x)$ is in rational canonical form.  Then deduce a contradiction).
\vskip .3in
\noindent 4. (20 points) Let $V$ be the infinite-dimensional module of the Klein four-group\hfil\break $K=C_2\times C_2$ over the field $F_2$ with two elements given in class, so that $V$ has a basis\hfil\break $\{v_i:i\in\Bbb Z\}\cup\{w_i:i\in\Bbb Z\}$ over $F_2$ and the commuting generators $x,y$ of $K$ are such that $xw_i = yw_i=w_i, xv_i=v_i+w_{i+1},yv_i=v_i+w_i$.  Show that $V$ is indecomposable.  (Assume for a contradiction that $V=V_1\oplus V_2$ for some nonzero submodules $V_1,V_2$.  Show that the projections $P_1,P_2$ of $V_1,V_2$ to the span $S$ of the $v_i$ intersect trivially  and that $S$ is their direct sum.   Then $(x-1)P_1+(x-1)P_2=(y-1)P_1+(y-1)P_2 = T$, the span of the $w_i$, forcing $(x-1)P_1=(y-1)P_1,(x-1)P_2=(y-1)P_2$.  Whenever a particular combination $\sum a_iv_i$ lies in $P_1$, this last fact forces a large number of other such combinations to lie in $P_1$, and similarly for $P_2$.  Deduce a contradiction unless one of $P_1,P_2$, say $P_1$, is all of $S$.  Then show that $V_1=V$.)
\end