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Abstract algebra
Introduction to groups - suggested problems - solutions
Note:
Sets: R = reals, Q = rationals, Z = integers, N = naturals
Operators: + = addition, − = subtraction, × = multiplication, ÷ = division
Which of the following are groups? Which groups are Abelian? If a set/operation pair is not a group,
which properties does it fail?
P1: < R, ÷ > (reals with division)
Not a group.
– Closure: Immediate failure right here - although for most reals, a real divided by a real
gives a real, there’s one big exception - anything divided by zero does not produce a real
number. Fails other properties fail as well.
– Associative: Fails. Counterexample:
12 ÷ (6 ÷ 2) = 12 ÷ 3 = 4
(12 ÷ 6) ÷ 2 = 2 ÷ 2 = 1
12 ÷ (6 ÷ 2) 6= (12 ÷ 6) ÷ 2
– Identity: You need a number e such that e ÷ x = x ÷ e for all x. Solving
e
x
=
x
e
gives me e2 = x2, or e = ±x, but that’s no good - the identity element has to stay fixed .
You can’t say e is 4 when x is 4, e is 2 when x is 2, and so on - the identity has to keep one
value for the whole set. So no identity element.
– Inverse: No identity means it’s meaningless to look for an inverse.
P2: < Q, + > (rationals with addition)
Group. Abelian group. Rational numbers and addition behave nicely - adding two rationals
gives a rational, addition is associative (and commutative, which is where the Abelian comes
from) for real numbers and all subsets of the reals, the additive identity 0 is in the rationals, and
additive inverses (negatives) are in the rationals.
P3: < Z, × > (integers with multiplication)
Not a group. Fails on multiplicative inverses - the reciprocals of integers are not necessarily
integers (i.e. 13 would be the multiplicative inverse of 3 in the reals, but it’s not here in the set
of integers). There’s also the divide by zero problem for zero.
P4: < N, + > (naturals with addition)
Not a group. Lacks identity - the additive identity would be 0, but 0 6∈ N. Also fails on inverses
- additive inverses of natural numbers would be negative (and not part of the naturals).
P5: < Q0 \ {0}, × > (irrationals, multiplication)
Note that that’s Q0, not Q - these are the irrationals! And removing 0 doesn’t do anything - 0
isn’t irrational to begin with!
The irrationals with multiplication is not a group.
√ √
– Closure: Fails. Counterexample ( 2)( 2) = 2 (a product of two irrationals which is rational - we’ve left the set).
– Associative: This one’s OK - the operation of multiplication is associative for any and all
reals, and the irrationals are part of the reals.
– Identity: Fails. Multiplicative identity would have to be 1 ... but 1 is rational, not irrational.
– Inverse: Meaningless (since no identity), but if you overlook that, you can sort of say they’re
there - the reciprocal of an irrational number is still an irrational number.
P6: < A, ∗ > where A is the set {a, b, c} and ∗ is defined by the table:
– Closure: Passes.
– Associative: Fails. Counterexample:
a ∗ (b ∗ c) = a ∗ c = a
(a ∗ b) ∗ c = b ∗ c = c
a ∗ (b ∗ c) 6= (a ∗ b) ∗ c for all elements
– Identity: Fails. b is the identity along the row, but there is no corresponding column.
– Inverse: Fails. Fails for every element, in fact - you need to have the same inverse on left
and right. So, considering, say a,
a ∗ b = b so a’s right inverse is b
c ∗ a = b so a’s left inverse is c
There is no single element a0 such that a ∗ a0 = a0 ∗ a = e (where e here is b - the identity).
So no, not a group.
P7: < A, ∗ > where A is the set {a, b, c} and ∗ is defined by the table:
– Closure: Passes.
– Identity: Fails. Notice there’s no column in the table that matches the starting set, in order.
And, a suggestion - the associative property is the most annoying one to check, because you have
to consider every combination looking for a counterexample... and if you don’t find one, you’re
still wondering if maybe there’s a combination you missed! If it fails being a group on some easy
to spot property (like identity), that’s it - it’s not a group.
P8: < A, ∗ > where A is the set {a, b, c, d} and ∗ is defined by the table:
I’m am pretty sure this is an Abelian group (how’s that for confidence?). Closed, identity and
inverse are easy to spot- the identity is c, every element has an inverse, AND ∗ commutative
(notice the symmetry over a diagonal fold). But did I check every possible case for associative?
Oh no ... Since there are many more of you looking at this than me, between us all we should
exhaust the possibilities - if anybody spots a counterexample, pass it along!