Normal Subgroups and Quotient Groups
A subgroup is normal exactly when multiplying its cosets is independent of the chosen representatives. The first isomorphism theorem identifies a homomorphism's image with its domain modulo its kernel.
What a learner can do afterwards
- Decide whether a subgroup is normal, by conjugation or by its index
- Write the multiplication table of a small quotient group
- Use the first isomorphism theorem to identify the image of a homomorphism
1 · Read
Recall that cosets partition a group. Write e for its identity and g^-1 for an inverse. A subgroup N is normal in G if gng^-1 belongs to N for every g in G and n in N. In an abelian group this conjugate equals n. An index-two subgroup is normal too: its left and right cosets are itself and its complement. Index counts left cosets; another index alone need not decide normality.
S4 is the group of permutations of four letters. The notation (34) means swap 3 and 4; products act rightmost first. Let H = {e, (34)}. Conjugating by (14) gives (14)(34)(14) = (13). Conjugation relabels the moved points: 3 stays 3 and 4 becomes 1. Since (13) is outside H, this subgroup is not normal. One escaped conjugate suffices to disprove normality.
For normal N, define (aN)(bN) = abN. Replacing a and b by an and bm gives ab(b^-1nb)m. Normality puts b^-1nb in N, so the product coset remains abN. Conversely, independence of representatives gives (nN)(gN) = (eN)(gN), hence ngN = gN and g^-1ng belongs to N. Thus this rule works exactly when N is normal. The quotient identity is N; the inverse of aN is a^-1N.
Z_n denotes residues modulo n under addition. In Z8, H = {0, 4} is normal because the group is abelian. Name its cosets H, A = 1 + H, B = 2 + H and C = 3 + H. Add representatives to calculate: A + B = C. The inverse of a + H is -a + H. Two cosets a + H and b + H agree exactly when a - b belongs to H.
Build a quotient table by combining each row representative with each column representative. Find the coset containing the result. For the Z8/H cosets just named, the table is:
| + | H | A | B | C |
|---|---|---|---|---|
| H | H | A | B | C |
| A | A | B | C | H |
| B | B | C | H | A |
| C | C | H | A | B |
For example, the row C and column C give B because 3 + 3 = 6 belongs to B.
For finite G and normal N, each coset has |N| elements, so |G/N| = |G| / |N|. Vertical bars count elements. Cyclic groups are abelian, so their subgroups are normal. In Z_n, list the subgroup residues, count them and divide n by that count. The centre Z(G) consists of elements commuting with every group element. It is normal because conjugation fixes each central element.
For a homomorphism phi: G -> K, its kernel N is normal: phi(gng^-1) = phi(g)e phi(g)^-1 = e for n in N. The first isomorphism theorem gives an isomorphism gN -> phi(g) from G/N to the image. For finite G, |image| = |G| / |N|. The image need not be all of K. If G is cyclic, its image is generated by the image of a generator.
The sign homomorphism sends even permutations to 1 and odd permutations to -1, with multiplication in {1, -1}. Even means expressible as an even number of swaps; odd means an odd number. This parity does not depend on the chosen expression. The identity is even and a single swap is odd, so both signs occur. Its kernel is the alternating subgroup of even permutations.
For finite G and normal N, the order of a quotient element gN is the least positive k with g^k in N. In additive notation, find the least positive k with ka in N. In Z20 with N = {0, 10}, the coset 5 + N has order 2: 5 is outside N, but 2 times 5 belongs to N. This counts repetitions of one quotient element, not the number of quotient elements.
Normality makes coset multiplication consistent, quotients divide orders, and every image is a quotient by its kernel.
The lesson, questions, worked solutions and base flashcards were written for this dataset.
The three readings and nine extra questions were written for this dataset. The flashcards include material from the imported fact collection.
More to read
What a normality test must cover
Normality concerns a subgroup inside a specified group. To disprove it, one escaped conjugate is enough. To prove it directly, the argument must cover every subgroup element and every conjugating group element.
For permutations, conjugation changes the names of the moved points. Take H = {e, (13)} inside S3. Conjugating (13) by (12) gives (23), which is outside H. This proves that H is not normal in S3, even though H itself is abelian.
A subgroup of an abelian group is different: the ambient group makes every conjugation leave each element unchanged. Index two gives another route to normality because the only cosets are the subgroup and its complement. Always check which group supplies the conjugating elements.
Reading a quotient table
A quotient element is an entire coset. A representative lets you calculate with that coset, but the answer must not depend on which representative you use.
In additive Z9, take H = {0, 3, 6}. The distinct cosets have representatives 0, 1 and 2. To fill a table entry, add its row and column representatives and identify the coset containing the sum. Row 2 and column 2 give 4 + H = 1 + H. Replacing both representatives 2 by 5 gives 10 + H, still 1 + H.
The row labelled H reproduces the column labels because H is the quotient identity. The entries equal to H identify inverse pairs. Here 1 + H and 2 + H are inverses. Normality is available because Z9 is abelian.
Kernel cosets and outputs
Let phi be a group homomorphism with kernel N. Inputs g and h have equal images exactly when h^-1g belongs to N. Equivalently, they belong to the same left coset of N. Each kernel coset therefore corresponds to one output value.
Reduction from Z14 to Z7 illustrates this. The kernel is {0, 7}. Inputs 0 and 7 give output 0; inputs 1 and 8 give output 1. Continuing gives seven cosets and seven outputs. Adding cosets agrees with adding the corresponding outputs.
The first isomorphism theorem expresses this agreement as G/N being isomorphic to the image. The image may be smaller than the codomain. The kernel records which inputs are identified, but does not specify the target group or every output label of the original map.
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Where it sits
17 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.