# 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.

The converse of Lagrange’s Theorem is not true. It is not true that for a finite group G, if d divides G, then there exists a subgroup H ≤ G of order d, e.g., |A4| = 4!/2 = 12, but even though 6 is a divisor of 12, A4 does not have a subgroup of order 6.

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 = akm = (am)k = ek = 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 ℤp.

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 ℤpThere 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 ℤ2p or the dihedral group Dp.

Proof.

If G has an element of order 2p ⇒ G is cyclic, and therefore G is isomorphic to ℤ2p 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 ≋ Dp.

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 ak, 0 ≤ k < p, or bak, 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, ···, ap-1, b, ba, ba2, ···, bap-1. Futhermore, G = ⟨a, b| ap = b2 = e, ab = bap-1⟩ where a-1 = ap-1 (ap = e ⇒ a·ap-1 = e ⇒[Uniqueness of inverses] a-1 = ap-1)

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

• akaj = aj+k (mod p)
• (aj)(bak) = (ajb)ak = ba-jak = bak-j (mod p)
• (baj)ak = baj + k (mod p)
• (baj)(bak) = b(ajb)ak = bba-jak = ak-j (mod p)

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

# 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

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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