Homomorphisms, Cosets and Lagrange's Theorem
Homomorphisms preserve a group's operation. Left cosets of a subgroup partition the group, as do right cosets separately. For a finite group, Lagrange's theorem says that each subgroup's order divides the group's order. Kernels connect these maps with quotient groups.
What a learner can do afterwards
- Show a given map is a homomorphism and find its kernel
- Partition a small group into cosets of a subgroup
- Use Lagrange's theorem to rule out a subgroup of a stated size
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A group homomorphism f satisfies f(ab) = f(a)f(b), using each group's operation. It sends the source identity to the target identity. Its kernel is the set of inputs sent to that identity. For additive groups, write f(a + b) = f(a) + f(b). Test a proposed map by computing both sides. For example, powers preserve multiplication on nonzero reals because (ab)^k = a^k b^k.
Z_m denotes residues modulo m under addition. If d divides m, reducing these residues modulo d defines a homomorphism to Z_d. Its kernel consists of multiples of d among 0 through m - 1. For Z18 to Z3, these are 0, 3, 6, 9, 12 and 15. More generally, f(a) = f(b) exactly when b inverse times a is in the kernel. Thus a trivial kernel makes f injective.
For a subgroup H of G, a left coset is aH = {ah : h in H}; a right coset is Ha = {ha : h in H}. Left cosets partition G, and right cosets partition G separately. Do not mix the two sides in this claim. In additive groups, write a + H. Multiplication by a is a bijection from H to aH, so each left coset has H's size. The right-sided map h to ha is also a bijection.
In additive Z14, H = {0, 7}. The coset 1 + H is {1, 8}. Representatives 0 through 6 produce seven disjoint cosets covering all fourteen residues. To recover H from a + H, subtract a from every member. Representatives a and b give the same additive coset exactly when a - b belongs to H.
Suppose c belongs to left cosets aH and bH. Write c = ah1 = bh2 with h1, h2 in H. Then b inverse times a = h2 h1 inverse belongs to H. Therefore a = bk for some k in H, so aH = bkH = bH. Here kH = H because H closes under products and inverses. Thus shared left cosets coincide. The corresponding argument applies separately to right cosets.
For finite G, its left cosets are disjoint, cover G and each have |H| elements. If there are k cosets, |G| = k|H|. Hence |H| divides |G| and k = |G|/|H|. This is Lagrange's theorem. A size that does not divide |G| is impossible. Divisibility alone does not guarantee that a subgroup exists. A group of order 22 cannot have a subgroup of order 4.
Check that maps preserve the operation, partition into cosets, and use divisibility to rule out subgroup sizes.
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