(a) By induction on $\left|Y\right|\text{.}$ Assume $\left|Y\right|>1\text{.}$ Since $Y$ is a $p\text{-group,}$ it has a normal subgroup $Y_1$ of index $p,$ and the subspace $V_1$ of invariants of $Y_1$ on $V$ is nonzero by the inductive assumption. Choose $y\in Y$ to generate $Y/Y_1,$ and $v$ in $V_1$ and nonzero. Then ${(1-y)}^{p}v=0\text{.}$ Now choose $r$ maximal so that ${(1-y)}^{r}v\ne 0\text{.}$ The resulting vector is fixed by $Y,$ whence (a).

(b) By (a).

(c) By inspection $\text{Rad}\hspace{0.17em}KY$ is an ideal, maximal because its codimension in $KY$ is $1\text{.}$ Bring the kernel of the trivial representation, which is the only irreducible one by (b), it is the unique maximal left ideal; and similarly for right ideals. On each factor of a composition series for the left regular representation of $KY$ on itself $Y$ acts trivially by (b), hence $\text{Rad}\hspace{0.17em}KY$ acts as $0,$ so that $\text{Rad}\hspace{0.17em}KY$ is nilpotent.

(d) By (b).

$\square$

(a) If $f\left(x\right)=1,$ we get the contradiction $xw\in B,$ while if $f\left(x\prime \right)=f\left(x\right),$ we see that $w^{-1}{x}^{-1}x\prime w\in B,$ so that $x\prime =x\text{.}$ Similarly for $g\text{.}$

(b) ${\left(\bar{X}w\right)}^2=\bar{X}{w}^2+\underset{x\in X-1}{\Sigma }\bar{X}f\left(x\right)h\left(x\right)wg\left(x\right)\text{.}$

Here $f\left(x\right)$ gets absorbed in $\bar{X}$ and $h\left(x\right)$ normalizes $\bar{X},$ whence (b).

$\square$

(a) By Lemma 80(b) the space of fixed points of $U=X$ on $V$ is nonzero. Since $H$ normalizes $U$ and is Abelian, that space contains a nonzero vector $v$ such that (a) and (b) of the definition of highest weight vector hold. Let $\bar{X}wv=v_1\text{.}$ If $v_1=0,$ then (c) holds with $\mu =0\text{.}$ If not, we replace $v$ by $v_1\text{.}$ Then (a) and (b) of the definition still hold with $w\lambda$ in place of $\lambda ,$ and by Lemma 81(b) so does (c) with $\mu =\Sigma \lambda \left(h\left(x\right)\right)\text{.}$ Now to prove that $KGv=KYv$ it is enough, because of the decomposition $G=YB\cup wB,$ to prove that $wv\in KYv\text{.}$ By the two parts of Lemma 81 we may write $\bar{X}wv=\mu v,$ after some simplification, in the form

with $y\left(x\right)=wx{w}^{-1}\in Y,$ whence our assertion.

(b) Let $V\prime =Kv$ and $V^{\prime \prime }=\text{Rad}\hspace{0.17em}KY·v\text{.}$ It follows from Lemma 80(c) that the sum $V=V\prime +V^{\prime \prime }$ is direct. Now assume there exists some $v_1\in V,$ $v_1\notin V\prime ,$ fixed by $X\text{.}$ We may assume that $v_1\in V^{\prime \prime }$ and also that $v_1$ is an eigenvector for $H$ since $H$ is Abelian. We have $\bar{X}w{v}_1=w\bar{Y}{v}_1=0\text{;}$ since $\bar{Y}(1-y)=0$ for any $y\in Y\text{.}$ Thus $v_1$ is a highest weight vector. By (a), $V=KY{v}_1\subseteq KY{V}^{\prime \prime }=V^{\prime \prime },$ a contradiction, whence (b).

(c) By (b) an irreducible $KG\text{-module}$ determines its highest weight $(\lambda ,\mu )$ uniquely. Conversely, assume that $V_1$ and $V_2$ are irreducible $KG\text{-modules}$ with highest weight vectors $v_1$ and $v_2$ of the same weight $(\lambda ,\mu )\text{.}$ Set $v=v_1+v_2\in V_1+V_2$ and then $V=KGv=KYv\text{.}$ Now $v_2\notin V,$ since otherwise we could write $v_2=cv+v^{\prime \prime }$ with $c\in K$ and $v^{\prime \prime }\in \text{Rad}\hspace{0.17em}KY·v$ and then projecting on $V_1$ and $V_2$ get that $c=0$ and $c=1,$ a contradiction. Thus we may complete the proof as in the proof of Theorem 39(e).

$\square$

(1) Proof of (b). In $KG$ let $H_{\lambda }=\underset{h\in H}{\Sigma }\lambda \left(h^{-1}\right)h,$ then $u=\bar{X}{H}_{\lambda }w\bar{X}$ and $v=\bar{X}{H}_{\lambda }\text{.}$ As in the proof of Theorem 44(a), asimple calculation yields $\bar{X}w(u+cv)=\underset{x\in X-1}{\Sigma }\lambda \left(h\left(x\right)\right)u+c\bar{X}w{H}_{\lambda }\bar{X}\text{.}$ Here $H$ acts, from the left, according to the characters $w\lambda ,\lambda ,w\lambda$ on the respective terms. Thus if $w\lambda \ne \lambda ,$ we may realize the weight $(\lambda ,0)$ by taking $c=0\text{.}$ If $w\lambda =\lambda ,$ we take $c=-\Sigma \lambda \left(h\left(x\right)\right)$ instead. Finally if $\lambda =1,$ then $\Sigma \lambda \left(h\left(x\right)\right)=-1$ and we get $\mu =-1$ by taking $c=0\text{.}$ To achieve $(\lambda ,\mu )$ in an irreducible module we simply take $KG(u+cv)$ modulo a maximal submodule.

(2) If $\text{dim}\hspace{0.17em}V=1,$ then $V$ is trivial and $(\lambda ,\mu )=(1,0)\text{.}$ Since $X$ and $Y$ are $p\text{-groups,}$ they act trivially by Lemma 80(b), whence (2).

(3) If $\text{dim}\hspace{0.17em}V\ne 1,$ then $\lambda$ determines $\mu \text{.}$ Write $V=V\prime +V^{\prime \prime }$ as in the proof of Theorem 44(b). We have $V^{\prime \prime }\ne 0\text{.}$ Since $Y$ ? fixes $V^{\prime \prime }$ it fixes some line in it, uniquely determined in $V,$ by Theorem 44(b) with $Y$ in place of $X\text{.}$ Since $Y$ clearly fixes $wv,$ we conclude that $wv\in V^{\prime \prime }\text{.}$ Projecting $(*)$ of the proof of Theorem 44(a) onto $V\prime ,$ we get

whence (3).

(4) Proof of (a). Combine (1), (2) and (3).

$\square$

(a) If $\lambda \ne 1,$ then $\mu =0$ by Theorem 45 (a), so that $\Sigma \lambda \left(h\left(x\right)\right)=0$ by $(**)$ above applied with $\lambda$ replaced by $w\lambda \text{.}$ Then $\Sigma n\left(h\right)\lambda \left(h\right)=0$ for every $\lambda \ne 1\text{.}$ By the orthogonality relations for the characters on $H$ (which are valid since $p\nmid \left|h\right|\text{),}$ we conclude that $n\left(h\right),$ as an element of $K,$ is independent of $h\text{.}$ If $n\left(h\right)$ were $0$ for some $h,$ we would get $\left|H\right|=\Sigma n\left(h\right)=0$ $\text{(mod}\hspace{0.17em}p\text{),}$ a contradiction.

(b) Let $V$ be an irreducible module whose dual $V^{*}$ has the highest weight $(\lambda ,\mu )\text{.}$ We consider, as in Theorem 40 the isomorphism $\phi$ of $V^{*}$ into the induced representation space of functions $a:G\to K$ such that $a\left(yb\right)=a\left(y\right){\lambda }^{*}\left(b\right)$ for all $y\in G,b\in B$ defined by $\left(\phi f\right)\left(x\right)=f\left(x{v}^{+}\right)$ with $v^{+}$ a highest weight vector for $V\text{.}$ Using the decomposition $V=Kw{v}^{+}+\text{Rad}\hspace{0.17em}KX·w{v}^{+},$ we may define $f^{+}$ as in Lemma 73, prove that it is a highest weight vector, and that $a^{+}=\phi f^{+}$ is given by the equations of Lemma 74, with ${\lambda }^{*}$ in place of $\lambda \text{.}$ Coverting functions on $G$ to elements of $KG$ in the usual way, $a\sim \Sigma a\left(x\right)x,$ we see that $a^{+}$ becomes the element of (b), whence (b). At the same time we see that $V$ may be realized in the induced module $B_{{\lambda }^{*}}\to G,$ as the unique irreducible submodule in case $\lambda \ne 1,$ as one of two in case $\lambda =1\text{.}$

(c) By (b).

(d) If $\mu =0,$ then $\bar{X}wv=0,$ whence $\bar{Y}v=0$ and $\text{dim}\hspace{0.17em}V<\left|X\right|$ by Theorem 44(a). Conversely if $\text{dim}\hspace{0.17em}V<\left|X\right|,$ then the annihilator of $v$ in $KY$ contains $\bar{Y}$ by Lemma 80(d), so that $\mu =0\text{.}$

(e) By a classical theorem of Brauer and Nesbitt (University of Toronto Studies, 1937) the number in question equals the number of irreducible $KG\text{-modules,}$ hence equals $\left|H\right|+1$ by Theorem 45(b).

$\square$

Since $\bar{w}={\bar{w}}_a{\bar{w}}_b\dots$ this easily follows by induction on $N\left(w\right)$ or by Appendix II.25.

$\square$

We shall prove this theorem in several steps.

This is proved as in Theorem 44(a).

		Proof.
	Let $x$ be any element of $U\text{.}$ Write $x=x_a^{\prime }{x}_a$ with $x_a\in X_a$ and $x_a^{\prime }\in X_a^{\prime },$ the subgroup of elements of $U$ whose $X_a$ components are $1\text{.}$ We recall that $X_a$ and ${\bar{w}}_a$ normalize $X_a^{\prime }$ (see Appendix I.11). Thus $x{v}_1=x_a^{\prime }{\bar{X}}_a{\bar{w}}_{a}v=v_1,$ since $Uv=v\text{.}$ Since $H$ normalizes $X_a$ and is normalized by ${\bar{w}}_a,$ we see that $v_1$ is also an eigenvector for $H\text{.}$ $\square$

		Proof.
	By Lemma 82 and (a2), $v_1$ is an eigenvector for $B\text{.}$ Let $a$ be any simple root. If $w^{-1}a>0,$ then $N\left(w_{a}w\right)=N\left(w\right)+1$ by Appendix II.19, so that ${\bar{X}}_{w_{a}w}\overline{w_{a}w}={\bar{X}}_a{\bar{w}}_a{\bar{X}}_w\bar{w}$ Lemma 82, and ${\bar{X}}_a{\bar{w}}_a{v}_1=0$ by the choice of $w\text{.}$ If $w^{-1}a<0,$ then we may choose a minimal expression $w=w_a{w}_b\dots$ starting with $w_a\text{.}$ Then $$\begin{array}{cccc}{\bar{X}}_a{\bar{w}}_a{v}_1& =& {\left({\bar{X}}_a{\bar{w}}_a\right)}^2{\bar{X}}_b{\bar{w}}_b\dots v& \text{by Lemma 82}\\ & =& \mu {\bar{X}}_a{\bar{w}}_a{\bar{X}}_b{\bar{w}}_b\dots v,& \text{with}\hspace{0.17em}\mu \in K\hspace{0.17em}\text{by Lemma 81(b)}\\ & =& \mu v_1\text{.}\end{array}$$ $\square$

By (a1) and (a3) we have the first statement in (a).

		Proof.
	We have $G={\bar{w}}_{0}G\subseteq \underset{W}{\cup }\bar{w}{U}^{-}B$ by Theorem 4. Thus it is enough to show each $\bar{w}$ fixes $K{U}^{-}v,$ and for this we may assume $w=w_a$ with $a$ simple. Assume $y\in U^{-}\text{.}$ Write $y=y_a{y}_a^{\prime }$ as above, but using negative roots instead. Then $y_a$ and ${\bar{w}}_a$ normalize $Y_a^{\prime },$ so that ${\bar{w}}_{a}yv={\bar{w}}_a{y}_a{y}_a^{\prime }v\in U^{-}{\bar{w}}_a{y}_{a}v\subseteq K{U}^{-}v$ by Theorem 44(a) applied to $⟨X_a,Y_a⟩\text{.}$ $\square$

		Proof.
	Write $y=y_a{y}_a^{\prime }$ as before. Then ${\bar{X}}_a{\bar{w}}_{a}yv=\underset{x\in X_a}{\Sigma }y\left(x\right)x{\bar{w}}_{a}v$ with $y\left(x\right)\in U^{-}\text{.}$ Thus ${\bar{X}}_a{\bar{w}}_a(y-1)v=\underset{x\in X_a}{\Sigma }(y\left(x\right)-1)x{\bar{w}}_{a}v\in V^{\prime \prime }\text{.}$ $\square$

(c) Same proof as for Theorem 44(c).

(d) By Theorem 45(a) applied to $⟨X_a,Y_a⟩\text{.}$

(e) Let $\pi$ be the set of simple roots $a$ such that ${\mu }_a=0,$ and $W_{\pi }$ the corresponding subgroup of $W\text{.}$ The reader should have no trouble in proving that the left ideal of $KG$ generated by $\underset{w\in W_{\pi }{w}_0}{\Sigma }\bar{U}{H}_{\lambda }\bar{w}{\bar{X}}_w$ is an irreducible $KG\text{-module}$ whose highest weight is $(\lambda ,{\mu }_a)\text{.}$

$\square$

(a) Clear.

(b) Write $\bar{U}{\bar{w}}_{0}v={\bar{X}}_{w_0}{\bar{w}}_{0}v={\bar{X}}_a{\bar{w}}_a{\bar{X}}_b{\bar{w}}_b\dots v,$ with $w_0=w_a{w}_b\dots$ as in Lemma 82, and then proceed as in the proof of Cor. (d) to Theorems 44 and 45.

(c) The given sum counts the number of possible weights $(\lambda ,{\mu }_a)$ according to the set $\pi$ of simple roots $a$ such that ${\mu }_a=-1\text{.}$

$\square$

(a) This is a refinement of Appendix IV.38 since the relations $w_{\alpha }^2=1$ are not required. It is an easy exercise to convert the proof of the latter result into a proof of the former, which we shall leave to the reader.

(b) Because of (a) we only have to prove (b) when $w$ has the form of the two sides of $(*)\text{.}$ For this we can refer to the proof of Lemma 56 since the extra restrictions there, that $G$ is untwisted and that ${\bar{w}}_a=w_a\left(1\right)$ for each $a,$ are not essential for the proof.

$\square$