Abstract Algebra Review

Everything from the section, organized so you can find it: the definitions, the named theorems, the reference groups, and the notation. Use it as a lookup table rather than reading straight through.

$(G, *)$ is a group when it is closed, associative, has an identity, and every element has an inverse. Commutativity is not required; groups that have it are abelian.

Consequences from the axioms alone:

• The identity is unique.
• Inverses are unique.
• Cancellation holds: $ab = ac \implies b = c$.
• $(ab)^{-1} = b^{-1}a^{-1}$ - the order reverses.
• $(a^{-1})^{-1} = a$.

Element orders in $\mathbb{Z}_n$:

$$|k| = \frac{n}{\gcd(k, n)}$$

so $k$ generates $\mathbb{Z}_n$ exactly when $\gcd(k,n) = 1$, and there are $\varphi(n)$ generators.

Test: nonempty, closed under the operation, closed under inverses. Or in one step, $ab^{-1} \in H$ for all $a,b \in H$. Never re-check associativity; it’s inherited.

A finite nonempty subset closed under the operation is automatically a subgroup.

A cyclic group of order $n$ has exactly one subgroup per divisor of $n$. Other groups can have many more: $D_4$ has ten subgroups against $\mathbb{Z}_8$‘s four.

Cosets partition $G$ into blocks all of size $|H|$, giving

$$|G| = [G:H] \cdot |H|$$

Corollaries: element orders divide $|G|$; $a^{|G|} = e$; every group of prime order is cyclic; Fermat’s little theorem and Euler’s theorem.

The converse is false. $A_4$ has order 12 with no subgroup of order 6.

$N \trianglelefteq G$ means $gng^{-1} \in N$ for all $g \in G$, $n \in N$. Automatic when $G$ is abelian, when $[G:N] = 2$, or for $Z(G)$.

Normality is exactly what makes $(aN)(bN) = abN$ well-defined, giving the quotient group $G/N$ of order $[G:N]$.

A group whose only normal subgroups are $\{e\}$ and itself is simple.

A homomorphism satisfies $\varphi(ab) = \varphi(a)\varphi(b)$. Then $\varphi(e) = e$ and $\varphi(a^{-1}) = \varphi(a)^{-1}$ come free.

• $\ker\varphi \trianglelefteq G$, and every normal subgroup is some kernel.
• $\operatorname{im}\varphi \le H$, generally not normal.
• $\varphi$ injective $\iff \ker\varphi = \{e\}$.
• Every fibre has size $|\ker\varphi|$.

First isomorphism theorem:

$$G/\ker\varphi \cong \operatorname{im}\varphi, \qquad |G| = |\ker\varphi|\cdot|\operatorname{im}\varphi|$$

To prove isomorphic, exhibit a map. To prove not isomorphic, name one differing invariant: order, abelian or not, the multiset of element orders, cyclic or not, number of subgroups, size of the center.

Counts of groups by order: 1 each for prime orders; 2 for order 4; 2 for order 6; 5 for order 8; 5 for order 12; 14 for order 16.

Fundamental theorem of finite abelian groups: every one is a direct product of cyclic groups of prime-power order, uniquely up to reordering.

$$\mathbb{Z}_m \times \mathbb{Z}_n \cong \mathbb{Z}_{mn} \iff \gcd(m,n) = 1$$

$S_n$ has $n!$ elements; $A_n$ has $n!/2$. Cycle notation: order is the lcm of the cycle lengths, parity is the sum of $(k_i - 1)$.

$D_n$ has $2n$ elements, presented as

$$\langle r, s \mid r^n = s^2 = e,\ sr = r^{-1}s\rangle$$

Every $r^ks$ is a reflection of order 2. The cube’s rotation group has 24 elements and is isomorphic to $S_4$.

Cayley’s theorem: every group embeds in a symmetric group.

• Ring - $(R,+)$ abelian group; $\times$ closed and associative; distributive.
• Zero divisor - nonzero $a$ with $ab = 0$ for some nonzero $b$. Zero divisors break cancellation.
• Integral domain - commutative, with unity, no zero divisors.
• Field - every nonzero element is a unit.
• Ideal - additive subgroup that absorbs multiplication by all of $R$.

Facts holding in every ring: $0a = 0$, $(-a)b = -(ab)$, $(-a)(-b) = ab$.

In $\mathbb{Z}_n$, an element $a$ is:

• a unit if $\gcd(a,n) = 1$ - these form $U(n)$, of size $\varphi(n)$
• a zero divisor otherwise (for $a \ne 0$)

$$\mathbb{Z}_n \text{ is a field} \iff n \text{ is prime}$$

$R/I$ is a ring when $I$ is an ideal, and absorption is exactly what makes the multiplication well-defined.

$$I \text{ prime} \iff R/I \text{ a domain}, \qquad I \text{ maximal} \iff R/I \text{ a field}$$

The key construction: for $F$ a field and $f$ irreducible, $F[x]/\langle f\rangle$ is a field. This builds $\mathbb{C}$ as $\mathbb{R}[x]/\langle x^2+1\rangle$ and every finite field.

A field has only two ideals, so there is nothing to quotient by.

Over an integral domain, $\deg(fg) = \deg f + \deg g$. $R[x]$ is a domain exactly when $R$ is, and never a field.

Division algorithm needs a field of coefficients. Factor theorem: $f(a) = 0 \iff (x-a) \mid f$. A degree-$n$ polynomial over a field has at most $n$ roots.

Irreducibility is relative to the field. Degree 2 or 3 is irreducible exactly when there is no root; from degree 4 that fails, as $x^4+4$ shows.

$GF(q)$ exists $\iff q = p^n$, and then it is unique up to isomorphism. Build it as $\mathbb{F}_p[x]/\langle f\rangle$ for irreducible $f$ of degree $n$.

$GF(p^n) \ne \mathbb{Z}_{p^n}$ once $n > 1$: $GF(4)$ has characteristic 2 and $\mathbb{Z}_4$ is not a field.

$GF(q)^*$ is cyclic of order $q-1$, so $x^q = x$ for every element.

$[K:F]$ is a vector-space dimension. $[F(\alpha):F] = \deg m(x)$ for $m$ the minimal polynomial.

Tower law: $[L:F] = [L:K]\cdot[K:F]$ - degrees multiply.

Constructible numbers have degree $2^k$ over $\mathbb{Q}$, which is why doubling the cube and trisecting the angle are impossible, and $\pi$ being transcendental settles squaring the circle.

The general quintic is unsolvable by radicals because $S_5$ is not solvable, since $A_5$ is simple and non-abelian.

Theorem	Statement
Lagrange	$\|H\|$ divides $\|G\|$, and $\|G\| = [G:H]\cdot\|H\|$
Cayley	Every group embeds in a symmetric group
Cauchy	If a prime $p$ divides $\|G\|$, then $G$ has an element of order $p$
First isomorphism	$G/\ker\varphi \cong \operatorname{im}\varphi$
Fundamental thm of finite abelian groups	A direct product of prime-power cyclic groups, uniquely
Fermat’s little	$a^{p-1} \equiv 1 \pmod p$ for $\gcd(a,p)=1$
Euler	$a^{\varphi(n)} \equiv 1 \pmod n$ for $\gcd(a,n)=1$
Classification of finite fields	A field of order $q$ exists iff $q$ is a prime power, and is then unique
Tower law	$[L:F] = [L:K]\cdot[K:F]$
Abel–Ruffini	No radical formula for the general quintic

• Writing a set and calling it a group. Always name the operation. “Is $\mathbb{Z}$ a group?” has no answer.
• Reading $a^{-1}$ as a reciprocal. In $(\mathbb{Z},+)$ the inverse of 5 is $-5$.
• Confusing associativity with commutativity. Parentheses versus order.
• Assuming the converse of Lagrange. A divisor need not be a subgroup order.
• Forgetting the order reverses in $(ab)^{-1} = b^{-1}a^{-1}$.
• Thinking a coset is a subgroup. Only $eH = H$ is.
• Reading normality as $gng^{-1} = n$. It’s $gng^{-1} \in N$.
• Concluding two groups are isomorphic because the invariants match. You must build the map.
• Assuming $GF(p^n) = \mathbb{Z}_{p^n}$. Different characteristic; the second usually isn’t a field.
• Calling a polynomial irreducible without naming the field.
• Cancelling in a ring with zero divisors. In $\mathbb{Z}_6$, $2\cdot1 = 2\cdot4$.
• Confusing a subring with an ideal. An ideal must absorb multiplication by everything.