# Contents

## Group Theory (including quasigroups, semigroups, etc)

1. A non-abelian group, all of whose subgroups are normal:

$Q≔⟨a,b\mid {a}^{4}=1,{a}^{2}={b}^{2},ab=b{a}^{3}⟩$Q \coloneqq \langle a, b | a^4 = 1, a^2 = b^2, a b = b a^3 \rangle
2. A finitely presented?, infinite, simple group

Thomson's group? T.

3. A group that is not the fundamental group of any 3-manifold.

${ℤ}^{4}$\mathbb{Z}^4
4. Two finite non-isomorphic groups with the same order profile.

${C}_{4}×{C}_{4},\phantom{\rule{2em}{0ex}}{C}_{2}×⟨a,b,\mid {a}^{4}=1,{a}^{2}={b}^{2},ab=b{a}^{3}⟩$C_4 \times C_4, \qquad C_2 \times \langle a, b, | a^4 = 1, a^2 = b^2, a b = b a^3 \rangle
5. A quasigroup that is not isomorphic to any loop.

$\left\{a,b,c\right\}$ with multiplication table:

$\begin{array}{cccc}*& a& b& c\\ a& a& c& b\\ b& c& b& a\\ c& b& a& c\end{array}$\begin{matrix} * & a & b & c \\ a & a & c & b \\ b & c & b & a \\ c & b & a & c \end{matrix}
6. A counterexample to the converse of Lagrange's theorem.

The alternating group? ${A}_{4}$ has order $12$ but no subgroup of order $6$.

7. A finite group in which the product of two commutators is not a commutator.

$G=⟨\left(ac\right)\left(bd\right),\left(eg\right)\left(fh\right),\left(ik\right)\left(jl\right),\left(mo\right)\left(np\right),\left(ac\right)\left(eg\right)\left(ik\right),\left(ab\right)\left(cd\right)\left(mo\right),\left(ef\right)\left(gh\right)\left(mn\right)\left(op\right),\left(ij\right)\left(kl\right)⟩\subseteq {S}_{16}$G = \langle (a c)(b d), (e g)(f h), (i k)(j l), (m o)(n p), (a c)(e g)(i k), (a b)(c d)(m o), (e f)(g h)(m n)(o p), (i j)(k l)\rangle \subseteq S_{16}
8. A finitely generated group? with a non-finitely generated subgroup.

The free group on two generators $x$ and $y$ has commutator subgroup freely generated by $\left[{x}^{n},{y}^{m}\right]$.

9. An Artinian but not Noetherian $ℤ$-module.

A Prüfer group. (The correct theorem is that an Artinian ring is Noetherian.)

## Ring Theory

1. A ring that is right Noetherian but not left Noetherian:

Matrices of the form $\left[\begin{array}{cc}a& b\\ 0& c\end{array}\right]$ where $a\in ℤ$ and $b,c\in ℚ$.

2. A ring that is local commutative Noetherian but not Cohen-Macaulay

$k\left[x,y\right]/\left({x}^{2},xy\right)$k[x,y]/(x^2, x y)
3. A number ring? that is a principal ideal domain that is not Euclidean.

$ℚ\left(\sqrt{-19}\right)$\mathbb{Q}(\sqrt{-19})
4. An epimorphism of rings that is not surjective.

$ℤ\to ℚ$\mathbb{Z} \to \mathbb{Q}
5. A ring whose spec has non-open connected components.

$\prod _{n=1}^{\infty }{𝔽}_{2}$\prod_{n=1}^\infty \mathbb{F}_2
6. A non-Noetherian ring $A$ such that all local rings on $\mathrm{Spec}\left(A\right)$ are Noetherian.

$\prod _{n=1}^{\infty }{𝔽}_{2}$\prod_{n=1}^\infty \mathbb{F}_2
7. A number field whose ring of integers is Euclidean but not norm-Euclidean.

$ℚ\left(\sqrt{69}\right)$\mathbb{Q}(\sqrt{69})

## Hopf Algebras

1. A non-commutative and non-cocommutative Hopf algebra

$\begin{array}{rlr}H& ≔& ⟨x,g\mid {g}^{2}=1,{x}^{2}=0,gxg=-x⟩\\ \Delta \left(g\right)& =& g\otimes g,\\ \Delta \left(x\right)& =& x\otimes 1+g\otimes x,\\ ϵ\left(g\right)& =& 1,\\ ϵ\left(x\right)& =& 0,\\ S\left(g\right)& =& g,\\ S\left(x\right)& =& -gx\end{array}$\begin{aligned} H &\coloneqq &\langle x, g | g^2 = 1, x^2 = 0, g x g = -x\rangle \\ \Delta(g) &= &g \otimes g, \\ \Delta(x) &= &x \otimes 1 + g \otimes x, \\ \epsilon(g) &=& 1, \\ \epsilon(x) &=& 0, \\ S(g) &= &g, \\ S(x) &= &- g x \end{aligned}

## Homological Algebra

1. An exact sequence that does not split:

$0\to ℤ\stackrel{×2}{\to }ℤ\to ℤ/2ℤ\to 0$0 \to \mathbb{Z} \stackrel{\times 2}{\to} \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z} \to 0

## Galois Theory

1. A polynomial, solvable in radicals, whose splitting field? is not a radical extension? of $ℚ$.

Take any cyclic cubic; that is, any cubic with rational coefficients, irreducible over the rationals, with Galois group cyclic of order $3$.

2. A composition of two normal extensions need not be normal:

$ℚ\subset ℚ\left({2}^{1/2}\right)\subset ℚ\left({2}^{1/4}\right)$\mathbb{Q} \subset \mathbb{Q}(2^{1/2}) \subset \mathbb{Q}(2^{1/4})

## References

The initial import of counterexamples in this entry was taken from this MO question.

Revised on January 31, 2012 18:22:33 by Todd Trimble (74.88.146.52)