Abstract Algebra Domain Guide

Concise answer

Abstract algebra on this exam is a constraint subject, not a computation subject. Lagrange's theorem forces every subgroup order and element order to divide the group order; cyclic groups are completely described by the divisors of $n$ ; homomorphisms transport structure and their kernels measure exactly what is lost. Rings add a second operation, and the interesting question becomes which elements can be cancelled — that is what separates a ring from an integral domain and an integral domain from a field.

Definitions

Group: A set $G$ with an associative operation, an identity $e$ , and an inverse for every element. Abelian means the operation also commutes.
Order: $|G|$ is the number of elements of the group; $|a|$ is the least $k > 0$ with $a^k = e$ . Keeping the two apart is half of most exam items.
Coset and index: $aH = \{ah : h \in H\}$ . The cosets of $H$ partition $G$ into $[G : H] = |G|/|H|$ blocks of equal size — the proof of Lagrange's theorem in one line.
Normal subgroup and quotient: $N \trianglelefteq G$ when $gNg^{-1} = N$ for all $g$ ; exactly then do the cosets themselves form a group $G/N$ .
Kernel: $\ker \varphi = \{g : \varphi(g) = e\}$ for a homomorphism $\varphi$ . It is always normal, and $\varphi$ is injective exactly when the kernel is trivial.
Zero divisor: A nonzero $a$ with $ab = 0$ for some nonzero $b$ . Zero divisors are what block cancellation: in $\mathbb{Z}_6$ , $2 \cdot 3 = 0$ .
Field: A commutative ring with identity in which every nonzero element has a multiplicative inverse — so division by anything nonzero is defined.

Intuition

A group is what is left when you keep only the ability to combine and to undo. That is why counting arguments dominate: with so little structure, the strongest available statements are about sizes. Lagrange's theorem is the archetype — it says that the way a subgroup tiles the group by cosets forces an exact divisibility relation, with no reference to what the elements are.

Cyclic groups are the fully solved case, and most exam items reduce to them. $\mathbb{Z}_n$ has exactly one subgroup for each divisor of $n$ , exactly $\varphi(d)$ elements of order $d$ for each divisor $d$ , and exactly $\varphi(n)$ generators. Once you can produce those three counts from the factorisation of $n$ , a large family of questions becomes arithmetic.

Rings are groups that have grown a second operation, and the whole subject organises around one question: can you cancel? In $\mathbb{Z}$ you can, because there are no zero divisors; in $\mathbb{Z}_6$ you cannot, because $2 \cdot 3 = 0$ with neither factor zero. An integral domain is a ring where cancellation works; a field is one where you can go further and divide.

Concept walkthrough

Start every group question by writing $|G|$ . Lagrange's theorem then says the order of any subgroup, and hence the order of any element, divides $|G|$ . This immediately answers 'can a group of order $15$ have an element of order $4$ ?' (no — $4 \nmid 15$ ) and it is the reason a group of prime order must be cyclic: any non-identity element generates a subgroup whose order divides $p$ and is not $1$ , so it is all of $G$ . The converse of Lagrange is false and is a favourite distractor: $A_4$ has order $12$ but contains no subgroup of order $6$ .

Cyclic groups carry most of the computational load. If $G = \langle a \rangle$ has order $n$ , then $|a^k| = n / \gcd(k, n)$ . From that one formula: the generators are the $a^k$ with $\gcd(k,n) = 1$ , so there are $\varphi(n)$ of them; for each divisor $d$ of $n$ there is exactly one subgroup of order $d$ , so the subgroup count is the number of divisors; and the number of elements of order $d$ is $\varphi(d)$ . Direct products extend this: in $G \times H$ the order of $(g,h)$ is $\operatorname{lcm}(|g|, |h|)$ , and $\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn}$ exactly when $\gcd(m,n) = 1$ .

Non-abelian structure appears through the symmetric group $S_n$ (order $n!$ ), the alternating group $A_n$ (order $n!/2$ for $n \ge 2$ ), and the dihedral group $D_n$ of symmetries of a regular $n$ -gon (order $2n$ ). Permutations are handled in cycle notation: a permutation's order is the least common multiple of its cycle lengths, and its sign is determined by parity — a $k$ -cycle is a product of $k-1$ transpositions, so it is even exactly when $k$ is odd. A common exam item asks for the largest possible order of an element of $S_n$ , which is a question about partitions of $n$ and their least common multiples.

Homomorphisms are where structure moves between groups. The kernel is always normal, the image is a subgroup of the target, and the first isomorphism theorem packages both: $G/\ker\varphi \cong \operatorname{im}\varphi$ . This is the fastest way to identify quotient groups — for instance the determinant map from $GL_n(\mathbb{R})$ onto the nonzero reals has kernel $SL_n(\mathbb{R})$ , so the quotient is $\mathbb{R}^{\times}$ without any coset arithmetic. To prove two groups are not isomorphic, compare invariants rather than searching for maps: abelian versus non-abelian, the number of elements of each order, whether the group is cyclic. $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$ both have order $4$ , but the first has an element of order $4$ and the second does not.

Rings shift attention to multiplication. A ring is an abelian group under addition with an associative multiplication that distributes; commutativity and an identity are separate assumptions, and this exam usually supplies both. An ideal is the ring analogue of a normal subgroup — the subobject you can quotient by — and $R/I$ is a field exactly when $I$ is maximal. The chain to memorise runs ring $\supset$ commutative ring with identity $\supset$ integral domain (no zero divisors, so cancellation holds) $\supset$ field (every nonzero element is a unit). Two facts close the loop: every field is an integral domain, and every finite integral domain is a field.

The number theory this exam uses is modular arithmetic read through that lens. $\mathbb{Z}_n$ is a commutative ring with identity; its units are the residues coprime to $n$ , so $|U(n)| = \varphi(n)$ , and $\mathbb{Z}_n$ is a field exactly when $n$ is prime — because for composite $n = ab$ with $1 < a, b < n$ the classes of $a$ and $b$ are nonzero with product zero. Euler's theorem ( $a^{\varphi(n)} \equiv 1 \bmod n$ for $\gcd(a,n) = 1$ ) and its special case Fermat's little theorem ( $a^{p-1} \equiv 1 \bmod p$ for $p \nmid a$ ) are then just Lagrange's theorem applied inside the group of units, and they are what make questions like 'find the last digit of $7^{2026}$ ' fast: $7$ has order $4$ modulo $10$ , $2026 \equiv 2 \bmod 4$ , so the answer is the last digit of $7^2$ , namely $9$ .

After this page, you should be able to

Use Lagrange's theorem to rule out subgroup and element orders, and state why its converse fails.
Describe every subgroup and count the elements of each order in a cyclic group from the divisors of $n$ .
Decide whether two groups of the same order are isomorphic by comparing invariants such as abelianness and order counts.
Apply the first isomorphism theorem to identify a quotient group without constructing its multiplication table.
Distinguish ring, commutative ring, integral domain, and field by exactly which axiom each one adds.
Compute confidently in $\mathbb{Z}_n$ : units, inverses, orders, and the reason $\mathbb{Z}_n$ is a field precisely when $n$ is prime.

Formulas and assumptions

Lagrange's theorem and index

|G| = [G:H]\,|H| \quad \Longrightarrow \quad |a| \mid |G| \ \text{and}\ a^{|G|} = e

Variables

|G|: the order of the group
H: a subgroup
[G : H]: the number of cosets of H in G

Assumptions

G must be finite.
The converse fails: a divisor of |G| need not be the order of any subgroup (A_4 has no subgroup of order 6).

Order of a power in a cyclic group

|a^{k}| = \frac{n}{\gcd(k,n)}; \qquad \langle a^{k} \rangle = G \iff \gcd(k,n) = 1

Variables

n: the order of the cyclic group
k: the exponent

Assumptions

The group must be cyclic of finite order n.
Applies verbatim to Z_n with multiplication of k replaced by addition.

Counts in a cyclic group of order n

\#\{\text{subgroups}\} = \tau(n), \qquad \#\{\text{generators}\} = \varphi(n), \qquad \#\{|x| = d\} = \varphi(d)

Variables

tau(n): the number of positive divisors of n
phi(n): Euler's totient, the count of integers in [1, n] coprime to n

Assumptions

The group must be cyclic; a non-cyclic group of the same order has different counts.
phi is multiplicative on coprime arguments, with phi(p^k) = p^k - p^(k-1).

Orders in a direct product

|(g,h)| = \operatorname{lcm}(|g|,|h|), \qquad \mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn} \iff \gcd(m,n) = 1

Variables

g, h: elements of the two factors

Assumptions

Both factors must have finite order for the lcm formula.

First isomorphism theorem

G/\ker\varphi \;\cong\; \operatorname{im}\varphi, \qquad \ker \varphi \trianglelefteq G

Variables

f: a group homomorphism
ker(f): the preimage of the identity

Assumptions

The map must be a homomorphism; nothing weaker suffices.
f is injective exactly when ker(f) is trivial.

Standard finite groups and their orders

|S_n| = n!, \quad |A_n| = \tfrac{n!}{2}, \quad |D_n| = 2n, \quad |\sigma| = \operatorname{lcm}(\text{cycle lengths})

Variables

S_n: the symmetric group on n letters
A_n: the even permutations
D_n: the symmetries of a regular n-gon

Assumptions

The cycle-length rule needs the cycles written in disjoint form.
A k-cycle is even exactly when k is odd, because it factors into k - 1 transpositions.

The ring hierarchy

\text{ring} \supset \text{comm. ring with } 1 \supset \text{integral domain} \supset \text{field}; \qquad \mathbb{Z}_n \text{ is a field} \iff n \text{ is prime}

Variables

zero divisor: a nonzero a with ab = 0 for some nonzero b
unit: an element with a multiplicative inverse

Assumptions

Cancellation is available exactly when there are no zero divisors.
Every finite integral domain is a field, so the two notions coincide in the finite case.

Euler and Fermat congruences

a^{\varphi(n)} \equiv 1 \pmod n \quad (\gcd(a,n)=1), \qquad a^{p-1} \equiv 1 \pmod p \quad (p \nmid a)

Variables

phi(n): Euler's totient
p: a prime

Assumptions

The coprimality condition is essential; without it a is not a unit modulo n.
These are Lagrange's theorem applied in the group of units of Z_n.

Worked example

Reading a cyclic group off its divisors

How many subgroups does $\mathbb{Z}_{36}$ have, and how many of its elements have order $6$ ?

1 $\mathbb{Z}_{36}$ is cyclic of order $36$ . In a cyclic group the subgroups correspond exactly to the divisors of the order — one subgroup for each divisor, and no others.
2Factor: $36 = 2^2 \cdot 3^2$ , so the number of divisors is $(2+1)(2+1) = 9$ , namely $1, 2, 3, 4, 6, 9, 12, 18, 36$ . Hence $\mathbb{Z}_{36}$ has exactly $9$ subgroups.
3For the element orders, use $|k| = 36/\gcd(k, 36)$ in additive notation. So $|k| = 6$ exactly when $\gcd(k, 36) = 36/6 = 6$ .
4Only multiples of $6$ can have $\gcd$ equal to $6$ , so test them: $\gcd(6,36) = 6$ , $\gcd(12,36) = 12$ , $\gcd(18,36) = 18$ , $\gcd(24,36) = 12$ , $\gcd(30,36) = 6$ . Exactly $k = 6$ and $k = 30$ qualify.
5Cross-check with the general count: a cyclic group of order $n$ has exactly $\varphi(d)$ elements of order $d$ for each divisor $d$ of $n$ , and $\varphi(6) = \varphi(2)\varphi(3) = 1 \cdot 2 = 2$ . The two answers agree.
6Note what this shows about subgroups: those two elements generate the same subgroup $\langle 6 \rangle = \{0, 6, 12, 18, 24, 30\}$ , which is the unique subgroup of order $6$ — consistent with 'one subgroup per divisor'.

$\mathbb{Z}_{36}$ has $9$ subgroups, and exactly $2$ elements of order $6$ (namely $6$ and $30$ ), which generate the unique order- $6$ subgroup $\{0, 6, 12, 18, 24, 30\}$ .

Common traps

Using the converse of Lagrange's theorem. A divisor of $|G|$ need not be the order of a subgroup: $A_4$ has order $12$ and no subgroup of order $6$ .
Confusing the order of the group with the order of an element — the two appear in the same sentence constantly and obey different constraints.
Assuming groups of equal order are isomorphic. $\mathbb{Z}_4$ and $\mathbb{Z}_2 \times \mathbb{Z}_2$ both have order $4$ ; only the first has an element of order $4$ .
Assuming every subgroup is normal. Index $2$ forces normality, but nothing smaller does — $S_3$ has non-normal subgroups of order $2$ .
Treating all of $\mathbb{Z}_n$ as a multiplicative group. Only the units — the residues coprime to $n$ — form a group under multiplication, and there are $\varphi(n)$ of them.
Cancelling in a ring with zero divisors. From $2x = 2y$ in $\mathbb{Z}_6$ you cannot conclude $x = y$ ; take $x = 1$ , $y = 4$ .
Confusing the number of generators of $\mathbb{Z}_n$ (which is $\varphi(n)$ ) with the number of subgroups (which is the number of divisors of $n$ ).
Assuming $\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn}$ in general. It holds exactly when $\gcd(m,n) = 1$ ; $\mathbb{Z}_2 \times \mathbb{Z}_2$ is not $\mathbb{Z}_4$ .

Counting Elements of a Given Order Group Theory Essentials GRE Mathematics overview Linear Algebra Domain Guide Real Analysis Domain Guide Sign in for current practice

Question depth and domain coverage vary by exam. Practice answers are checked after submission.

Sources

GRE Subject Test Content and Structure — ETS. Accessed 2026-07-06. Use as a cited source for exam facts; do not imply affiliation or reproduce protected test material.
Abstract Algebra: Theory and Applications (Judson), Section 3.8: Definitions and Examples — LibreTexts Mathematics. Accessed 2026-08-15. Judson's Abstract Algebra on LibreTexts is released under the GNU Free Documentation License 1.3; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Abstract Algebra: Theory and Applications (Judson), Section 3.9: Subgroups — LibreTexts Mathematics. Accessed 2026-08-15. Judson's Abstract Algebra on LibreTexts is released under the GNU Free Documentation License 1.3; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Abstract Algebra: Theory and Applications (Judson), Section 4.1: Cyclic Subgroups — LibreTexts Mathematics. Accessed 2026-08-15. Judson's Abstract Algebra on LibreTexts is released under the GNU Free Documentation License 1.3; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Abstract Algebra: Theory and Applications (Judson), Section 6.2: Lagrange's Theorem — LibreTexts Mathematics. Accessed 2026-08-15. Judson's Abstract Algebra on LibreTexts is released under the GNU Free Documentation License 1.3; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Abstract Algebra: Theory and Applications (Judson), Section 9.1: Definition and Examples (Isomorphisms) — LibreTexts Mathematics. Accessed 2026-08-15. Judson's Abstract Algebra on LibreTexts is released under the GNU Free Documentation License 1.3; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Abstract Algebra: Theory and Applications (Judson), Section 16.3: Rings — LibreTexts Mathematics. Accessed 2026-08-15. Judson's Abstract Algebra on LibreTexts is released under the GNU Free Documentation License 1.3; attribute the author and platform and avoid verbatim reuse beyond short cited references.
Abstract Algebra: Theory and Applications (Judson), Section 16.4: Integral Domains and Fields — LibreTexts Mathematics. Accessed 2026-08-15. Judson's Abstract Algebra on LibreTexts is released under the GNU Free Documentation License 1.3; attribute the author and platform and avoid verbatim reuse beyond short cited references.

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Abstract Algebra Domain Guide

Concise answer

Definitions

Intuition

Concept walkthrough

After this page, you should be able to

Formulas and assumptions

Lagrange's theorem and index

Order of a power in a cyclic group

Counts in a cyclic group of order n

Orders in a direct product

First isomorphism theorem

Standard finite groups and their orders

The ring hierarchy

Euler and Fermat congruences

Worked example

Reading a cyclic group off its divisors

Common traps

Related pages and practice

Sources

Sources and corrections