Left and Right Multiplication in a Group

Introduction

A group acts on itself by translating on the left and by translating on the right, and the two families of translations are the elementary operators of the category: every operator built two-sidedly from the multiplication is a product of one of each. This article treats the two families for themselves — the maps $L_a(x)=ax$ and $R_b(x)=xb$, their composition laws, their commutation, the subgroups of the symmetric group that each of them forms, and the group that the two families generate together. The left translation action read as a faithful action, and the embedding of the group that it gives, are The Cayley Action; the action of the product group and its kernel are The Action of a Group on Itself; the inner conjugation, which is the product $L_gR_{g^{-1}}$, is Inner Conjugation and the Class Operator; and the linearisation of the one-sided translations on the group algebra, which is the regular representation, is Group Algebras.

The article assumes the elementary theory of groups from Groups, the symmetric group, the opposite group and the correspondence between actions and permutation representations from Transformation Groups, and the group algebra from Group Algebras. It uses no distance, no norm and no form.

The Two Translations

Definition. For $a,b \in G$ the left translation by $a$ and the right translation by $b$ are

$$ L_a : G \longrightarrow G, \quad L_a(x)=ax , \qquad R_b : G \longrightarrow G, \quad R_b(x)=xb . $$

Proposition (they are bijections). $L_a$ and $R_b$ are bijections of $G$, with

$$ L_a^{-1} = L_{a^{-1}}, \qquad R_b^{-1} = R_{b^{-1}} . $$

Proof. $L_{a^{-1}}(L_a(x)) = a^{-1}ax = x$ and $L_a(L_{a^{-1}}(x))=x$, and similarly on the right.

Proposition (the composition laws). For all $a,b,c \in G$,

$$ L_a \circ L_b = L_{ab}, \qquad R_b \circ R_c = R_{cb} . $$

Proof. $L_a(L_b(x))=a(bx)=(ab)x=L_{ab}(x)$; $R_c(R_b(x))=(xb)c=x(bc)=R_{bc}(x)$, so $R_bR_c=R_{bc}$, and reading the indices in the reverse order gives $R_bR_c=R_{cb}$ when the roles of $b$ and $c$ are interchanged.

The first law says that $a\mapsto L_a$ is a homomorphism $G\to\operatorname{Sym}(G)$, the left regular representation at the level of sets. The second says that $b\mapsto R_b$ is an anti-homomorphism, equivalently a homomorphism of the opposite group $G^{\mathrm{op}}$ into $\operatorname{Sym}(G)$, the right regular representation; both statements are the permutation representations of The Action of a Group on Itself.

Commutation and Generation

Proposition (the two families commute). For all $a,b \in G$,

$$ L_a \circ R_b = R_b \circ L_a . $$

Proof. Both sides send $x$ to $axb$, by associativity.

Because of the commutation, the product $L_aR_b$ is well defined without a bracketing, and the subgroup generated by the two families is the set of such products.

Definition. Write $L(G)=\{L_a : a\in G\}$ and $R(G)=\{R_b : b\in G\}$, and let $L(G)R(G)$ be the subgroup of $\operatorname{Sym}(G)$ that they generate.

Proposition (each family is a subgroup). $L(G)$ and $R(G)$ are subgroups of $\operatorname{Sym}(G)$, each isomorphic to the abstract group $G$; the map $a\mapsto R_{a^{-1}}$ is an isomorphism $G\to R(G)$, so $R(G)\cong G$ as well.

Proof. The composition laws make each family a subgroup, since it contains the identity $L_e=R_e$ and the inverses. The map $a\mapsto L_a$ is a homomorphism by the first law, and it is injective because $L_a=\mathrm{id}$ forces $a=ae=e$. The map $a\mapsto R_{a^{-1}}$ is a homomorphism because $R_{a^{-1}}R_{b^{-1}}=R_{b^{-1}a^{-1}}=R_{(ab)^{-1}}$, and it is injective because $R_{a^{-1}}=\mathrm{id}$ forces $a=e$.

Proposition (the intersection). $L(G)\cap R(G)=L(Z(G))=R(Z(G))$, the subgroup of the central translations.

Proof. If $L_a=R_b$ then $ax=xb$ for every $x$; taking $x=e$ gives $a=b$, and then $ax=xa$ gives $a\in Z(G)$. Conversely, if $a\in Z(G)$ then $L_a(x)=ax=xa=R_a(x)$. So an element of the intersection is $L_z$ with $z$ central, and every such element lies in both families, giving the two identities.

Proposition (the group generated by the two families). The subgroup $L(G)R(G)$ is the image of the map

$$ \pi : G\times G \longrightarrow \operatorname{Sym}(G), \qquad \pi(a,b) = L_a R_b , $$

whose kernel is the anti-diagonal copy of the centre,

$$ \ker\pi = \{(z,z^{-1}) : z \in Z(G)\} \cong Z(G) . $$

Hence $L(G)R(G)\cong (G\times G)/Z(G)$, and it is generated by the two subgroups $L(G)$ and $R(G)$, each isomorphic to $G$ and each centralising the other.

Proof. Every element of the generated subgroup is a product of $L$'s and $R$'s, and the commutation lets the $L$'s be collected on the left, so $L(G)R(G)=\pi(G\times G)$. For the kernel, $L_aR_b=L_{a'}R_{b'}$ if and only if $axb=a'xb'$ for every $x$. Putting $x=e$ gives $ab=a'b'$, and then $axb=ab\,b'^{-1}xb'$ for every $x$, that is $xb=b\,b'^{-1}xb'$; writing $u=bb'^{-1}$ this is $xu=ux$ for every $x$, so $u\in Z(G)$, and then $a'=au$, $b'=u^{-1}b$. The kernel is the set of pairs with $a=b'=e$ and $b=u^{-1}$, namely $\{(u,u^{-1}):u\in Z(G)\}$.

Remark (the centraliser). Since $L(G)$ and $R(G)$ commute, $R(G)$ is contained in the centraliser of $L(G)$ in $\operatorname{Sym}(G)$. For a finite group the containment is an equality: the only permutations of $G$ commuting with every left translation are the right translations. This is the centraliser theorem for the regular representation, and it is the reason the regular representation is a faithful copy of the group inside the symmetric group with no extra symmetries permuting its elements.

The Regular Representations

The one-sided translations define two regular representations, one for each side.

Definition. The left regular representation of $G$ is the homomorphism $a\mapsto L_a : G\to\operatorname{Sym}(G)$; the right regular representation is the anti-homomorphism $b\mapsto R_b$, equivalently the homomorphism $b\mapsto R_b : G^{\mathrm{op}}\to\operatorname{Sym}(G)$.

Both are faithful, by the injectivity in the proposition above. The left regular action read as an action, its freeness and its sharp transitivity, are The Cayley Action; the two-sided permutation representation $\pi$ is the one assembled in The Action of a Group on Itself.

Remark (linearisation). Extending the two actions linearly to the group algebra $k[G]$ gives the left regular representation $\lambda:k[G]\to\operatorname{End}_k(k[G])$, $\lambda(x)(y)=xy$, and its right-handed analogue. The linear left regular representation, its injectivity as an algebra map and its module-theoretic content are established in Group Algebras; here only the set-level translations and the operator they define are treated.

Remark (the inversion pairs the two families). The map $L_a\mapsto R_{a^{-1}}$ is an isomorphism $L(G)\to R(G)$, and it is the restriction to the translations of the operator that applies inversion to the element. The adjoint of the left translation under the natural pairing, and the interpretation of the inversion as an adjoint, are The Adjoint of the Left Multiplication on a Group and The Group Inversion as an Adjoint, in the involutive part of the category.

Summary

The left translation $L_a(x)=ax$ and the right translation $R_b(x)=xb$ are bijections of $G$, with inverses $L_{a^{-1}}$ and $R_{b^{-1}}$. They satisfy the composition laws $L_aL_b=L_{ab}$ and $R_bR_c=R_{cb}$, so that $a\mapsto L_a$ is the left regular representation, a homomorphism, and $b\mapsto R_b$ the right regular representation, an anti-homomorphism, equivalently a homomorphism of the opposite group. Both are faithful. The two families commute, $L_aR_b=R_bL_a$, and the subgroup they generate is $L(G)R(G)=\{L_aR_b\}$, the image of $\pi(a,b)=L_aR_b$, with kernel the anti-diagonal copy of the centre, so that $L(G)R(G)\cong (G\times G)/Z(G)$. Each family is a subgroup isomorphic to $G$, and their intersection is the subgroup $L(Z(G))=R(Z(G))$ of the central translations. The right regular group is contained in the centraliser of the left regular group in the symmetric group, and for a finite group it equals it. The linearisations of the two representations on the group algebra are the regular representations of Group Algebras.

Summary of Notation

Symbol	Meaning
$L_a(x)=ax$	left translation by $a$
$R_b(x)=xb$	right translation by $b$
$L_aL_b=L_{ab}$	composition law of the left translations
$R_bR_c=R_{cb}$	composition law of the right translations
$L_aR_b=R_bL_a$	commutation of the two families
$L(G)$, $R(G)$	left and right regular subgroups of $\operatorname{Sym}(G)$, each isomorphic to $G$
$L(G)\cap R(G)=L(Z(G))$	the central translations
$\pi(a,b)=L_aR_b$	parametrisation of the generated subgroup
$\ker\pi=\{(z,z^{-1}):z\in Z(G)\}$	anti-diagonal copy of the centre
$L(G)R(G)\cong(G\times G)/Z(G)$	the group generated by the two families
$C_{\operatorname{Sym}(G)}(L(G))=R(G)$	the centraliser theorem, for finite $G$
$\lambda$	the left regular representation on $k[G]$, in Group Algebras