FP2 1.1 - Binary operations and group axioms

FP2 1.1 - Binary operations and group axioms

Test a set and an operation against the group axioms (FP2 1.1). You will check closure, associativity, an identity and inverses, and distinguish changing brackets from changing order.

Specify the operation and check closure

Can −1-1 do the job that zero normally does? On the integers, define a new rule by a⋆b=a+b+1a\star b=a+b+1. The extra 11 changes what “do nothing” and “undo” mean. We will find out whether this rule really makes a group.

You already know

Integer arithmetic includes zero and negative integers. Even integers have form 2k2k, and odd integers have form 2k+12k+1, where k∈Zk\in\mathbb Z. A modular operation returns a representative of its residue class. Matrix multiplication uses rows by columns and preserves the order of the factors.

An operation needs a set as well as a rule. Subtraction, for instance, always returns an integer from two integers, but can leave the nonnegative integers.

Binary operation and closure

A binary operation on a set GG assigns one element of GG to each ordered pair of elements of GG.

For a proposed rule, check that it is defined uniquely for every ordered pair, and that it is closed on GG: whenever a,b∈Ga,b\in G, the result a⋆ba\star b is also in GG.

For example, division is not a binary operation on all real numbers because division by zero is undefined. A rule giving two possible outputs would also fail to define an operation. The group terminology and rules in this lesson are not in the formula booklet.

Worked example 1: closure on all integers

Question. Show that a⋆b=a+b+1a\star b=a+b+1 defines a binary operation on Z\mathbb Z.

Plan. Take arbitrary integer inputs and show that the uniquely specified result is an integer.

a,b∈Zarbitrary permitted inputsa+b+1∈Zintegers are closed under additiona⋆b=a+b+1one specified output\begin{aligned} a,b&\in\mathbb Z&&\text{arbitrary permitted inputs}\\ a+b+1&\in\mathbb Z&&\text{integers are closed under addition}\\ a\star b&=a+b+1&&\text{one specified output} \end{aligned}

Why this step? Checking arbitrary inputs proves closure for the entire set; a few successful sums would not do that.

Answer. The rule is defined uniquely for every pair and is closed on Z\mathbb Z, so it is a binary operation on Z\mathbb Z.

Check. 3⋆(−5)=3−5+1=−13\star(-5)=3-5+1=-1, still an integer.

Change only the set to the even integers E={2k:k∈Z}E=\{2k:k\in\mathbb Z\}. The same rule can now fail.

Worked example 2: a set can break closure

Question. Does a⋆b=a+b+1a\star b=a+b+1 define a binary operation on EE?

Plan. Find two members whose output leaves EE.

0,0∈Eboth inputs are even0⋆0=0+0+1=1apply the stated rule1∉Ethe output is odd\begin{aligned} 0,0&\in E&&\text{both inputs are even}\\ 0\star0&=0+0+1=1&&\text{apply the stated rule}\\ 1&\notin E&&\text{the output is odd} \end{aligned}

Why this step? One failed pair disproves closure, even though proving closure requires a reason covering every pair.

Answer. No. The rule is not closed on the even integers.

Check. More generally, 2r⋆2s=2(r+s)+12r\star2s=2(r+s)+1, always odd.

A familiar operation name does not supply closure automatically; the specified set decides which outputs are allowed.

Watch out

The line 0⋆0=1∈E0\star0=1\in E is wrong: 11 is odd. The correct conclusion is 1∉E1\notin E, so closure fails on EE, even though it holds on Z\mathbb Z.

Change the allowed set once more and use general odd inputs.

Associativity changes brackets, not order

Three inputs can be combined in two bracketings. Associativity says that they give the same result while the inputs stay in their original order.

Associativity and commutativity

An operation is associative when, for all a,b,c∈Ga,b,c\in G,

(a⋆b)⋆c=a⋆(b⋆c)(a\star b)\star c=a\star(b\star c)

It is commutative when, for all a,b∈Ga,b\in G,

a⋆b=b⋆aa\star b=b\star a

These are different properties. Associativity is a group axiom; commutativity is not required.

Ordinary addition is associative. That does not automatically make every newly defined rule associative: substitute the rule into both bracketings.

Worked example 3: prove associativity for the shifted sum

Question. Show that a⋆b=a+b+1a\star b=a+b+1 is associative on Z\mathbb Z.

Plan. Expand each bracketing for arbitrary integers a,b,ca,b,c.

(a⋆b)⋆c=(a+b+1)+c+1apply the outer operation=a+b+c+2collect the constantsa⋆(b⋆c)=a+(b+c+1)+1apply the outer operation=a+b+c+2the same expression\begin{aligned} (a\star b)\star c&=(a+b+1)+c+1 &&\text{apply the outer operation}\\ &=a+b+c+2&&\text{collect the constants}\\ a\star(b\star c)&=a+(b+c+1)+1 &&\text{apply the outer operation}\\ &=a+b+c+2&&\text{the same expression} \end{aligned}

Why this step? The equality holds for arbitrary integers, so it proves the universal condition.

Answer. The operation is associative on Z\mathbb Z.

Check. With a=2,b=3,c=4a=2,b=3,c=4, both bracketings give 1111.

Change only the sign of the second input in the rule: a⋄b=a−b+1a\diamond b=a-b+1. Closure on the integers survives, but associativity need not.

Worked example 4: disprove associativity

Question. Is a⋄b=a−b+1a\diamond b=a-b+1 associative on Z\mathbb Z?

Plan. Test both bracketings for a=2,b=3,c=4a=2,b=3,c=4; unequal outputs give a counterexample.

2⋄3=2−3+1=0first inner operation(2⋄3)⋄4=0−4+1=−3left bracketing3⋄4=3−4+1=0second inner operation2⋄(3⋄4)=2−0+1=3right bracketing\begin{aligned} 2\diamond3&=2-3+1=0&&\text{first inner operation}\\ (2\diamond3)\diamond4&=0-4+1=-3&&\text{left bracketing}\\ 3\diamond4&=3-4+1=0&&\text{second inner operation}\\ 2\diamond(3\diamond4)&=2-0+1=3&&\text{right bracketing} \end{aligned}

Why this step? A single unequal pair of bracketed results disproves a property required for every triple.

Answer. The operation is not associative.

Check. The general outputs are a−b−c+2a-b-c+2 and a−b+ca-b+c, which need not agree.

Matrix multiplication is an important reminder: it is associative even though reversing factors can change a product. A group is allowed to behave that way.

Watch out

The line (2⋄3)⋄4=2⋄(3⋄4)(2\diamond3)\diamond4=2\diamond(3\diamond4) is false: the sides are −3-3 and 33. Checking only a⋄ba\diamond b against b⋄ab\diamond a would test commutativity, not associativity.

For another proposed rule, compare the bracketings directly.

Find the identity before the inverses

An identity leaves every element unchanged on both sides. An inverse reverses an element back to that identity. Find the identity first: it may not be the familiar number zero or one.

Continue this lesson.

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