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A.3 · Groups, cosets and quotient maps

GRE · GRE Subject Test · GRE Mathematics · Topic 9

Train
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Scope and prerequisites

Undergraduate GRE preparation. Local objectives within the reviewed ETS scope; this is original teaching, not an official test or score predictor.

Prerequisites: Group axioms, functions, integer divisibility and permutations.

  • Verify group structure and compute element orders and subgroup indices
  • Use kernels and images to identify quotient groups
  • Distinguish normal subgroups from arbitrary subgroups
  • Classify permutation conjugacy by cycle type

coset 陪集: A translate of a subgroup that forms one part of the coset partition.

normal subgroup 正规子群: A subgroup invariant under conjugation by every group element.

Vocabulary Train
English
coset/ˈkɒset/
normal subgroup/ˈnɔːml ˈsʌbɡruːp/
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Choose and justify a method

Before computing an order, check closure, associativity, an identity and an inverse for every element under the stated operation. A subset can inherit associativity yet fail closure or omit the identity. In a finite group, the order of an element is the least positive power giving the identity. In the additive group Z/nZ, it is the least positive multiple giving zero; the order of residue a is n/gcd(a,n). Lagrange's theorem says subgroup orders divide the group order. The converse is not a general existence theorem, and the order of a group is not the order of each element.

A left coset gH is a translate of a subgroup H. Cosets have equal size and partition the group, so the index is |G|/|H| in a finite group. In an additive group write g+H. Membership in the same coset means the difference lies in H. A coset usually is not itself a subgroup because it may omit the identity.

A homomorphism preserves the operation. Its kernel consists of elements sent to the identity, and its image consists of values actually reached. Every kernel is normal. The first isomorphism theorem identifies G/ker(phi) with im(phi); do not replace the image with the whole codomain unless the map is onto.

Quotient multiplication is well defined only when H is normal. All subgroups of an abelian group are normal. In a nonabelian group test gHg^−1=H; a subgroup of index two is normal. For permutations compose in the stated convention, here rightmost first. Disjoint cycle lengths give the permutation order by their least common multiple. Conjugation hσh⁻¹ relabels the elements in σ’s cycles, so it preserves cycle lengths; conversely permutations with the same cycle lengths can be related by a relabelling. Thus conjugacy classes in S_n correspond to partitions of n, including fixed-point cycles. In S4 the types are 1+1+1+1, 2+1+1, 2+2, 3+1 and 4: five classes, not one class for each possible element order. The types 2+1+1 and 2+2 both have order two but are not conjugate.

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Worked reasoning

Define phi from Z/12Z to Z/3Z by reducing residues modulo 3. It is onto and preserves addition. Its kernel H is {0,3,6,9}, so |H|=4 and the index is 12/4=3. The other cosets are {1,4,7,10} and {2,5,8,11}. Thus (Z/12Z)/H is isomorphic to Z/3Z. The element 3 in the original group has order 12/gcd(3,12)=4, not 3.

Groups, cosets and quotient maps: course example
Original course illustration; its values belong to the worked example, not the later practice.
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Conditions and counterexamples

A quotient has one element per coset, not one per element of its kernel. A homomorphism need not be onto its stated codomain.

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Guided application

In the additive group $\mathbb Z/18\mathbb Z$, let $H=\langle6\rangle$. List H and its cosets. Determine the order of 6 and identify the quotient.

Worked solution

$H=\{0,6,12\}$ has size three. The six cosets are $r+H=\{r,r+6,r+12\}$ for $r=0,1,2,3,4,5$. They partition all 18 residues. The element 6 has order three, while the quotient has order six. Reduction modulo six is onto with kernel H, so $(\mathbb Z/18\mathbb Z)/H\cong\mathbb Z/6\mathbb Z$. Normality holds because the original group is abelian.

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Independent transfer

In $S_3$, let $H=\{e,(12)\}$. Check whether $H$ is normal, and decide whether coset multiplication defines a quotient group. Separately find the order of $(123)(45)$ in $S_5$.

Check after attempting

Conjugation by $(123)$ sends $(12)$ to $(23)$, which is outside H. Thus H is not normal. Left cosets still partition $S_3$, but multiplication of those cosets is not well defined independently of representatives. The disjoint cycles in $(123)(45)$ have lengths three and two, so its order is $\operatorname{lcm}(3,2)=6$. Element order and subgroup index are different quantities.

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