If $a, b \in \overline{H}$ then $ab^{-1} \in \overline{H}$, because the mapping $(x, y) \to xy^{-1}$ is continuous on $G \times G$ and transforms $H \times H$ into $H$ (Chapter I, § 2, no. 1, Theorem 1). In the same way, the continuity of the mapping $x \to axa^{-1}$ shows that if $H$ is normal then $\overline{H}$ is normal.

In particular, the closure $N$ of the set $\{ e \}$ consisting only of the identity element of $G$, is a normal subgroup of $G$; and $N = \{ e \}$ if and only if $G$ is Hausdorff ($\S$ 1, no. 2, Proposition 2).

By reason of Proposition 1, we may limit ourselves to the case where $H$ is dense in $G$. The continuous functions $xy$ and $yx$ are equal on $H \times H$, and are therefore equal on $G \times G$, by virtue of the principle of extension of identities (Chapter I, § 8, no. 1, Proposition 2, Corollary 1).

For $M'$ is the intersection of the sets $F_m (m \in M)$, where $F_m$ is the set of all $x \in G$ such that $xm = mx$; and $F_m$ is closed (Chapter I, § 8, no. 1, Proposition 2).

By translation, $H$ is locally closed at each of its points, i.e. $H$ is locally closed in $G$. Let $V$ be a symmetric open neighbourhood of $e$ in $G$ such that $V \cap H$ is closed in $V$. If $x \in \overline{H}$, then $xV$ meets $H$; if $y \in xV \cap H$, we have $x \in yV$, and $y(V \cap H) = (yV) \cap H$ is closed in $yV$. But $x$ lies in the closure of $(yV) \cap H$, hence $x \in H$.

If a subgroup $H$ has an interior point, then by translation all the points of $H$ are interior and so $H$ is open. The second assertion of the corollary is a particular case of Proposition 4.

If $H$ is discrete, every point of $H$ is isolated. Conversely if $H$ has an isolated point, then by translation every point of $H$ is isolated and therefore $H$ is discrete. If $H$ is discrete and $G$ is Hausdorff, then there is a neighbourhood $V$ of $e$ such that $V \cap H = \{e\}$; $\{e\}$ is closed in $G$ and therefore a fortiori in $V$, so that $H$ is locally closed at $e$. Hence $H$ is closed in $G$ by Proposition 4.

The converse of this proposition is in general false, as we shall see in Chapter IV (§ 2, no. 5). If a topological group $G$ is generated by each neighbourhood of the identity element, the most we can say is that $G$ contains no open subgroup other than $G$.

\* As an example of a non-connected group $G$ which has an open subgroup distinct from $G$, we may cite the multiplicative group $\mathbf{R}^*$ of non-zero real numbers, in which the subgroup $\mathbf{R}_+^*$ of real numbers $> 0$ is both open and closed (see Chapter IV, § 3, no. 2). \*

If $a \in K$, then $a^{-1}K$ is connected and contains $e$; hence $K^{-1}K \subset K$, which shows that $K$ is a subgroup of $G$. This subgroup is invariant under all automorphisms of $G$, and in particular under all inner automorphisms, therefore $K$ is normal in $G$; also $K$ is closed (Chapter I, , no. 5, Proposition 9). Finally, the left translation $y \to xy$ is a homeomorphism of $G$ which sends $e$ to $x$, and hence the component of $x$ is $x.K$

The component of the identity element $e$ of $G$ is called the identity component of $G$.

For the mapping $(z, x) \to zxz^{-1}$ is continuous on $G \times G$, and maps $H \times K$ into $K$; hence (Chapter I, § 2, no. 1, Theorem 1) it maps $G \times \overline{K} = \overline{H} \times K$ into $\overline{K}$.

Let $V$ be any symmetric neighbourhood of $e$ in $G$. Then $V \cap H$ is a neighbourhood of $e$ in $H$, and hence generates $H$. It follows that $V$ generates a subgroup $H'$ which contains $H$; but $H'$ is open and closed (no. 1, Corollary to Proposition 4), and therefore contains $\overline{H} = G$.

For this mapping is a continuous bijection whose inverse $x \to s^{-1}.x$ is also continuous.

We recall that for each $x \in X$ the set $G.x$ of transforms $s.x$ of $x$ by the elements $s$ of $G$ is called the orbit of $x$ (with respect to the group of operators $G$), and that the set of all $s \in G$ such that $s.x = x$ is a subgroup of $G$ called the stabilizer of $x$. The relation $R(x, y) : "y$ belongs to the orbit of $x"$ is an equivalence relation on $X$, called the equivalence relation defined by $G$; the equivalence classes with respect to this relation are the orbits of the points of $X$. The topological space $X/R$ is called the orbit space of $X$ (with respect to $G$), or the quotient space of $X$ by the group $G$, and is denoted by $X/G$; and the topology of $X/G$ is said to be the quotient of the topology of $X$ by $G$.

For the saturation with respect to $R$ of an open subset $U$ of $X$ is the set $\bigcup_{s \in G} s.U$, and each $s.U$ is open by Lemma 1.

Now let $S$ be an equivalence relation on $X$ such that, for each $s \in G$, the mapping $x \to s.x$ is compatible with $S$ [in other words, such that the relation $x \equiv y \pmod{S}$ implies $s.x \equiv s.y \pmod{S}$]; for the sake of brevity we shall say that the relation $S$ is compatible with the group $G$. If $\psi$ is the canonical mapping of $X$ onto $X/S$, and if $s.\psi(x)$ denotes the class mod $S$ of $s.x$, then $X/S$ has $G$ as groups of operators with respect to the external law $(s, \psi(x)) \to s.\psi(x) = \psi(s.x)$. Moreover:

Since the relation of equality on $G$ and the relation $S$ on $X$ are open, it is enough to show that the mapping $(s, x) \to s.\psi(x) = \psi(s.x)$ of $G \times X$ into $X/S$ is continuous (Chapter I, § 5, no. 3, Corollary to Proposition 8); but this follows from the continuity of $\psi$ and of the mapping

Since the equivalence relation $x^{-1}y \in H$ is open (no. 4, Lemma 2) this is a particular case of Proposition 11 of no. 4.

H is an equivalence class for the relation $x^{-1}y \in H$ and therefore, if $G/H$ is Hausdorff, H is closed in G. Conversely, if H is closed, then the graph of this relation is closed in $G \times G$, since it is the inverse image of H under the continuous mapping $(x, y) \to x^{-1}y$. Since the relation $x^{-1}y \in H$ is open, it follows from Chapter I, § 8, no. 3, Proposition 8 that $G/H$ is Hausdorff.

For the inverse images in G of the points of $G/H$ under the canonical mapping are the cosets $xH$ ($x \in G$); and these sets are open in G if and only if H is open in G.

Let X be a topological space on which a topological group G operates continuously and transitively; X is then (in the algebraic sense) a homogeneous space of G. Let x be a point of X, H_x its stabilizer. The continuous surjection s → s.x of G onto X factorizes canonically as

where f_x is the canonical mapping of G onto the homogeneous space G/H_x, and g_x is the bijection s.H_x → s.x of G/H_x onto X; moreover (Chapter I, § 3, no. 4, Proposition 6) g_x is a continuous mapping. But g_x is not necessarily a homeomorphism of G/H_x onto X (Exercise 29). If g_x is a homeomorphism for each x ∈ X, then we say that X is a topological homogeneous space (of the topological group G); for this to be so it is necessary and sufficient that, for each x ∈ E, the mapping s → s.x of G into X should be open.

Every x ∈ X can be written as x = t.x_0 for some t ∈ G. If V is a neighbourhood of e, then V.x = (Vt).x_0 is a neighbourhood of x, for we can write (Vt).x_0 = t((t^{-1}Vt).x_0), and the assertion follows from the facts that t^{-1}Vt is a neighbourhood of e in G and that y → t.y is a homeomorphism of X onto itself (no. 4, Lemma 1). It follows that if U is any open subset of G and if x is any point of X, then U.x is open in X; for if t ∈ U, then t^{-1}U is a neighbourhood of e, hence (t^{-1}U).x is a neighbourhood of x, and t.((t^{-1}U).x) = U.x is a neighbourhood of t.x. Hence U.x is open in X, so that the mapping s → s.x of G into X is open. This completes the proof.

If x → x̂ is the canonical mapping of G onto G/H, then we have to show that (x̂, ŷ) → x̂ŷ^{-1} is a continuous mapping of (G/H) × (G/H) into G/H. Since the equivalence relation x^{-1}y ∈ H is open (no. 4, Lemma 2), it is enough to show that (x, y) → x̂ŷ^{-1} is a continuous mapping of G × G into G/H (Chapter I, § 5, no. 3, Corollary to Proposition 8, and § 3, no. 4, Proposition 6). But (x, y) → x̂ŷ^{-1} is the composition of the continuous mappings x → x̂ and (x, y) → xy^{-1}, hence is continuous.

Whenever in the sequel we consider a quotient group $G/H$ of a topological group $G$ as a topological group, it is always to be understood that the topology of $G/H$ is the quotient by $H$ of the topology of $G$, unless the contrary is expressly stated.

This is a particular case of Proposition 5 of Chapter I, § 5, no. 3.

Propositions 13 and 14 give in particular, for quotient groups:

If $G$ is a topological group and $N$ is the closure of $\{e\}$ in $G$, then $N$ is a closed normal subgroup of $G$ (no. 1, Proposition 1), hence $G/N$ is Hausdorff; $G/N$ is called the Hausdorff group associated with $G$.

Let $V$ be a neighbourhood of $e$ in $G$ which does not contain any point of $H$ other than $e$, and let $W$ be a symmetric open neighbourhood of $e$ in $G$ such that $W^2 \subset V$. Then the restriction to $W$ of the canonical mapping $\varphi$ of $G$ onto $G/H$ is injective; for if $x, y \in W$ and $\varphi(x) = \varphi(y)$, then $x^{-1}y \in W^2 \subset V$ and $x^{-1}y \in H$, so that $x = y$. By Proposition 17 it follows that the restriction of $\varphi$ to $W$ is a homeomorphism of $W$ onto $\varphi(W)$; since moreover $\varphi(xy) = \varphi(x)\varphi(y)$ for all $x, y \in W$, we conclude that $G$ and $G/H$ are locally isomorphic (§ 1, no. 3, Proposition 3).

This follows from the preceding remarks, and from Proposition 10 of no. 4.

The canonical bijection of $A/(A \cap H)$ onto $\varphi(A)$ is a continuous homomorphism, since it arises from the restriction of $\varphi$ to $A$ by passing to the quotients; but in general the topological groups $A/(A \cap H)$ and $AH/H$ are not isomorphic (cf. § 4, no. 1, Proposition 1, Corollary 3).

\* For example, take $G$ to be the additive group $\mathbf{R}$ of real numbers, take $H$ to be the group $\mathbf{Z}$ of integers, and take $A$ to be the group $\theta \mathbf{Z}$ of integer multiples of an irrational number $\theta$. Then $A \cap H = \{0\}$, so that $A/(A \cap H)$ is a discrete group, isomorphic to $\mathbf{Z}$; on the other hand, $A + H$ is dense in $\mathbf{R}$ (as we shall see in Chapter V, § 1, no. 1, Proposition 1), hence $(A + H)/H$, which is locally isomorphic to $A + H$ (no. 6, Proposition 19), is not a discrete group and consequently is not isomorphic to $A/(A \cap H)$. \*

Nevertheless, we have the following proposition:

Since $H_0 = H \cap G_0$, it is enough to show that if $U_0$ is any open subset of $G_0$ which is saturated with respect to the relation $x^{-1}y \in H_0$, then $U_0$ is the intersection of $G_0$ and an open subset of $G$ which is saturated with respect to the relation $x^{-1}y \in H$ (Chapter I, § 3, no. 6, Proposition 10). Let $U$ be an open subset of $G$ such that $U_0 = U \cap G_0$. Since $U_0 = U_0 H_0$, it is easily seen that $U_0 = U H_0 \cap G_0$; but $U H_0$ is open in $G$, so we may suppose that $U = U H_0$. The set $U H$ is open in $G$ and is saturated with respect to the relation $x^{-1}y \in H$; hence the proposition will be proved if we show that $U H \cap G_0 = U_0$. Now, if $u \in U$ and $h \in H$ are such that $u h \in G_0$, then there is a symmetric neighbourhood $V$ of $e$ in $G$ such that $u V \subset U$; since $V h$ is a neighbourhood of $h$ in $G$, there exists $z \in V$ such that $z h \in H_0$. But then we have $u z^{-1} \in U$, and $u h = (u z^{-1})(z h)$; therefore $U H \cap G_0 \subset U H_0$. But $U H_0 = U$, hence $U H \cap G_0 \subset U \cap G_0 = U_0$. This completes the proof.

Let $G$ be a topological group operating continuously on a topological space $X$, and let $K$ be a normal subgroup of $G$ which is contained in the stabilizer of each point of $X$. For each $x \in X$, the relation $s \equiv t \pmod{K}$ therefore implies $s.x = t.x$, and passing to the quotient we define a mapping $s \to s.x$ of $G/K$ into $X$. It is immediately verified that with respect to the external law $(s', x) \to s'.x$, X has G/K as a group of operators. Moreover, G/K operates continuously on X with respect to this law; for the relation of equality on X and the relation $s \equiv t \pmod{K}$ on G are both open equivalence relations, and therefore the result follows from the continuity of the mapping

of $G \times X$ into X (Chapter I, § 5, no. 3, Corollary to Proposition 8, and § 3, no. 4, Proposition 6).

Now let G be a topological group operating continuously on a topological space X, and let H be any normal subgroup of G; then H operates continuously on X. Let S be the equivalence relation defined by H on X; then S is open (no. 4, Lemma 2). The relation S is compatible with the group G (no. 4); for if $y \equiv x \pmod{S}$ then there exists $t \in H$ such that $y = t.x$; hence for all $s \in G$ we have $s.y = (sts^{-1}).s.x$, and $sts^{-1} \in H$ since H is normal in G; thus $s.y \equiv s.x \pmod{S}$. If $\psi$ is the canonical mapping of X onto X/S, the group G therefore operates continuously on X/S with respect to the external law

(no. 4, Proposition 11). Moreover, the group H is contained in the stabilizer of each of the points of X/S; as we have seen above, G/H operates continuously on $X/S = X/H$ with respect to the external law $(s', \psi(x)) \to \psi(s.x)$. If R denotes the equivalence relation on X defined by G, then the relation S implies R, and the equivalence relation R/S on X/S is that defined by the group G/H. Hence (Chapter I, § 3, no. 4, Proposition 7):

We know already that this bijection is an isomorphism of groups, and Proposition 22 (applied to the group K operating on the right on G) shows that it is a homeomorphism.

Suppose $f$ is continuous at a point $a \in G$; then if $V'$ is any neighbourhood of $f(a)$, $V = \overline{f^{-1}}(V')$ is a neighbourhood of $a$. Hence if $x$ is any point of $G$, we have

and therefore $f$ is continuous at $x$.

A continuous homomorphism of a topological group $G$ into a topological group $G'$ is also called a morphism of $G$ into $G'$ for the topological group structures.

Let $f$ be a continuous homomorphism of a topological group $G$ into a topological group $G'$; the inverse image $H = \overline{f^{-1}}(e')$ of the identity element $e'$ of $G'$ is a normal subgroup of $G$, and $f(G)$ is a subgroup of $G'$. Consider the canonical factorization $f = \psi \circ \dot{f} \circ \varphi$, where $\varphi$ is the canonical mapping $G \to G/H$, $\psi$ is the canonical injection $f(G) \to G'$ and lastly $\dot{f}$ is a continuous bijective homomorphism of the quotient group $G/H$ onto the subgroup $f(G)$ (Chapter I, § 3, no. 5); $\dot{f}$ is said to be the bijective homomorphism associated with $f$. In general, $\dot{f}$ is not an isomorphism of topological groups.

For example, let $G'$ be a non-discrete topological group, and $G$ the topological group obtained by giving $G'$ the discrete topology; then the identity mapping of $G$ into $G'$ is a continuous bijective homomorphism, but is not bicontinuous.

In view of Lemma 2 of no. 4, the equivalence of a) and b) follows immediately from the definitions (Chapter I, § 5, no. 3, Proposition 5). The equivalence of b) and c) is a particular case of Proposition 15 of no. 5, if we observe that $G$ operates continuously on $f(G)$ by the external law $(s, f(t)) \to f(st)$.