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The evaluation map is the counit representing application in an exponential object. For every $Z$, composition with evaluation gives the natural bijection $\mathrm{Hom}(Z,Y^X)\cong \mathrm{Hom}(Z\times X,Y)$. Its inverse is currying. Evaluation must be a morphism in the ambient category, such as an equivariant map in a category of group actions.

For a group, evaluation at the identity identifies the monoid exponential with all set functions $X\to Y$. Its inverse is $e_f(g,x)=f(x{g}^{-1})g$, and the resulting function action is the displayed formula. Fixed functions are precisely equivariant maps. The inverse on the input is necessary for equivariance of evaluation and for the right-action law.

Give $M\times X$ the diagonal right monoid action. On its equivariant maps into $Y$, define $(ek)(g,x)=e(kg,x)$. The evaluation map of an exponential object is $e(1,x)$ and currying of $h:Z\times X\to Y$ is $\hat{h}(z)(g,x)=h(zg,x)$. Both maps are equivariant, and they are inverse under evaluation. The left multiplication $kg$ in this right action is essential in a noncommutative monoid.

Take the symmetric group $G=S_3$ and the subgroup $H=⟨(12)⟩=\{e,(12)\}$ of group order two. With the rightmost permutation applied first, its three left cosets areThe subscripts record the image of $3$: every member of $C_i$ sends $3$ to $i$. Because $H$ is the stabilizer subgroup of $3$ in the usual group action, $aH\mapsto a(3)$ is a bijection onto $\{1,2,3\}$. It is an equivariant map, since $(ga)(3)=g(a(3))$. The coset action therefore becomes the familiar group action on three letters. In particular,

The three stabilizer subgroups areTheir intersection is $\{e\}$, so the subgroup core and the group kernel are trivial. This proves that the coset action is a faithful group action even though every point has a nontrivial stabilizer subgroup. The orbit-stabilizer theorem reads $6=3\cdot 2$, while the induced group homomorphism gives six distinct permutations of the three cosets. For example, $(12)$ and $e$ agree at $C_3$, but disagree at $C_1$; inspecting one label would miss their difference.

The subgroup need not be a normal subgroup for the coset action to exist: $(123)H(123{)}^{-1}=\{e,(23)\}\ne H$. Here the group action is informative because the three conjugate stabilizer subgroups have no common nonidentity element. For the normal subgroup $A_3$ instead, nonfaithful coset action of the symmetric group on three letters shows that only the sign of a permutation survives.

For a linear group representation, the common fixed vectors of a subgroup form an invariant subspace for any equivariant map restricted to it. Its dimension is central to symmetry-based branching theorems. It need not be invariant under the entire group; conjugate subgroups have conjugate fixed-point subspaces.

An intertwining operator between representations $\rho$ and $\sigma$ is a linear equivariant map $T$ satisfying $T\rho (g)=\sigma (g)T$ for every group element $g$.

Let $M\times X$ have the diagonal right monoid action $(g,x)m=(gm,xm)$, and put $E={\mathrm{Hom}}_M(M\times X,Y)$. Define the right action on this set of equivariant maps of monoid sets byIt really stays in $E$: $(ek)(gm,xm)=e(kgm,xm)=e(kg,x)m$. Also $e1=e$ and $((ek)l)(g,x)=e(klg,x)=(e(kl))(g,x)$, so it satisfies the right-action law. Left multiplication in the first argument is intentional; no commutativity of $M$ has been assumed.

Define the evaluation map of an exponential objectIt is equivariant, becauseFor any right $M$-set $Z$ and equivariant map $h:Z\times X\to Y$, define its currying byFor fixed $z$, this is equivariant in the diagonal variables: $h(zgm,xm)=h(zg,x)m$. The map $z\mapsto \hat{h}(z)$ is also equivariant, sinceIt satisfies $\mathrm{ev}(\hat{h}(z),x)=h(z,x)$.

Conversely, given an equivariant $u:Z\to E$, set $h(z,x)=u(z)(1,x)$. Evaluation makes this equivariant, and currying recovers $u$:These constructions are inverse and natural in $Z$. Therefore they establish the exponential object universal propertyThis is the exponential of right monoid actions, not the set of ordinary maps $X\to Y$ with an arbitrarily guessed action.

When $M=G$ is a group, evaluation at its identity gives a bijection from the previous equivariant-map set to all functions $X\to Y$. For an arbitrary function $f$, the inverse construction isIndeed,so $e_f$ is an equivariant map. Conversely, if $e$ is equivariant, $(g,x)=(1,x{g}^{-1})g$ implies $e(g,x)=e(1,x{g}^{-1})g$, so $e=e_f$ for $f(x)=e(1,x)$.

Transport the right action from part (a) through this bijection. At $x$ its value isThus the exponential of right group actions isFor clarity, the right-action law holds even in a nonabelian group:Evaluation is equivariant because $(fk)(xk)=f(x)k$. The underlying functions need not be equivariant; the fixed points of this exponential action are exactly the equivariant functions.

Fix $\alpha \in \Omega$. If $\beta =g\alpha$, then normality of $H$ givesThus the transitive group action of $G$ permutes the orbits of a group action of $H$, so all $H$-orbits have the same cardinality $d$. They partition the prime-sized set $\Omega$, hence $d$ divides $|\Omega |$. The only possibilities are $d=1$ and $d=|\Omega |$. The first would make every element of $H$ fix every point, contrary to the assumption that $H$ acts nontrivially. Therefore $d=|\Omega |$ and

Now suppose $H\cap G_{\alpha }=\{1\}$. Define the orbit mapIt is surjective by transitivity of $H$. If $\theta (h_1)=\theta (h_2)$, then $h_2^{-1}{h}_1\in H\cap G_{\alpha }$, so $h_1=h_2$; hence $\theta$ is bijective. For $k\in G_{\alpha }$,Thus $\theta$ is an equivariant map: conjugation by $G_{\alpha }$ on $H$ corresponds exactly to the given action of $G_{\alpha }$ on $\Omega$.

A group action on a nonempty set $X$ is simply transitive when exactly one group element sends any chosen point to any other:Equivalently it is both a transitive group action and a free group action; this is also a regular group action. Transitivity supplies a transport, and freeness makes it unique. Conversely, putting $y=x$ in the uniqueness condition shows that every stabilizer subgroup is $\{e\}$, since the identity element already fixes $x$. In a transitive group action, checking that just one stabilizer subgroup is trivial suffices: all the others are conjugate subgroups.

The cyclic group $C_3$ in the diagram under transitive group action is simply transitive. More generally, the left regular action of any group $G$ on itself is simply transitive: the equation $gx=y$ has the unique solution $g=y{x}^{-1}$, using the inverse element of $x$. This shows why the property is useful: every position has a unique label relative to a chosen starting position.

Choose a point $x_0\in X$. Its orbit mapis a bijection and an equivariant map from the left regular action: ${\varphi }_{x_0}(hg)=h\cdot {\varphi }_{x_0}(g)$. This identifies $X$ with a copy of $G$, but $X$ does not come with a preferred identity element. If $x_1=a\cdot x_0$, then ${\varphi }_{x_1}(g)={\varphi }_{x_0}(ga)$: changing the chosen point composes this identification with right multiplication by $a$. An old label $g$ therefore becomes the new label $g{a}^{-1}$, which also specifies the side of multiplication for a noncommutative group.

The transport from $x$ to $y$ remains independent of the chosen point. Writing $x=b\cdot x_0$ and $y=c\cdot x_0$, it is $c{b}^{-1}$. With $x_1=a\cdot x_0$, the new labels are $b{a}^{-1}$ and $c{a}^{-1}$, andFor a finite group, simple transitivity implies $|X|=|G|$. Conversely, a transitive group action with $|X|=|G|<\infty$ is simply transitive by the orbit-stabilizer theorem. Equality of cardinalities alone is insufficient without transitivity: the cyclic group $C_2$ can fix both points of a two-point set, giving two group orbits and two possible self-transports at each point.