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Subgroups, Cosets and Lagrange's Theorem

Source lecture(s): PHY622 Ch. 6

Intuition

Lagrange's theorem is the first genuinely powerful result in group theory, and it is startlingly restrictive: the order of a subgroup must divide the order of the group. No dynamics, no computation — just counting. A group of order 12 simply cannot contain a subgroup of order 5, and you know that before examining a single element.

Cosets

Given a subgroup \(H \le G\) and \(g \in G\), the left coset is

\[gH = \{gh : h\in H\}\]

Three facts do all the work:

  1. Every coset has \(|H|\) elements — the map \(h\mapsto gh\) is a bijection.
  2. Two cosets are either identical or disjoint. If \(g_1H \cap g_2H \neq \emptyset\) then \(g_1H = g_2H\).
  3. The cosets cover \(G\) — every \(g\) lies in \(gH\), since \(e \in H\).

So the cosets partition \(G\) into equal-sized blocks, and immediately:

\[|G| = [G:H]\cdot|H| \qquad\Longrightarrow\qquad |H| \ \big|\ |G|\]

with \([G:H]\) the index, the number of cosets. That is Lagrange's theorem, and the proof is just "equal-sized blocks tile the set".

Immediate corollaries

  • The order of any element divides \(|G|\). The powers of \(g\) form a cyclic subgroup of order equal to the order of \(g\).
  • Every group of prime order is cyclic, and has no proper subgroups. There is essentially only one group of order 7.
  • \(g^{|G|} = e\) for all \(g\). Specialised to modular arithmetic this is Fermat's little theorem — the same counting argument underlies RSA.

Normal subgroups and quotients

Left and right cosets need not agree. When they do, \(gH = Hg\) for all \(g\), the subgroup is normal (\(H \trianglelefteq G\)), and something valuable happens: the cosets themselves form a group, the quotient group \(G/H\), under \((g_1H)(g_2H) = g_1g_2H\). This is well defined only for normal subgroups.

Normal subgroups are exactly the kernels of homomorphisms, which is why they matter to physics:

  • \(SU(2) \to SO(3)\) has kernel \(\{\pm I\} \cong \mathbb{Z}_2\), so \(SO(3) \cong SU(2)/\mathbb{Z}_2\). That \(\mathbb{Z}_2\) is the whole content of the 720° rotation of a spinor — the double cover is a quotient by a normal subgroup.
  • Gauge theories are built on quotients: the physical configuration space is field space modulo the gauge group.
  • Crystallography classifies space groups by their translation subgroups, which are normal.

Worked example: \(S_3\)

The symmetric group on 3 objects — equivalently the symmetries of an equilateral triangle — has order 6: the identity, three transpositions (reflections), two 3-cycles (rotations).

By Lagrange, subgroups can only have order 1, 2, 3 or 6. And indeed:

Order Subgroup Normal?
1 \(\{e\}\) yes
2 \(\{e, (12)\}\) and two others no
3 \(\{e, (123), (132)\}\) — the rotations yes (index 2)
6 \(S_3\) yes

No subgroup of order 4 or 5 — forbidden by divisibility, without any search.

The order-3 subgroup is normal because any subgroup of index 2 is (left and right cosets are both "\(H\) and everything else"), and \(S_3/A_3 \cong \mathbb{Z}_2\) — the even/odd permutation grading, which is where the sign of a determinant comes from.

The order-2 subgroups are not normal: \(S_3\) is the smallest non-abelian group, and its non-normal subgroups are the reason.

Common mistakes

  • Reading Lagrange backwards. It says a subgroup's order divides \(|G|\). It does not say that every divisor has a subgroup — the converse is false (\(A_4\) has order 12 and no subgroup of order 6).
  • Forming \(G/H\) for non-normal \(H\). The coset multiplication is not well defined; you get a set of cosets, not a group.
  • Assuming left and right cosets coincide. True only for normal subgroups — i.e. always in an abelian group, and not in general otherwise.

Knowledge graph position

Prerequisites: symmetry groups. Leads to: quotient groups, representations, SU(2)→SO(3), crystallographic classification.

Quiz

Q1 (conceptual). A group has 15 elements. What subgroup orders are possible?

Answer

Only divisors of 15: 1, 3, 5 and 15. Orders 2, 4, 6, … are impossible by Lagrange. (In fact every group of order 15 is cyclic, but that needs more than Lagrange.)

Q2 (conceptual). Why can you only form a quotient group \(G/H\) when \(H\) is normal?

Answer

The product \((g_1H)(g_2H) = g_1g_2H\) must not depend on which representatives you pick. That well-definedness requires \(gH = Hg\) for all \(g\) — precisely normality. Without it, the cosets are a set with no consistent multiplication.

Q3 (MCQ). Lagrange's theorem implies that a group of prime order \(p\):

  • (a) is abelian and cyclic, with no proper subgroups
  • (b) has exactly \(p\) subgroups
  • (c) is non-abelian
  • (d) has a subgroup of every order less than \(p\)
Answer

(a). Subgroup orders must divide \(p\), so only 1 and \(p\) are available. Any non-identity element therefore generates the whole group, making it cyclic and hence abelian.