Substitute the finite formula and write $n=m+2j$. Under the allowed rearrangements,Hence $G(t,z)=e^{2tz-z^2}$, with the normalization of the physicists' Hermite polynomials used in the source.

Differentiating the Rodrigues identity gives $H_0=1$ and $H_{n+1}=2t{H}_n-H_n^{\prime }$. Induct with the proposed finite sum. The coefficient at $t^{n+1-2j}$ in this recurrence is the contribution from $2t{H}_n$ minus the derivative contribution with previous index $j-1$:Using a common denominator, this equals $(n+1)!(-1{)}^j{2}^{n+1-2j}/[j!(n+1-2j)!]$. At an endpoint omit terms whose factorial indices would be negative; the same formula then covers the sole remaining term. ThusThe initial value and recurrence identify these polynomials with the given Rodrigues definition.

Expand both squared norms using sesquilinearity and conjugate symmetry. Their mixed terms $⟨u,v⟩+⟨v,u⟩$ appear with opposite signs and cancel. Hencethe parallelogram identity.

A rational map from irreducible $X$ to $Y$ is an equivalence class of morphisms $F_U:U\to Y$ defined on nonempty open subsets $U\subseteq X$. Two representatives are equivalent if they agree on a nonempty open subset of their intersection.

Irreducibility makes all these nonempty opens dense and ensures their intersections are nonempty, so the equivalence expresses agreement at general points. A rational map need not extend across all of $X$, and it need not be dominant. An injective homomorphism of function fields additionally requires a dominant morphism of affine varieties, or the corresponding condition on a rational map.

A morphism $F:X\to Y$ is a continuous map in the Zariski topologies such that for every open $V\subseteq Y$ and every regular function $h$ on $V$, the pullback $h\circ F$ is regular on $F^{-1}(V)$.

Equivalently, near every point, choose affine charts of source and target; the target coordinates of $F$ must then be regular functions of the source coordinates. These local conditions agree on overlaps. In particular a morphism is defined at every point of its source, not merely on a dense open subset.

All 63 involutions lie in Sylow subgroups for $2$. Each such subgroup has exactly seven nonidentity elements and distinct ones have no nonidentity overlap. Hence

If Sylow subgroups $S,T$ share $t\ne 1$, they are abelian by the preceding deduction, so $S,T\le C_G(t)$. But $|C_G(t)|=8$ and both subgroups have order eight. Thus $S=C_G(t)=T$. Consequently distinct Sylow $2$-subgroups intersect only in the identity.

A Sylow subgroup for $2$ has order eight. Every one of its nonidentity elements lies in the only even-order class, so has order two. In a group of exponent two,It is therefore an elementary abelian group and $\boxed{S\cong C_2^3}$.

On the same basis, $Y=D^2$ kills $v,u$ and sends $w$ to $16w$. Explicitly,Thus all three basis lines are submodules after restriction of scalars, andEach summand is a simple module, hence an indecomposable module. Squaring has destroyed the length-two nilpotent Jordan chain.

A direct determinant computation givesThe zero eigenvalue has algebraic multiplicity two but a one-dimensional eigenspace, so it gives one Jordan block of size two. The other block is $J_1(4)$. For an explicit Jordan basis, takeThese are linearly independent and satisfy $Dv=0$, $Du=v$, $Dw=4w$. ThusThe first summand has the chain $u\mapsto v\mapsto 0$ and is an indecomposable module; the other is simple.

Apply the decomposition to the module on which $X$ acts as $\alpha$. On $\mathbb{C}[X]/((X-\lambda {)}^m)$ put $t=X-\lambda$ and use the ordered basisMultiplication by $X=\lambda +t$ has diagonal entries $\lambda$ and superdiagonal entries $1$, so its matrix is the Jordan block $J_m(\lambda )$. Concatenating the bases of the direct sum gives a basis of $V$ in whichThis proves existence of Jordan normal form; uniqueness of the primary cyclic decomposition also proves uniqueness of its blocks up to order.

Every finite-dimensional $\mathbb{C}[X]$-module is a finitely generated torsion module over the principal ideal domain $\mathbb{C}[X]$. The structure theorem for finitely generated modules over a principal ideal domain giveswhere the $p_j$ are uniquely determined monic nonconstant polynomials. Equivalently, factoring over $\mathbb{C}$ and using the Chinese remainder theorem givesThere are finitely many summands, and the multisets of positive integers $m_{\lambda j}$ are uniquely determined. Each displayed primary cyclic summand is an indecomposable module.

The Jacobson radical is $J=\mathbb{C}{e}_{12}$. Indeed $J^2=0$, so $J$ is contained in the Jacobson radical, and $A/J\cong \mathbb{C}\times \mathbb{C}$ is a semisimple algebra, giving the reverse inclusion. The radical of a module of the left regular module is thereforeIts socle is the sum of its simple modules. The displayed left action shows that its $S_1$-submodules span $\mathbb{C}{e}_{11}\oplus \mathbb{C}{e}_{12}$. There is no $S_2$-submodule: if $av=tv$ for every $a$, applying $e_{11}$ forces the first row of $v$ to vanish, and applying $e_{12}$ then forces the remaining $e_{22}$ coefficient to vanish. HenceA homomorphism from $S_1$ is determined by any image $\alpha e_{11}+\beta e_{12}$ of $s_1$. Thus $V_{S_1}$ is the two-dimensional row right module with actionIts line $\mathbb{C}{e}_{12}$ is a nonzero proper submodule, so $V_{S_1}$ is not simple. Since $S_2$ has no copy in the socle, any homomorphism $S_2\to A$ is zero. Thus $V_{S_2}=0$ is not simple either, since a simple module is required to be nonzero.

Choose bases $S_1=\mathbb{C}{s}_1$, $S_2=\mathbb{C}{s}_2$ and setBoth are simple modules. They are nonisomorphic because $e_{11}$ acts as the identity on $S_1$ and as zero on $S_2$. In the ascending composition series of a module above, the factors are, in order,The last factor is represented by $e_{22}$ modulo the span of $e_{11},e_{12}$.

Write $e_{ij}$ for the matrix units. Left multiplication by $a=\left(\begin{array}{cc}r& s\\ 0& t\end{array}\right)$ givesThus a composition series of a module isEach quotient is one-dimensional, hence a simple module. The required direct sum isThe first summand is simple. The second has a unique nonzero proper submodule, $\mathbb{C}{e}_{12}$: any vector $u{e}_{12}+v{e}_{22}$ with $v\ne 0$ generates $e_{12}$ by applying $e_{12}$ and then generates $e_{22}$. A decomposition into two nonzero summands would require a second one-dimensional submodule, which is impossible. Hence this summand is an indecomposable module.

By the Artin–Wedderburn theorem, a finite-dimensional complex simple algebra is $M_d(\mathbb{C})$, with a unique simple module up to isomorphism, namely the column module $S={\mathbb{C}}^d$. Its left regular module is $S^{\oplus d}$. The Schur lemma gives ${\mathrm{End}}_A(S)=\mathbb{C}$, soThe right action can be seen explicitly. For a row vector $r$, the map $f_r(m)=mr$ is $A$-linear, and all such maps are obtained this way, since the row space and $V_M$ both have dimension $d$. Moreover $f_r\cdot x=f_{rx}$. Any nonzero row generates the whole row space under right multiplication by matrices: if $r_j\ne 0$, suitable matrix units send it to each coordinate row. Thus $V_M$ is a simple right $A$-module.