**Counterexamples in Algebra**. So, I thought, why not make a small sort-of compendium of some of these.

- A group of infinite order, which has subgroups of every order.
- Two isomorphic subgroups of a finite group, whose quotient groups are not isomorphic.
- A field that is properly contained in the Real field and which properly contains the field of Rational numbers.
- A Euclidean domain which is not a field.
- A Principal Ideal domain that is not Euclidean.
- A Unique Factorization domain that is not a Principal Integral domain.
- An integral domain that is not a Unique Factorization Domain, i.e. a reducible element which can be factored in two non-trivial ways.
- A non-abelian group of order p^n where n is a triangular number and p is prime.
- A group which is isomorphic to a proper subgroup of itself. (Hint: example 1)
- A non-abelian group G, wherein the relations (ab)^n = (a^n)(b^n) and (ab)^(n+1) = (a^(n+1)).(b^(n+1)) holds true for every a and b in G

- The group of all complex roots of unity, under multiplication, S = { z, a complex number | z^n =1 for some natural number, n}
- Let G = Z2 x Z4 = {0,1} x {0,1,2,3}. Let H = Z2 x 0 = { (0,0), (1,0) } and K = 0 x <2> = { (0,0), (0,2) }, Then, H and K, being groups of order 2 are isomorphic but G/H is Z4 and G/K is Z2 x Z2 and clearly both are not isomorphic.
- Q(√D) = { a+b√D where a and b are rational numbers and D ≡ 1 (mod 4) is positive square-free integer}. This is properly contained in R and properly contains Q. One more interesting example is of the field of algebraic numbers, i.e. real numbers which are the roots of a monic polynomial with rational coefficients. (Real numbers which are not algebraic are trancendental, e.g. e, Π)
- Simplest example- Z, the set of integers. Also, polynomials in one variable over integers, i.e. Z[x].
- The quadratic integer ring Z(√D) = { a + b√D where a, b are integers and D = (1+√-19) / 2) } is a PID but is not Euclidean. For a proof, see page 282 of Dummit & Foote’s book here.
- Z[x] is a Unique Factorization domain (since Z is) but is not Principal Ideal domain. This is because the ideal (2,x) is properly contained in Z[x] but ideal generated by the gcd of 2 and x is (1) = Z[x], which is the whole ring.
- The element 6 in Z[√-5] can be factored in two ways as shown:- 6 = (2).(3) = (1+√-5).(1-√-5) Hence Z[√-5] is not UFD.
- n is a triangular number, so let n = k.(k+1)/2. Consider the set of all k x k upper triangular matrices with 1 on the main diagonal and remaining entries coming from (Z/pZ) = {0, 1, …, p-1}. Order of the group is number of ways of choosing n numbers from Z/pZ allowing repititions, which is p^n.
- Since the group is isomorphic to its proper subgroup, its order is infinite. The complex roots of unity will do the job for us. Fix a prime, p. Let S be the set of roots of unity, whose some ‘p’th power is unity. Then the function Φ : S –> S defined by Φ(w) = w^p is a group homomorphism with a non-trivial kernel (consisting of all ‘p’th roots of unity).
- I don’t have an answer to this one. If you have, please let me know. For three consecutive integers, the group G can be proved to be abelian.

## 2 comments

Comments feed for this article

October 16, 2009 at 13:22

venkateshthere is no such abelian group where those relations hold(answer for question 10).

October 16, 2009 at 14:32

abhishekparabThe question has a glitch: I should have told the problem as:

“Find a non-abelian group such that there is an with the following two equations holding for every elements :