Utente:Grasso Luigi/sandbox4/Indice di un sottogruppo

In matematica, in particolare nella teoria di gruppo, l'indice di un sottogruppo H in un gruppo G indica il numero di cosets sinistre di H in G, o in modo equivalente, il numero di cosets destri di H in G. Vengono utilizzati simboli diversi:

|G:H|

o

[G:H]

o

(G:H)

.

Poiché G è l'unione disgiunta dei coset di sinistra e poiché ogni coset di sinistra ha la stessa dimensione di H, l'indice è correlato all'ordine dei due gruppi in base alla formula

|G|=|G:H||H|

(interpretare le quantità come numeri cardinali se qualcuno ha valore infinito).

Così l'indice $|G:H|$ misura la dimensione relativa di G rispetto H.

Per esempio, sia $G=\mathbb {Z}$ il gruppo degli interi con l' operazione +, e sia $H=2\mathbb {Z}$ il sottogruppo degli interi pari. Allora $2\mathbb {Z}$ ha due classi laterali in $\mathbb {Z}$ , l'insieme degli interi pari e quello degli interi dispari, per cui l'indice $|\mathbb {Z} :2\mathbb {Z} |$ vale 2. Generalizzando , $|\mathbb {Z} :n\mathbb {Z} |=n$ per ogni intero positivoany n.

Quando G è un gruppo finito , la relazione diventa $|G:H|=|G|/|H|$ , è ciò implica il teorema di Lagrange che $|H|$ divide $|G|$ .

When G is infinite, $|G:H|$ is a nonzero cardinal number that may be finite or infinite. For example, $|\mathbb {Z} :2\mathbb {Z} |=2$ , but $|\mathbb {R} :\mathbb {Z} |$ is infinite.

If N is a normal subgroup of G, then $|G:N|$ is equal to the order of the quotient group $G/N$ , since the underlying set of $G/N$ is the set of cosets of N in G.

Proprietà[modifica | modifica wikitesto]

If H is a subgroup of G and K is a subgroup of H, then

|G:K|=|G:H|\,|H:K|.

If H and K are subgroups of G, then

|G:H\cap K|\leq |G:H|\,|G:K|,

with equality if

HK=G

. (If

|G:H\cap K|

is finite, then equality holds if and only if

HK=G

.)

Equivalently, if H and K are subgroups of G, then

|H:H\cap K|\leq |G:K|,

with equality if

HK=G

. (If

|H:H\cap K|

is finite, then equality holds if and only if

HK=G

.)

If G and H are groups and $\varphi \colon G\to H$ is a homomorphism, then the index of the kernel of $\varphi$ in G is equal to the order of the image:

|G:\operatorname {ker} \;\varphi |=|\operatorname {im} \;\varphi |.

Let G be a group acting on a set X, and let x ∈ X. Then the cardinality of the orbit of x under G is equal to the index of the stabilizer of x:

|Gx|=|G:G_{x}|.\!

This is known as the orbit-stabilizer theorem.

As a special case of the orbit-stabilizer theorem, the number of conjugates $gxg^{-1}$ of an element $x\in G$ is equal to the index of the centralizer of x in G.
Similarly, the number of conjugates $gHg^{-1}$ of a subgroup H in G is equal to the index of the normalizer of H in G.
If H is a subgroup of G, the index of the normal core of H satisfies the following inequality:

|G:\operatorname {Core} (H)|\leq |G:H|!

where ! denotes the factorial function; this is discussed further below.

As a corollary, if the index of H in G is 2, or for a finite group the lowest prime p that divides the order of G, then H is normal, as the index of its core must also be p, and thus H equals its core, i.e., it is normal.
Note that a subgroup of lowest prime index may not exist, such as in any simple group of non-prime order, or more generally any perfect group.

Esempi[modifica | modifica wikitesto]

The alternating group $A_{n}$ has index 2 in the symmetric group $S_{n},$ and thus is normal.
The special orthogonal group $\operatorname {SO} (n)$ has index 2 in the orthogonal group $\operatorname {O} (n)$ , and thus is normal.
The free abelian group $\mathbb {Z} \oplus \mathbb {Z}$ has three subgroups of index 2, namely

\{(x,y)\mid x{\text{ is even}}\},\quad \{(x,y)\mid y{\text{ is even}}\},\quad {\text{and}}\quad \{(x,y)\mid x+y{\text{ is even}}\}

.

More generally, if p is prime then $\mathbb {Z} ^{n}$ has $(p^{n}-1)/(p-1)$ subgroups of index p, corresponding to the $(p^{n}-1)$ nontrivial homomorphisms $\mathbb {Z} ^{n}\to \mathbb {Z} /p\mathbb {Z}$ .^{[senza fonte]}
Similarly, the free group $F_{n}$ has $(p^{n}-1)/(p-1)$ subgroups of index p.
The infinite dihedral group has a cyclic subgroup of index 2, which is necessarily normal.

Indice infinito[modifica | modifica wikitesto]

If H has an infinite number of cosets in G, then the index of H in G is said to be infinite. In this case, the index $|G:H|$ is actually a cardinal number. For example, the index of H in G may be countable or uncountable, depending on whether H has a countable number of cosets in G. Note that the index of H is at most the order of G, which is realized for the trivial subgroup, or in fact any subgroup H of infinite cardinality less than that of G.

Indice finito[modifica | modifica wikitesto]

A subgroup H of finite index in a group G (finite or infinite) always contains a normal subgroup N (of G), also of finite index. In fact, if H has index n, then the index of N will be some divisor of n! and a multiple of n; indeed, N can be taken to be the kernel of the natural homomorphism from G to the permutation group of the left (or right) cosets of H. Let us explain this in more detail, using right cosets:

The elements of G that leave all cosets the same form a group. Template:Collapse top If Hca ⊂ Hc ∀ c ∈ G and likewise Hcb ⊂ Hc ∀ c ∈ G, then Hcab ⊂ Hc ∀ c ∈ G. If h₁ca = h₂c for all c ∈ G (with h₁, h₂ ∈ H) then h₂ca⁻¹ = h₁c, so Hca⁻¹ ⊂ Hc. Template:Collapse bottom

Let us call this group A. Let B be the set of elements of G which perform a given permutation on the cosets of H. Then B is a right coset of A.

Template:Collapse top First let us show that if bTemplate:Sub∈B, then any other element bTemplate:Sub of B equals abTemplate:Sub for some a∈A. Assume that multiplying the coset Hc on the right by elements of B gives elements of the coset Hd. If cb₁ = d and cb₂ = hd, then cb₂b₁⁻¹ = hc ∈ Hc, or in other words bTemplate:Sub=abTemplate:Sub for some a∈A, as desired. Now we show that for any b∈B and a∈A, ab will be an element of B. This is because the coset Hc is the same as Hca, so Hcb = Hcab. Since this is true for any c (that is, for any coset), it shows that multiplying on the right by ab makes the same permutation of cosets as multiplying by b, and therefore ab∈B. Template:Collapse bottom

What we have said so far applies whether the index of H is finite or infinte. Now assume that it is the finite number n. Since the number of possible permutations of cosets is finite, namely n!, then there can only be a finite number of sets like B. (If G is infinite, then all such sets are therefore infinite.) The set of these sets forms a group isomorphic to a subset of the group of permutations, so the number of these sets must divide n!. Furthermore, it must be a multiple of n because each coset of H contains the same number of cosets of A. Finally, if for some c ∈ G and a ∈ A we have ca = xc, then for any d ∈ G dca = dxc, but also dca = hdc for some h ∈ H (by the definition of A), so hd = dx. Since this is true for any d, x must be a member of A, so ca = xc implies that cacTemplate:Sup ∈ A and therefore A is a normal subgroup.

The index of the normal subgroup not only has to be a divisor of n!, but must satisfy other criteria as well. Since the normal subgroup is a subgroup of H, its index in G must be n times its index inside H. Its index in G must also correspond to a subgroup of the symmetric group STemplate:Sub, the group of permutations of n objects. So for example if n is 5, the index cannot be 15 even though this divides 5!, because there is no subgroup of order 15 in STemplate:Sub.

In the case of n = 2 this gives the rather obvious result that a subgroup H of index 2 is a normal subgroup, because the normal subgroup of H must have index 2 in G and therefore be identical to H. (We can arrive at this fact also by noting that all the elements of G that are not in H constitute the right coset of H and also the left coset, so the two are identical.) More generally, a subgroup of index p where p is the smallest prime factor of the order of G (if G is finite) is necessarily normal, as the index of N divides p! and thus must equal p, having no other prime factors. For example, the subgroup ZTemplate:Sub of the non-abelian group of order 21 is normal (see List of small non-abelian groups and Frobenius group#Examples).

An alternative proof of the result that a subgroup of index lowest prime p is normal, and other properties of subgroups of prime index are given in Template:Harv.

Esempi[modifica | modifica wikitesto]

The group O of chiral octahedral symmetry has 24 elements. It has a dihedral D₄ subgroup (in fact it has three such) of order 8, and thus of index 3 in O, which we shall call H. This dihedral group has a 4-member D₂ subgroup, which we may call A. Multiplying on the right any element of a right coset of H by an element of A gives a member of the same coset of H (Hca = Hc). A is normal in O. There are six cosets of A, corresponding to the six elements of the symmetric group S₃. All elements from any particular coset of A perform the same permutation of the cosets of H.

On the other hand, the group T_h of pyritohedral symmetry also has 24 members and a subgroup of index 3 (this time it is a D_2h prismatic symmetry group, see point groups in three dimensions), but in this case the whole subgroup is a normal subgroup. All members of a particular coset carry out the same permutation of these cosets, but in this case they represent only the 3-element alternating group in the 6-member S₃ symmetric group.

Sottogruppi normali dell'indice numero principale[modifica | modifica wikitesto]

Normal subgroups of numero principale index are kernels of surjective maps to p-groups and have interesting structure, as described at Focal subgroup theorem: Subgroups and elaborated at focal subgroup theorem.

There are three important normal subgroups of prime power index, each being the smallest normal subgroup in a certain class:

E^p(G) is the intersection of all index p normal subgroups; G/E^p(G) is an elementary abelian group, and is the largest elementary abelian p-group onto which G surjects.
A^p(G) is the intersection of all normal subgroups K such that G/K is an abelian p-group (i.e., K is an index $p^{k}$ normal subgroup that contains the derived group $[G,G]$ ): G/A^p(G) is the largest abelian p-group (not necessarily elementary) onto which G surjects.
O^p(G) is the intersection of all normal subgroups K of G such that G/K is a (possibly non-abelian) p-group (i.e., K is an index $p^{k}$ normal subgroup): G/O^p(G) is the largest p-group (not necessarily abelian) onto which G surjects. O^p(G) is also known as the p-residual subgroup.

As these are weaker conditions on the groups K, one obtains the containments

\mathbf {E} ^{p}(G)\supseteq \mathbf {A} ^{p}(G)\supseteq \mathbf {O} ^{p}(G).

These groups have important connections to the Sylow subgroups and the transfer homomorphism, as discussed there.

Struttura geometrica[modifica | modifica wikitesto]

An elementary observation is that one cannot have exactly 2 subgroups of index 2, as the complement of their symmetric difference yields a third. This is a simple corollary of the above discussion (namely the projectivization of the vector space structure of the elementary abelian group

G/\mathbf {E} ^{p}(G)\cong (\mathbf {Z} /p)^{k}

,

and further, G does not act on this geometry, nor does it reflect any of the non-abelian structure (in both cases because the quotient is abelian).

However, it is an elementary result, which can be seen concretely as follows: the set of normal subgroups of a given index p form a projective space, namely the projective space

\mathbf {P} (\operatorname {Hom} (G,\mathbf {Z} /p)).

In detail, the space of homomorphisms from G to the (cyclic) group of order p, $\operatorname {Hom} (G,\mathbf {Z} /p),$ is a vector space over the finite field $\mathbf {F} _{p}=\mathbf {Z} /p.$ A non-trivial such map has as kernel a normal subgroup of index p, and multiplying the map by an element of $(\mathbf {Z} /p)^{\times }$ (a non-zero number mod p) does not change the kernel; thus one obtains a map from

\mathbf {P} (\operatorname {Hom} (G,\mathbf {Z} /p)):=(\operatorname {Hom} (G,\mathbf {Z} /p))\setminus \{0\})/(\mathbf {Z} /p)^{\times }

to normal index p subgroups. Conversely, a normal subgroup of index p determines a non-trivial map to $\mathbf {Z} /p$ up to a choice of "which coset maps to $1\in \mathbf {Z} /p,$ which shows that this map is a bijection.

As a consequence, the number of normal subgroups of index p is

(p^{k+1}-1)/(p-1)=1+p+\cdots +p^{k}

for some k; $k=-1$ corresponds to no normal subgroups of index p. Further, given two distinct normal subgroups of index p, one obtains a projective line consisting of $p+1$ such subgroups.

For $p=2,$ the symmetric difference of two distinct index 2 subgroups (which are necessarily normal) gives the third point on the projective line containing these subgroups, and a group must contain $0,1,3,7,15,\ldots$ index 2 subgroups – it cannot contain exactly 2 or 4 index 2 subgroups, for instance.