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The left regular action of a group on its own underlying set is left multiplication. It is faithful, because an element fixing the identity is itself the identity. For a finite group it gives the permutation representation used in Cayley theorem.

A fixed point of a group action is an element fixed by every group element. The set $A^G=\{a:g\cdot a=a\text{ for all }g\in G\}$ is the categorical limit of the action viewed as a functor from the one-object category associated with $G$ to the Category of sets.

The symmetric group acts on multivariate polynomials by replacing each variable $x_i$ by $x_{\sigma (i)}$. With $(\sigma \tau )(i)=\sigma (\tau (i))$, substitution gives $\sigma \cdot (\tau \cdot f)=(\sigma \tau )\cdot f$. Thus it is a left group action. A polynomial encoding a combinatorial partition has a group orbit encoding all relabelings and a stabiliser subgroup encoding its label symmetries.

If every element of a group squares to the identity, every element equals its inverse. Then $ab=(ab{)}^{-1}=b^{-1}{a}^{-1}=ba$, so the group is an abelian group. The argument does not require a finite group.

The trivial group has exactly one element, its identity. Every group homomorphism to it is trivial, and a group is trivial precisely when all elements in a generating set equal the identity.

An infinite group is a group with infinitely many elements. An element of infinite order proves that its group is infinite, but the converse need not hold: an infinite group can have all elements of finite order.

The group of braids on $n$ strands, composed by stacking, with adjacent-crossing generators ${\sigma }_i$ and the braid group relations. It retains over/under crossings and is not a permutation group.

The Hopf function encodes the departure of a plane-parallel grey atmosphere from a purely linear source profile. With physical bolometric radiative flux $F$ and standard mean intensity $J$, it is defined by $J=3F[\tau +q(\tau )]/(4\pi )$. In thermal radiative equilibrium, $T^4=3T_{\mathrm{eff}}^4[\tau +q(\tau )]/4$. The Eddington closure approximation with its approximate surface condition gives constant $q=2/3$; exact angular transfer generally gives a depth-dependent function.

Consider $I=\alpha [\beta +\tau +\mu +\gamma e^{-\delta \tau }/(1+ϵ\mu )]$ with real $|ϵ|<1$, $\alpha \gamma \ne 0$, and $\delta >0$. Its angular mean includes $A_0=\mathrm{atanh}ϵ/ϵ$. Substitution into $\mu I_{\tau }=I-J$ requires $1-A_0+(\delta -A_0ϵ)\mu =0$. Thus $A_0=1$ and $\delta =ϵ$. For real $ϵ$, $A_0=1$ only at zero, giving $\delta =0$, a contradiction. Independently, its flux contains a varying term proportional to $A_1=(1-A_0)/ϵ$, which is nonzero for nonzero real $ϵ$. Exact conservative transfer therefore needs a more general angular/depth structure than this one smooth exponential correction.

A viscous current with density between those of two ambient fluids spreads along their interface. Vertical buoyancy divides its thickness above and below the undisturbed interface. When its internal shear and in-plane extension are both negligible, resistance from the two ambient Stokes flows controls its spreading.

For steady flow with entrainment coefficient $E$, bottom-drag coefficient $c$ and slope angle $\theta$, a constant buoyancy flux $B$ gives $h=h_0+Ex$, $g^{\prime }=B/(uh)$ and $u^3=B(\sin \theta -E\cos \theta /2)/(c+E)$. The half factor comes from hydrostatic pressure, while the denominator includes both bed drag and acceleration of entrained fluid.

For an isolated cold initially resting razor-thin disk, the displayed pointwise bound makes the absolute gravity-equivalent magnetic surface density smaller than its actual surface density. The positive-kernel comparison of thin-disk fields then makes the integrand in the horizontal virial balance of a cold magnetized fluid negative above and below the disk. Consequently its radial second moment begins to decrease. This establishes initial contraction in the global virial sense, not inward acceleration of every individual fluid element or unlimited subsequent collapse.

For a homogeneous inviscid razor-thin disk approximation with solid-body rotation and barotropic sound speed $c$, the compressive wave branch obeys ${\omega }^2=c^2{k}^2-2\pi G{\Sigma }_0|k|+4{\Omega }^2$. The razor-thin disk Poisson kernel supplies the self-gravity term, and the radial epicyclic frequency is $2|\Omega |$. Growing modes have ${\omega }^2<0$; a separate zero-frequency balanced mode can also exist.

For a free Dirac field with action $\int \bar{\psi }K\psi$ and independent odd sources coupled as $\bar{\psi }J+\bar{J}\psi$, translation invariance of the Berezin integral and completion of the square give $Z[J,\bar{J}]=Z[0,0]\exp (-i\bar{J}{K}_F^{-1}J)$. Here $K_F^{-1}$ has Feynman i-epsilon prescription boundary conditions, and each bilinear includes its spacetime integrations. A left Grassmann derivative with respect to $\bar{J}$, followed by a right Grassmann derivative with respect to $J$, has normalized value $-i{K}_F^{-1}$ at zero sources. Multiplying by $-1$ removes the two insertion factors $i^2$ and yields the Dirac propagator $S_F=i{K}_F^{-1}$. Keeping source order fixed prevents a spurious fermionic sign.

If every vertex subset of size at least $m$ contains an independent set of size $r$, repeatedly colour and remove an $r$-element independent set while at least $m$ vertices remain. Giving each remaining vertex its own colour uses at most $|V|/r+m$ colours.

Two graph triangles sharing a vertex $B$ force their two opposite edges to use the same two colours in a graph colouring with three colours. A palette graph triangle $T,F,B$ and another graph triangle $u,\bar{u},B$ therefore force the literal pair to have opposite Boolean colours. This gadget is used in a reduction from 3-SAT to the three-colourability problem.

Take auxiliary graph triangles $a,b,c$ and $d,t,e$, join $a$ to $d$, and add edges $bx,cy,ez$. When $t$ has colour $T$ and the inputs $x,y,z$ have colours in $\{T,F\}$, this graph admits an extension to a graph colouring with three colours exactly when at least one input is $T$. All-false inputs force $a=d=F$, contradicting their edge. Conversely, use $(a,b,c,d,e)=(T,F,B,F,B)$ if $x=T$, $(T,B,F,F,B)$ if $x=F,y=T$, and $(F,T,B,B,F)$ if $x=y=F,z=T$. The extension property is necessary for both directions of a polynomial-time many-one reduction.

This decision problem asks whether a finite undirected graph has a graph colouring with three colours. It is NP-complete: a colouring certifies membership in NP, while 3-SAT reduces using Boolean-pair colouring gadgets and three-colour clause gadgets.

A graph is $r$-partite when its vertices can be partitioned into $r$ classes, each an independent set. Edges may be missing between different classes. If all possible edges between different classes occur, it is a complete multipartite graph.

A blow-up replaces each vertex of a graph by an independent set, and each original edge by all edges between the corresponding two sets. Replacing every vertex of $K_r$ by $t$ vertices gives the balanced complete multipartite blow-up $K_r(t)$.