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The group ring $R[G]$ consists of finite formal sums ${\sum }_{g\in G}{r}_{g}g$ with coefficients in a ring $R$. Addition is coefficientwise and multiplication extends the group operation and multiplication in $R$ distributively. A group algebra is the case in which $R$ is a field.

The multiplicative algebraic group has ${\mathbb{G}}_m(K)=K^{\times }$, with multiplication as its group operation. Its coordinate ring is $K[z,z^{-1}]$. The formal multiplicative group describes its law near the identity $z=1$.

Let $C$ be the projective cubic of a Weierstrass equation of an elliptic curve, allowing its elliptic-curve discriminant to vanish, and let $O=[0:1:0]$. Use only the smooth locus of a variety $C_{\mathrm{sm}}(K)$: the singular point, if present, is excluded. For $P,Q$, intersect their chord with $C$, using the tangent if $P=Q$ and counting intersection multiplicity. If the third intersection is $R$, define $P+Q=-R$, wherein general Weierstrass equation of an elliptic curve coordinates. The line through $R$ and $O$ gives this reflection. A vertical chord gives $P+(-P)=O$, and the tangent at $O$ meets $C$ three times at $O$, so $O$ is the identity. The construction is symmetric in $P,Q$. It stays in the nonsingular locus: a line through a singular point has intersection multiplicity at least two there, and therefore cannot also contain two smooth intersections counted with multiplicity.

For the nonsingular case, prove associativity by transporting a known abelian group law. The Abel-Jacobi map of a genus-one curveis bijective over an algebraic closure. Indeed, the Riemann-Roch theorem in genus one says that every divisor class of degree one has a unique effective representative consisting of one point: existence follows from $\ell (D)=1$, and uniqueness follows because two distinct representatives would produce a degree-one map to the projective line, impossible for a genus one curve. Subtracting $O$ gives the claimed bijection.

Any line section represents the same divisor class as $3O$, including tangencies. Thus $P+Q+R\sim 3O$ as divisors on an algebraic curve, while the vertical line gives $R+(-R)+O\sim 3O$. HenceAddition in the Picard group is associative, soThe chord-and-tangent group law is defined over $K$, so the group operation restricts to the $K$-rational points.

For completeness, a singular Weierstrass equation of an elliptic curve produces the smooth-locus group of a singular Weierstrass cubic, not an elliptic curve. Over an algebraic closure, its normalization of an algebraic curve is the projective line; deleting the two preimages of a nodal crossing gives the multiplicative algebraic group, while deleting the single preimage of a cusp gives the additive group. For example, on $y^2=x^3$, the coordinate $u=x/y$, with $u(O)=0$, makes the smooth-locus law addition. On $y^2=x^2(x+1)$ in characteristic different from two, put $t=y/x$ and $z=(t+1)/(t-1)$, with $z(O)=1$; the chord relation gives $z(P)z(Q)z(R)=1$, so the group law is multiplication of $z$. A nonsplit node gives the corresponding form of the multiplicative algebraic group over $K$. These descriptions also establish the singular-case group laws.

For a Matrix Lie group $G$, the left Maurer-Cartan form identifies tangent vectors with the Lie algebra by translating them to the identity:It is a Lie algebra-valued differential form of degree one. For a fixed $h\in G$, replacing $g$ by $hg$ gives $(hg{)}^{-1}d(hg)=g^{-1}dg$, so it is a left-invariant differential form.

Differentiate $g^{-1}g=I$ to get $d(g^{-1})=-g^{-1}(dg)g^{-1}$. The exterior derivative then givesHere the exterior product of matrix-valued differential forms includes matrix multiplication; the order of the matrices matters. Hence the Maurer-Cartan equation is

Write $\rho ={\sigma }^{\alpha }{T}_{\alpha }$. Antisymmetry of the exterior product impliesComparison of the Lie algebra components yields the Maurer-Cartan equation in a Lie-algebra basis:Thus the canonical antisymmetric choice is $f^{\gamma }{}_{\alpha \beta }=-\frac{1}{2}{c}^{\gamma }{}_{\alpha \beta }$. The factor one half is required because the sum includes both ordered pairs $(\alpha ,\beta )$ and $(\beta ,\alpha )$. If only $\alpha <\beta$ is summed, its coefficient is $-c^{\gamma }{}_{\alpha \beta }$. Strictly, the equality of differential forms determines only the antisymmetric part of $f$: one may add any tensor symmetric in $\alpha ,\beta$ without changing it. This is the symmetric-part ambiguity in Maurer-Cartan coefficients.

A faithful matrix representation of the orientation-preserving real affine group isIts action on $(x,1{)}^T$ has first component $ax+b$. The group operation and inverse areDifferent transformations have different matrix entries, proving faithfulness. With the Lie algebra basisthe left Maurer-Cartan form isThe requested left-invariant differential forms therefore form the Maurer-Cartan coframe of the real affine group:For a direct invariance check, left translation by $(a_0,b_0)$ sends $(a,b)$ to $(a_{0}a,b_0+a_{0}b)$, and the pullbacks of the two displayed differential forms are unchanged. Their dual left-invariant vector fields are $a{\partial }_a$ and $a{\partial }_b$, whose bracket is $a{\partial }_b$, in agreement with $[D,T]=T$.

A subgroup $H$ of a group $G$ is a normal subgroup when $gH{g}^{-1}=H$ for every $g\in G$, equivalently $gH=Hg$. The quotient group $G/H$ has the left cosets as elements, with $(gH)(kH)=gkH$, identity $H$ and inverse $(gH{)}^{-1}=g^{-1}H$. Normality makes this multiplication independent of representatives.

The first isomorphism theorem for groups says that for a group homomorphism $\phi :G\to K$, its kernel of a group homomorphism is normal andThe image is taken with the induced group operation; surjectivity onto all of $K$ is not required.

For the given upper triangular matrices, take diagonal entries:The target is a direct product of groups, with multiplication of nonzero real numbers in each coordinate. Indeed the product of two such matrices has diagonal $(a{a}^{\prime },d{d}^{\prime })$, proving the group homomorphism property. Any pair of nonzero diagonal entries is attained by a diagonal matrix, so $\phi$ is surjective. Its kernel of a group homomorphism consists precisely of the matrices with $a=d=1$, namely $H$. Hencewith coordinatewise multiplication, not addition on ${\mathbb{R}}^2$.

An elliptic curve has good reduction over $K$ if it admits an integral Weierstrass equation of an elliptic curve with unit elliptic-curve discriminant. Equivalently, the reduction of its Minimal Weierstrass equation is a nonsingular cubic over the residue field $k$. Its smooth integral projective model is a group scheme with the point $O$ as identity.

Choose homogeneous coordinates in the valuation ring ${\mathcal{O}}_K$ with at least one coordinate a unit, and reduce them modulo its maximal ideal ${\mathfrak{m}}_K$. This defines $\mathrm{red}:E(K)\to \tilde{E}(k)$. The chord-and-tangent group law extends over the smooth integral model, so reduction is a group homomorphism. For any point of $\tilde{E}(k)$, choose an affine projective chart containing it. Smoothness means some partial derivative of its equation is nonzero in $k$. Lift the other coordinate arbitrarily and apply the Hensel lemma to the coordinate with unit derivative. This constructs a point of $E(K)$ above the prescribed point, proving the surjectivity of good reduction over a local field.

The kernel of reduction of an elliptic curve consists precisely of points reducing to $O$. The parameter $t=-x/y$ at $O$ gives a bijection between this kernel and ${\mathfrak{m}}_K$: the integral formal-coordinate series solve the equation uniquely near $O$. Addition in these coordinates is the formal group of an elliptic curve, not ordinary addition in ${\mathfrak{m}}_K$. ThusHere the group operation on the last expression is its formal group law.

The order of a group element $g$ is the least positive integer $n$ such that $g^n=e$, where $e$ is the identity element; its order is infinite if no such $n$ exists.

Let $g$ have finite order $n$. A group homomorphism preserves the group operation and the identity, soThe order of an element divides every positive exponent that gives the identity. Hence

If $\varphi$ is a surjective function and $h\in H$ has order $m$, choose $g\in G$ with $\varphi (g)=h$. The first result gives $m\mid n$, where $n=\mathrm{ord}(g)$. The elementthen has order $m$, because the order of $g^k$ is $n/\gcd (n,k)$.

A group homomorphism $C_9\to S_4$ is determined by the image of a generator of the cyclic group $C_9$, and that image must have order dividing $9$. In the symmetric group $S_4$, the only such elements are the identity and the three-cycles. There arethree-cycles. Therefore the number of homomorphisms is

If an elliptic curve has good reduction over a local field $K$, its integral projective model is smooth. Every point over the residue field lifts by the Hensel lemma, applied in a chart with a unit partial derivative. Thus reduction gives an exact sequence $0\to E_1(K)\to E(K)\to \tilde{E}(k)\to 0$. The kernel of reduction of an elliptic curve is the maximal ideal of the valuation ring, with the formal group of an elliptic curve as its group operation.