Lagrange's Thereom

Proposition 7.8.1 has shown that all cosets have the same cardinality; This result is particularly interesting in the case of finite groups.

7.9.1 Definition. Let H be a subgroup of G. The number of distinct left cosets of H in G is called the index of H in G and is denoted by [G: H].

7.9.2 Lagrange's Theorem. Let H a subgroup of the finite group G, then the order of H divides the order G. In particular,

|G| = |H| ⋅ [G : H]

Proof. We know that G is partitioned into left cosets of H and from Proposition 7.8.1, we known they all share the same cardinality |H|.   □

We continue to derive consequences of Lagrange's Theorem.

7.9.3 Corollary. Every group of prime order is cyclic and isomorphic to ℤp.

Proof. Let G be a finite group of prime order p. Let a ∈ G different from the identity, and let H = ⟨a⟩ the cyclic subgroup generated by a. From Lagrange's theorem, the order of H divides the order of G. Hence |H| = 1, or p, since p is a prime. If |H| = 1, then a = e, contradicting that a ≠ e. Therefore, |H| = p and H = G and G is cyclic.   □

The argument above show that a finite group of prime order can have no nontrivial proper subgroups.

We have studied that in a finite group every element g has necessarily finite order: we indicate it as o(g). Another consequence of Lagrange's theorem is the following corollary:

7.9.4 Corollary. Let G be finite group. Then the order o(g) of every g ∈ G divides the |G|.

Proof. It suffices to observe that o(g) = |⟨g⟩|.   □

7.9.5 Corollary. If G is finite and |G| = n, then

gn = e,   ∀g ∈ G.

Proof. The order of g in G is the size (order) of the subgroup ⟨g⟩, which is a divisor of |G|. So gd = e for some natural d that divides |G|. So g|G| = (gd)(|G|/d) = g|G| = e  □

As a consequence of the last corollary we find Euler's theorem

7.9.6 Euler's Theorem. Let φ be Euler's function, if (a,n) = 1, then

aφ(n) ≡ 1  (mod n).

Proof. Let Un the set of invertible elements in ℤn, that is the set whose elements are the classes [a] such that (a,n) = 1. Un is a group such that |Un| = φ(n). Thus

[aφ(n)] = [a]φ(n) = [1]

that is aφ(n) ≡ 1  (mod n).  □

7.9.7 Theorem. The product HK of two subgroups H, K of a group G is a itself a subgroup if and only if H and K commute, that is, if and only if HK = KH.

Proof. The statement HK = KH does not demand that this is so elementwise. In other words, it is not required that hk = kh for all h ∈ H and all k ∈ K; all that is required is that for any h ∈ H and k ∈ K, hk = h1k1, for some elements h1 ∈ H and k1 ∈ K and similarly for kh.

For all subgroups H and K, the set K contains 1, because 1 = 1 ⋅ 1. Also if h ∈ H and k ∈ K then (hk)−1 = k−1h−1 ∈ KH = HK, and hence HK is closed under inverses Now suppose that HK is a subgroup of G. If x ∈ HK, then x−1 ∈ HK. By the preceding observation x = (x−1)−1 ∈ KH. Thus HK ⊆ KH. On the other hand, if x ∈ KH then x−1 ∈ HK, and since the latter is a subgroup, x ∈ HK too. Hence KH ⊆ HK. Therefore HK = KH.

For the converse, assume HK = KH, to show that HK is a subgroup of G. We need to show closure and the existence of inverses. Assume that x,y ∈ HK ⇒ x = h1k1, and y = h2k2. Thus

xy = (h1k1) (h2k2) = h1(k1h2)k2 = h1(h'k')k2 = (h1h')(k'k2) ∈ HK

Thus closure is proved, and we have to show the existence of inverses. Let x ∈ HK. Then x = hk for h ∈ H and k ∈ K. This implies

x−1 = (hk)−1 = k−1h−1 = h'k' ∈ HK.

where we used the fact that any product of the form k−1h−1 can be written in the form h'k' since HK = HK.  ■

Every non-trivial group has at least two subgroups, itself and the trivial subgroup: which groups have these two subgroups and no more? The question is easily answered using (4.1.7) and Lagrange’s Theorem.

7.9.8 Theorem. A group G has just two subgroups if and only if G ≃ ℤp for some prime p.

Proof. Assume that G has only the two subgroups {e} and G. Let e ≠ x ∈ G; then e ≠ ⟨x⟩ ≤ G, so G = ⟨x⟩ and G is cyclic. Now G cannot be infinite; for then it would have infinitely many subgroups by (Proposition 7.2.9). Thus G has finite order say n. Now if n is not a prime, it has a divisor d where 1 < d < n, and ⟨xn/d⟩ is a subgroup of order d, which is impossible. Therefore G has prime order p and G ≃ ℤp by (Corollary 7.9.3). Conversely, if G ≃ ℤp, then |G| = p and Lagrange’s Theorem shows that G has no non-trivial proper subgroups.  □

Exercises

  1. Let σ = (1 3 4) and τ = (1 3) two permutations of 𝓢4; Determine the subgroup H generated by σ and τ, the right and left cosets.

  2. Let G the group of rotations of the plane that leave a point O fixed. Let Φ the rotation of π radians, consider the subgroup H generated by Φ. Determine left and right cosets.

  3. Determine the subgroups of D4 and study respect to them left and right cosets.

Solutions

  1. σ2 = (134)(134) = (143); στ = (14); τσ = (34). Thus H = {e, (134), (143), (13), (34), (14)}. We want to partition 𝓢4 into cosets.

    Left cosets

    eH = {e, (134), (143), (13), (34), (14)} = H.

    (12)H = {(12), (1342), (1432), (132), (12)(34), (142)}

    (23)H = {(23), (1234), (1423), (123), (234), (14)(23)}.

    (24)H = {(24), (1324), (1243), (13)(24), (243), (124)}.

    Right cosets

    He = {e, (134), (143), (13), (34), (14)} = H.

    H(12) = {(12),(1234), (1243), (123), (34)(12), (124)}

    H(23) = {(23), (1324), (1432), (132), (243), (134))}

    H(24) = {(24), (1342), (1423), (13)(24), (234), (142))}

  2. G is an abelian group. We have π2 = id, so H contains only two elements id and π. So we have Hx = {pi°x,x} with x any angle in radians. We've that Hx = xH.

  3. The subset of D4 which form subgroups of D4 are:

    {e}
    {e, r, r2, r3}
    {e, r2}
    {e, s}
    {e, sr}
    {e, sr2}
    {e, sr3}
    {e, r2, s, sr2}
    {e,r2, sr, sr3}

    To see why these form the subgroups of D4 just look at Cayley table for D4

«Cosets Index Group Isomorphism»