Let e be the identity element of M; then $ea = a$ for every element $a \in A$ and therefore e belongs to the fixer of A. Let $\alpha$ and $\beta$ be elements of E which stabilize A. Then $(\alpha \beta)A = \alpha(\beta A) \subset \alpha A \subset A$ and therefore the stabilizer of A is a submonoid of M. Similarly for the strict stabilizer and fixer of A, whence (a). If $\alpha A = A$, then $A = \alpha^{-1}(\alpha A) = \alpha^{-1}A$. If for all $a \in A$, $\alpha a = a$, then $a = \alpha^{-1}(\alpha a) = \alpha^{-1}a$, whence (b).

The first assertion follows from the proposition and F is the kernel of the homomorphism of S into $\mathfrak{S}_A$ associated with the operation of S on A.

A group G operates faithfully on a set E if and only if the fixer of E consists of the identity element of G. The fixer of E is the kernel of the given homomorphism of G into $\mathfrak{S}_E$; this homomorphism is injective if and only if its kernel consists of the identity element (§ 4, no. 5, Theorem 3).

Let M be a monoid, E an M-set and $a$ an element of E. The fixer, strict stabilizer and stabilizer of $\{a\}$ are equal; this monoid is called equally the fixer or stabilizer of $a$. The fixer of a subset A of E is the intersection of the fixers of the elements of A. $a$ is called an invariant element of E if the fixer of $a$ is the monoid M. M is said to operate trivially on E if every element of E is invariant.

If $s \in S_x$, then $\alpha s \alpha^{-1}(\alpha x) = \alpha s x = \alpha x$, whence $\alpha S_x \alpha^{-1} \subset S_{\alpha x}$. As $x = \alpha^{-1}(\alpha x)$, $\alpha^{-1} S_{\alpha x} \alpha \subset S_x$, whence $S_{\alpha x} \subset \alpha S_x \alpha^{-1}$.

It is seen similarly that, if A and B are two subsets of E and T is the transporter (resp. strict transporter) of A to B, then the transporter (resp. strict transporter) of $\alpha A$ to $\alpha B$ is equal to $\alpha T \alpha^{-1}$.

If $x, y$ and $z$ are elements of G, then $(xyx^{-1})(xz x^{-1}) = xyz x^{-1}$ and hence Int(x) is an endomorphism of G. For $x$ and $y$ elements of G,

for all $z \in G$, $x(yzy^{-1})x^{-1} = (xy)z(xy)^{-1}$. On the other hand, $\operatorname{Int}(e)$ is the identity mapping of G. The mapping Int is therefore a monoid homomorphism from G to the monoid End(G) of endomorphisms of the group G. As the elements of G are invertible, the mapping Int takes its values in the set Aut(G) of invertible elements of End(G) (§ 2, no. 3). Now $xyx^{-1} = y$ if and only if $x$ and $y$ commute and hence $\operatorname{Int}(x)$ is the identity mapping of G if and only if $x$ is a central element. Finally, let $\alpha$ be an automorphism of G and let $x \in G$; then

For $y \in G$,

Hence $\alpha \cdot \operatorname{Int}(G) \cdot \alpha^{-1} \subset \operatorname{Int}(G)$.

The restriction to H of an automorphism (resp. inner automorphism) of G is an automorphism of H and therefore leaves K invariant.

Let G be a group, A $\subset$ G and $b \in G$. b is said to normalize A if $bAb^{-1} = A$; b is said to centralize A if, for all $a \in A$, $bab^{-1} = a$. Let A and B be subsets of G; B is said to normalize (resp. centralize) A if every element of B normalizes (resp. centralizes) A.

The set of $g \in G$ which normalize (resp. centralize) A is called the normalizer (resp. centralizer) of A (cf. § 1, no. 5, Definition 9); it is often denoted by $N_G(A)$ or simply $N(A)$ (resp. $C_G(A)$ or $C(A)$). It is a subgroup of G. When A is a subgroup of G, $N_G(A)$ may be characterized as the largest subgroup of G which contains A and in which A is normal.

The orbit in $G/H$ of the element $e = H$ is $G/H$, whence the first assertion. We now prove the second. If $n \in N$ defines by right translation the identity mapping on $G/H$, then $e.n = e$, that is $H.n = H$, whence $n \in H$. Therefore

N/H operates faithfully on G/H on the right and $\phi$ is injective. The left operation of G and right operation of N/H on G/H commute and hence the operators of N/H define G-morphisms of G/H into itself, which are necessarily G-automorphisms since they are bijective. Therefore $\phi$ takes its values in the group $\Phi$ of G-automorphisms of G/H. We show that the image of $\phi$ is $\Phi$. Let $f \in \Phi$. By transporting the structure, the stabilizer of $f(\dot{e})$ in G is equal to the stabilizer of $\dot{e}$ and hence to H. Let $n \in G$ be such that $f(\dot{e}) = n\dot{e}$. The stabilizer of $n\dot{e}$ in G is $nHn^{-1}$ (no. 2, Proposition 2), whence $nHn^{-1} = H$ and $n \in N$. For every element $xH$ of G/H, $f(xH) = f(x.\dot{e}) = x.f(\dot{e}) = xnH = xHn$ and $f$ coincides with the mapping defined by $n$.

If f is a solution, then $f(x.K) = x.a$ for all x in G, whence the uniqueness; we show the existence. The orbital mapping defined by a is compatible with the equivalence relation $y \in xK$ on G. For, if $y = xk, k \in K$, then

A mapping f is thus derived of G/K into H which satisfies $f(x.K) = x.a$ for all x in G. This mapping is a G-morphism and $f(K) = a$. This mapping is surjective for its image is a non-empty stable subset of E. Suppose now that K = H and let us show that f is injective. If $f(x.H) = f(y.H)$, then $x.a = y.a$, whence $x^{-1}y.a = a$ and $x^{-1}y \in H$, whence $x.H = y.H$.

As a homogeneous G-set is non-empty, assertion (a) follows from Proposition 6. We show (b). Let $f : G/H \to G/H'$ be a G-set isomorphism. The subgroup H is the stabilizer of H and hence, by transport of structure, the stabilizer of an element of $G/H'$. The subgroups $H$ and $H'$ are therefore conjugate (no. 2, Proposition 2). If $H' = \alpha H \alpha^{-1}$, $H'$ is the stabilizer of the element $\alpha . H$ of $G/H$ (no. 2, Proposition 2) and hence $G/H'$ is isomorphic to $G/H$ (Proposition 6).

Let $\zeta$ and $\zeta'$ be two cycles whose supports are disjoint. If

then $\zeta \zeta'(x) = \zeta' \zeta(x) = x$. If $x$ belongs to the support of $\zeta$, then $\zeta'(x) = x$, and $\zeta(x)$ belongs to the support of $\zeta$, whence $\zeta \zeta'(x) = \zeta' \zeta(x) = \zeta(x)$. Similarly when $x$ belongs to the support of $\zeta'$, then $\zeta' \zeta(x) = \zeta \zeta'(x) = \zeta'(x)$. Hence $\zeta \zeta' = \zeta' \zeta$. Therefore the $\zeta_i$ are pairwise permutable and, for $i \in I$ and $x \in S_i$, $\sigma(x) = \zeta_i(x) \in S_i$. The mappings $\sigma$ and $\zeta_i$ coincide on $S_i$, hence $S_i$ is stable under $\sigma$ and the subgroup of $\mathfrak{S}_{S_i}$ generated by the restriction of $\sigma$ to $S_i$ operates transitively on $S_i$; therefore $S_i$ is a $\bar{\sigma}$-orbit. As the $S_i$ are non-empty and are pairwise disjoint, the mapping $i \mapsto S_i$ is injective. As $\bigcup_i S_i$ is the set of $x$ such that $\sigma(x) \neq x$, every $\bar{\sigma}$-orbit not consisting of a single element is one of the $S_i$.

(a) the supports of the elements of C are pairwise disjoints;

(b) $\sigma = \prod_{\zeta \in C} \zeta$ (the elements of C being pairwise permutable by Lemma 1).

Let $\bar{\sigma}$ be the subgroup generated by $\sigma$ and let S be the set of $\bar{\sigma}$-orbits not consisting of a single element. For $s \in S$, write $\zeta_s(x) = \sigma(x)$ if $x \in s$ and $\zeta_s(x) = x$ if $x \notin s$. For all $s \in S$, $\zeta_s$ is a cycle whose support is $s$ and $\sigma = \prod_{s \in S} \zeta_s$, as is seen by applying the two sides to any element of E. The uniqueness of C follows from Lemma 1.

For every permutation $\sigma$ let $F_\sigma$ be the set of $x \in E$ such that $\sigma(x) = x$. We show by descending induction on $p$ that every permutation $\sigma$ such that $\mathrm{Card}(F_\sigma) = p$ is a product of transpositions. If $p \geq \mathrm{Card}(E)$, the permutation $\sigma$ is the identity mapping of E; it is the product of the empty family of transpositions. If $p < \mathrm{Card}(E)$, suppose that the property is proved for every permutation $\sigma'$ such that $\mathrm{Card}(F_{\sigma'}) > p$. Now $E - F_\sigma \neq \varnothing$; let $x \in E - F_\sigma$ and $y \in \sigma(x)$. Then $y \neq x$ and $y \in E - F_\sigma$. Let $\sigma' = \tau_{x,y} \cdot \sigma$. The set $F_{\sigma'}$ contains $F_\sigma$ and $x$ and hence $\mathrm{Card}(F_{\sigma'}) > \mathrm{Card}(F_\sigma) = p$. By the induction hypothesis $\sigma'$ is a product of transpositions and hence $\sigma = \tau_{x,y} \cdot \sigma'$ is a product of transpositions.

By virtue of Proposition 8, it suffices to show that every transposition $\tau_{p,q}$, $1 \leq p < q \leq n$, belongs to the subgroup H generated by the $\tau_{i, i+1}$, $1 \leq i \leq n-1$. We show this by induction on $q-p$. For $q-p = 1$, it is obvious. If $q-p > 1$, then (formula (4)) $\tau_{p,q} = \tau_{q-1,q} \tau_{p,q-1} \tau_{q-1,q}$. By the induction hypothesis $\tau_{p,q-1} \in H$ and therefore $\tau_{p,q} \in H$.

If $\sigma \in \mathfrak{S}_n$, every ordered pair $(i, j)$ of elements of $\{1, n\}$ such that $i < j$ and $\sigma(i) > \sigma(j)$ is called an inversion of $\sigma$. Let $\nu(\sigma)$ denote the number of inversions of $\sigma$.

Let P be the additive group of mappings from $\mathbf{Z}^n$ to $\mathbf{Z}$. For $f \in P$ and $\sigma \in \mathfrak{S}_n$, let $\sigma f$ be the element of P defined by

(5)

The action of $\mathfrak{S}_n$ on P thus defined is an operation; for all $\sigma, \tau \in \mathfrak{S}_n$ and $f \in P$, $\sigma f = f$ and

Formula (5) shows that $\sigma(-f) = -\sigma f$ for $\sigma \in \mathfrak{S}_n$ and $f \in P$.

Let $p$ be the element of P defined by

$p(1, 2, \ldots, n) = \prod_{i < j} (j - i) \neq 0$ and hence $p \neq 0$. On the other hand, if $\sigma \in \mathfrak{S}_n$, then

Let C be the set of ordered pairs $(i, j)$ such that $1 \leq i \leq n,\ 1 \leq j \leq n,\ i < j$. A permutation $\theta$ is defined on C by setting $\theta(i, j) = (\sigma(i), \sigma(j))$ if $(i, j)$ is not an inversion, $\theta(i, j) = (\sigma(j), \sigma(i))$ if $(i, j)$ is an inversion. This implies $\sigma p = (-1)^{\nu(\sigma)} p$.

The uniqueness follows from Proposition 8. We show the existence. By transporting the structure, it may be assumed that $E = \{1, n\}$. Using the above notation, let $\varepsilon(\sigma) = (-1)^{\nu(\sigma)}$. Then (Lemma 2)

On the other hand,

As $p \neq -p$, it follows that $\varepsilon(\sigma \sigma') = \varepsilon(\sigma) \varepsilon(\sigma')$ and thus $\varepsilon$ is a homomorphism. We now show that, for every transposition $\tau$, $\varepsilon(\tau) = -1$. $\nu(\tau_{n-1, n}) = 1$, whence $\varepsilon(\tau_{n-1, n}) = -1$. As every transposition $\tau$ is conjugate to $\tau_{n-1, n}$ and the group $\{-1, +1\}$ is commutative, $\varepsilon(\tau) = \varepsilon(\tau_{n-1, n}) = -1$.