Lagrange's Theorem

Numbers are the basic building blocks of reality, and this is how they should be taught: […] Four. Chairs and tables with four legs are a terrible idea. Five. You can have five senses and still make no sense at all, Apocalypse, Anawim, #justtothepoint.

Recall: Let H ≤ G, the index of H in G, written or denoted as [G:H] is the number of left cosets of H in G, that is [G:H] = [G/H], e.g., [ℤ₆:⟨3⟩] = 3 (there are 3 cosets: H = {0, 3}, 1+H, 2+H), and [S₃:A₃] = 2 (there are two left cosets: H = {1, (123), (132)} and (12)H).

Lagrange’s Theorem. Let G be a finite group and let H be a subgroup of G. Then, the order of H is a divisor of the order of G, .i.e., |H| | (divides) |G| or, more precisely, |G| = |H| |[G:H]|.

A group of order 12 may have subgroups of order 12 (the group itself), 6, 4, 3, 2, and 1 (the trivial subgroup). A group of order 120 (=2³·3·5) may have subgroups of order 120, 60, 40, 30, 24, 20, 15, 12, 10, 8, 6, 5, 4, 3, 2, and 1.

Proof.

Let |G| = n and |H| = m. Recall: aH = H ↭ a ∈ H, ∀a, b ∈ G, |aH| = |bH| ⇒ eH = H (H ≤ G ⇒ e ∈ H) and every coset has the same number of elements ⇒ every coset of H also has m elements.

Let r be the number of cells in the partition of G into left cosets of H. Then, |G| = n = r (number of cells in the partition of G) * m (number of elements in each and every partition) = [G:H]·|H|. In particular, |H| | |G|.

Proposition. The group A₄ has no subgroup of order 6.

Proof.

Recall that A₄ is the alternating group on 4 letters, A₄ = {(1), (12)(34), (13)(24), (14)(23), (123), (132), (124), (142), (134), (143), (234), (243)}

For the sake of contradiction, let’s assume that there is a subgroup H ≤ A₄, |H| = 6 ⇒ [By Lagrange, |G| = [G:H]·|H|] [A₄:H] = |G|/|A₄| = 12/6 = 2 ⇒ there are only two cosets of H in A₄ ⇒ As one of the cosets is H itself, then right and left cosets must coincide, that is, ∀g ∈ G, gH = Hg ⇒ [Cosets aH = Ha ↭ H = aHa^-1] gHg^-1 = H ∀g ∈ A₄.

Besides, there are eight 3-cycles in A₄ (|A₄| = 12, |H| = 6), so at least one 3-cycle must be in H. Without lose of generality, we may assume that (123) is in H ⇒ (123)^-1 = (132) ∈ H.

Since ghg^-1 ∈ H ∀g ∈ A₄, h ∈ H, (124)(123)(124)^-1 = (124)(123)(142) = (243) ∈ H ⇒ (243)^-1 = (234) ∈ H
(243)(123)(243)^-1 = (243)(123)(234) = (142) ∈ H ⇒ (142)^-1 = (124) ∈ H ⇒ H = {(1), (123), (132), (243), (234), (142), (124), ···} ⊥ |H| = 6.

Corollary. The order of an element of a finite group divides the order of the group. In particular, a^|G| = e.

Proof: The order of an element a is the order of the cyclic subgroup that it generates, |⟨a⟩| and by Lagrange’s Theorem it divides the order of the group. In particular, |a| = |⟨a⟩| = m, ∃k: km = n = |G|, aⁿ = a^km = (a^m)^k = e^k = e∎

Corollary. Every group of prime order is cyclic. Futhermore, any arbitrary no trivial element of G, g ∈ G: g ≠ e, is a generator of G. Besides, there is only one group structure, up to isomorphism, of any given prime number, namely ℤ_p.

Proof. For any subgroup H in G (H ≤ G), by Lagrange’s Theorem, we have that the order of H divides the order of G, i.e., |H| divides p ⇒ [p prime ⇒ there are only two options, namely 1 and p itself] H = {e} or H = G.

Let a ∈ G, a ≠ e ⇒ ⟨a⟩ is a subgroup of G different from {e} ⇒ ⟨a⟩ = G, G is cyclic and “a” is a generator of G.

Futhermore, since we have previously demonstrated that every cyclic group of order p is isomorphic to ℤ_p ⇒ There is only one group structure, up to isomorphism, of any given prime number.

Example. Let G = U₃₀ = {1, 7, 11, 13, 17, 19, 23, 29} and let H = {1, 11}.

We know that |G| = |H|·|[G:H]| ⇒ |[G:H]| = |G| / |H| = 8 / 2 = 4, i.e., there are four and only four different right (or left) cosets, i.e., G/H = [1H, 7H, 13H, 19H]

1H = {1, 11}. 7H = {7, 77 mod 30} = {7, 17}
We don't need to calculate 11H or 17H, we already know that 1H = 11H and 7H = 17H. 13H = {13, 23}. 19H = {19, 29}.

Corollary. Let H, K ≤ G, G finite group such that K ≤ H ≤ G. Then, [G : K] = [G : H][H : K]

Proof:

[G : K] = [By Lagrange’s theorem] $\frac{|G|}{|K|}=\frac{|G|}{|K|}\frac{|H|}{|H|}=\frac{|G|}{|H|}\frac{|H|}{|K|}=[G : H][H : K]$∎

Classification of Groups of Order 2p

Classification of Groups of Order 2p. Let G be a group, let the order of G be 2p, where p is a prime, p ≥ 2. Then, G is either isomorphic to ℤ_2p or the dihedral group D_p.

Proof.

If G has an element of order 2p ⇒ G is cyclic, and therefore G is isomorphic to ℤ_2p and we are done.

Let’s assume G does not have an element of order 2p. We claim that G is isomorphic to the dihedral group, G ≋ D_p.

By the previous corollary, we know that the order of an element, say d, divides the order of the group. In particular, ∀a ∈ G, a^|G| = e ⇒ [d|2p, p prime ⇒ d = 1, 2 or p]. There are two possibilities: a = e (d = 1) or a ≠ e, |a| = 2 or p, where p is a prime. In other words, any nonidentity element of G must have order 2 or p.

Let’s suppose that any nonidentity element of G have order 2 ⇒ ∀a ∈ G, a ≠ e, a² = e ⇒[a² = e ⇒ a·a = e ⇒ a = a^-1 because of uniqueness of inverses] a = a^-1.

∀a, b ∈ G ⇒ ab ∈ G and ab =[By assumption, any nonidentity element of G have order 2] (ab)^-1 =[The Socks and Shoes Principle] b^-1a^-1 = ba, and therefore, G is Abelian. Then, ∀a, b ∈ G, a ≠ b, a ≠ e, b ≠ e, the set {e, a, b, ab} is closed, so it is a subgroup of G of order 4 (it is closed, it contains the identity element, and is finite), and this clearly contradicts Lagrange’s Theorem ⊥ (4 ≠ 1, 2 or p)

|{e, a, b, ab}| = 4 because a ≠ b, a ≠ e, b ≠ e. If ab = e ⇒[Uniqueness of inverses] b = a^-1 = a, but a ≠ b ⊥

Therefore, there is an element of order p. Let’s be super duper original, and call it a. Let b be any element of G that is not in ⟨a⟩, such an element exists because |G| = 2p. Then, we claim that |b| = 2.

b is not in ⟨a⟩ ⇒ [aH = H ↭ a ∈ H] b⟨a⟩ ≠ ⟨a⟩ and G can be partitioned into ⟨a⟩ and b⟨a⟩, so every element is of the form a^k, 0 ≤ k < p, or ba^k, 0 ≤ k < p. Notice that |[G:H]| = |G|/|H| = 2p/p = 2, so there are only two distinct cosets of ⟨a⟩ into G, namely, ⟨a⟩ and b⟨a⟩ ⇒ b²⟨a⟩ = ⟨a⟩ or b²⟨a⟩ = b⟨a⟩.

If b⟨a⟩ = b²⟨a⟩ ⇒ [aH = bH iff a^-1b ∈ H] b^-1b² ∈ ⟨a⟩ ⇒ b^-1b² =[Associativity] (b^-1b)b = b ∈ ⟨a⟩ ⊥

Therefore, b²⟨a⟩ = ⟨a⟩ ⇒[aH = H ↭ a ∈ H] b² ∈ ⟨a⟩ where |⟨a⟩| = p, prime ⇒[By Lagrange’s Theorem] |b²| = 1 or p. Let’s assume |b²| = p, and [we have started by assuming that G does not have an element of order 2p] |b| ≠ 2p ⇒ |b| = p, so |b| = |b²| = p and they are the same cyclic group ⟨b⟩ = ⟨b²⟩. ⟨b⟩ = ⟨b²⟩ and b² ∈ ⟨a⟩ ⇒ b ∈ ⟨a⟩ ⊥ Therefore, |b| = 2, that is, any element of G (b was taken arbitrarily) that is not in ⟨a⟩ has order 2.

Notice: Another way of proving that |b| = 2 is as follows. ⟨a⟩ is cyclic of order p prime ⇒ its only subgroups are ⟨e⟩ and ⟨a⟩ itself. b ∉ ⟨a⟩, ⟨a⟩∩⟨b⟩ = {e} and |⟨a⟩∩⟨b⟩| = 1 ⇒ [subgroups $|HK|=\frac{|H||K|}{|H∩K|}$] |⟨a⟩⟨b⟩|=|⟨a⟩||⟨b⟩| = p|⟨b⟩|. If ⟨b⟩=p, then |⟨a⟩⟨b⟩| = p² > 2p = |G|, p > 2 ⊥

Next, let us consider ab, ab ∉ ⟨a⟩ ⇒ |ab| = 2 (∀g ∉ ⟨a⟩, |g| = 2) ⇒ (ab) = (ab)^-1 =[The Socks and Shoes Principle] b^-1a^-1 =[|b| = 2] ba^-1.

ab ∉ ⟨a⟩ = H because |⟨a⟩| = p, |b| = 2, b ∉ ⟨a⟩, then we know that G =[[G:H] = |G|/|H| = 2] H ∪ Hb = ⟨a⟩ ∪ ⟨a⟩b.

Observe that this relation completely determines the multiplication table for G, e.g., a³(ba⁴) = a²(ab)a⁴ =[ab = ba^-1] a²(ba^-1)a⁴ = a(ab)a³ =[ab = ba^-1] a(ba^-1)a³ = (ab)a² = ba^-1a² = ba. The multiplication table for all noncyclic groups of order 2p is uniquely determined by the relation (ab) = ba^-1, so all of them must be isomorphic to each other.

More generally, aⁱb ∉ ⟨a⟩, so aⁱb = (aⁱb)^-1 = b^-1a^-i = ba^-i. The elements of G (G = H ∪ bH where H = ⟨a⟩) are e, a, a², ···, a^p-1, b, ba, ba², ···, ba^p-1. Futhermore, G = ⟨a, b| a^p = b² = e, ab = ba^p-1⟩ where a^-1 = a^p-1 (a^p = e ⇒ a·a^p-1 = e ⇒[Uniqueness of inverses] a^-1 = a^p-1)

Notice that aⁱb = ba^-i. In particular, ab = ba^-1 =[a^-1 = a^p-1] ab = ba^p-1.

• a^ka^j = a^{j+k (mod p)}
• (a^j)(ba^k) = (a^jb)a^k = ba^-ja^k = ba^{k-j (mod p)}
• (ba^j)a^k = ba^{j + k (mod p)}
• (ba^j)(ba^k) = b(a^jb)a^k = bba^-ja^k = a^{k-j (mod p)}

D_p, the dihedral group of order 2p is one such a group, so all non-cyclic groups of order 2p are isomorphic to D_p.

Examples

• S₃ is the symmetric group on a set of three elements, that is, the group of all permutations of a three-element set. S₃ = {() (1, 2) (2, 3) (1, 3) (1, 2, 3) (1, 3, 2)}, |S₃| = 6 = 2·3, 3 prime ⇒ G ≋ ℤ₆ or G ≋ D₃.

It is not Abelian (e.g., (12)(13) = (132) ≠ (13)(12) = (123)), and so it cannot be cyclic (all cyclic groups are Abelian), and therefore it is isomorphic to the dihedral group D₃, S₃ ≋ D₃ that is, the dihedral group obtained by composing the six symmetries of an equilateral triangle.

• GL(2, ℤ₂) = {$[\bigl(\begin{smallmatrix}1 & 0\\ 0 & 1\end{smallmatrix}\bigr)], [\bigl(\begin{smallmatrix}1 & 1\\ 1 & 0\end{smallmatrix}\bigr)], [\bigl(\begin{smallmatrix}0 & 1\\ 1 & 1\end{smallmatrix}\bigr)], [\bigl(\begin{smallmatrix}0 & 1\\ 1 & 0\end{smallmatrix}\bigr)], [\bigl(\begin{smallmatrix}1 & 0\\ 1 & 1\end{smallmatrix}\bigr)], [\bigl(\begin{smallmatrix}1 & 1\\ 0 & 1\end{smallmatrix}\bigr)]$} is the general linear group of degree two over ℤ₂. |GL(2, ℤ₂)| = 6 In other words, the collection of 2 x 2 matrices with entries in ℤ₂ which have non-zero determinant.

$(\begin{smallmatrix}0 & 1\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}1 & 1\\ 1 & 0\end{smallmatrix})=(\begin{smallmatrix}1 & 0\\ 1 & 1\end{smallmatrix})≠(\begin{smallmatrix}1 & 1\\ 1 & 0\end{smallmatrix})(\begin{smallmatrix}0 & 1\\ 1 & 0\end{smallmatrix})=(\begin{smallmatrix}1 & 1\\ 0 & 1\end{smallmatrix})$

Since it is not Abelian (hence not cyclic), and |GL(2, ℤ₂)| = 6 = 2·3, 3 prime ⇒ GL(2, ℤ₂) ≋ D₃.

Exercises

• Suppose (G, ·) group, a ∈ G, |a| = 8, |b| = 5. What is the minimum possible order of G?

By Corollary. The order of an element of a finite group divides the order of the group ⇒ 5 | |G| and 9 | |G|. The minimum possible order of G is lcm(5, 9) = 45.

• Suppose (G, ·) group, |G| = 360 with nested groups K ≤ H ≤ G where |K| = 18. What are the possible values of |H|?

Recall the previous corollary. Let H, K ≤ G, G finite group such that K ≤ H ≤ G. Then, [G : K] = [G : H][H : K]

The possible options |H| given that |G| = 360 = 2³·3²·5, |K| = 18 = 2·3² are as follows:

• Let G be a group of order 121. Show that G is cyclic or that g¹¹ = e ∀g ∈ G.

Solution. If G is cyclic ⇒ we are done.

Suppose G is not cyclic ⇒ ∀g ∈ G: |g| | 121 ⇒ ∀g ∈ G: |g| = 1 or 11. Otherwise, |g| = 121 ⇒ G = ⟨g⟩ ⇒ G is cyclic ⊥

If |g| = 1 ⇒ g= e ⇒ g¹¹ = g = e. Therefore, ∀g∈ G: g¹¹ = e∎

• Let G be a group of order pqr, where p, q, and r are distinct primes. If H, K ≤ G, |H| = pq, |K| = qr, then | H ∩ K | = q

Proof

H, K ≤ G ⇒ H ∩ K ≤ H, H ∩ K ≤ K ⇒[By Lagrange] |H ∩ K| | |H | = pq and |H ∩ K| | |K| = qr ⇒ [p, q, and r are distinct primes] |H ∩ K| = 1 or q (we are done).

Let’s suppose for the sake of contradiction |H ∩ K| = 1. We know that |HK| = $\frac{|H||K|}{|H ∩ K|} = |H||K|$ = pq²r > pqr = |G| ⊥ since HK ≤ G

Bibliography

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5. Michael Penn (Abstract Algebra), and MathMajor.
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