A.2 · Abstract algebra and number theory

A group needs closure, associativity, identity and inverses. Commutativity is an additional condition.

A homomorphism preserves the operation. Its kernel is a normal subgroup and identifies elements mapping to the identity.

A field permits division by every nonzero element. Integers form a ring but not a field.

Work with congruences modulo n. A residue a has a multiplicative inverse exactly when gcd(a,n)=1.

Modulo 8, 3 has inverse 3 because 3·3=9≡1. But 2 has no inverse because gcd(2,8)=2. Modulo a prime, every nonzero residue has an inverse; this gives a finite field.

Do not cancel a factor in a modular equation without checking that it is invertible.

Compare the additive groups and multiplicative structures of $\mathbb Z/5\mathbb Z$ and $\mathbb Z/6\mathbb Z$. List all multiplicative units in each. Explain why removing zero does not always leave a multiplicative group.

Worked solution

Both whole residue sets are additive groups: closure, associativity and identity come from integer addition, and $-a$ is an additive inverse. A multiplicative residue is a unit precisely when its greatest common divisor with the modulus is one. The units modulo five are $1,2,3,4$; modulo six they are $1,5$. Modulo six, $2\cdot3=0$ although both factors are nonzero. The nonzero set is not even closed under multiplication. Modulo five every nonzero residue has an inverse; this ring is a field. A ring, its additive group and its group of units are different structures.

The reduction map $\phi:\mathbb Z\to\mathbb Z/6\mathbb Z$ is given by $\phi(n)=[n]$. Determine its kernel, image and quotient. Solve $2x\equiv2\pmod6$ completely and explain why cancelling 2 modulo 6 fails.

Check after attempting

The map preserves addition and multiplication. Its kernel is $6\mathbb Z$ and its image is all six residues. The quotient identifies integers with the same remainder, so $\mathbb Z/\ker\phi\cong\mathbb Z/6\mathbb Z$. The congruence means $6\mid2(x-1)$, equivalently $3\mid x-1$. Hence $x\equiv1\pmod3$, giving residues $1$ and $4$ modulo six. Both substitute correctly. Cancelling 2 while retaining modulus six would discard 4; 2 has no multiplicative inverse there.