Groups
Converse of Lagrange’s theorem. A finite group need not have a subgroup of each order dividing its order. The alternating group \(A_4\) has order \(12\) and no subgroup of order \(6\); \(A_5\) has order \(60\) and no subgroup of order \(30\).
Groups of order \(p^3\). A group of order \(p^3\) need not be abelian: the dihedral group \(D_4\) and the quaternion group \(Q_8\) have order \(8\). Every group of order \(p^2\) is abelian.
Groups whose proper subgroups are abelian. A group all of whose proper subgroups are abelian need not be abelian: \(A_4\) is nonabelian, and each of its proper subgroups has order \(1\), \(2\), \(3\), or \(4\), hence is abelian.
Transitivity of normality. In \(A_4\), \(\langle (1\,2)(3\,4)\rangle \normal V_4\) and \(V_4 \normal A_4\), but \(\langle (1\,2)(3\,4)\rangle\) is not a normal subgroup of \(A_4\).
Quotients embedding as subgroups. A quotient of \(G\) need not be isomorphic to a subgroup of \(G\). The quotient \(Q_8/\{\pm 1\}\cong \ZZ/2\times\ZZ/2\) does not embed in \(Q_8\), whose only element of order \(2\) is \(-1\); the quotient \(\ZZ/2\) of \(\ZZ\) does not embed in \(\ZZ\). For a finite abelian group every quotient is isomorphic to a subgroup.
Rings
The classes of rings, each with a ring separating it from the next, are on Which kind of ring is this?.
Irreducible elements. An irreducible element need not be prime; in \(\ZZ[\sqrt{-5}]\), \(3\) is irreducible, and \(9 = 3\cdot 3 = (2+\sqrt{-5})(2-\sqrt{-5})\) with \(3\) dividing neither factor on the right.
Subrings of a PID. A subring of a principal ideal domain need not be one: \(\ZZ[x] \subseteq \QQ[x]\).
Polynomial rings over a PID. If \(R\) is a principal ideal domain, \(R[x]\) need not be: the ideal \(\gens{2,x}\normal\ZZ[x]\) is not principal.
Modules
Torsion-free modules. A torsion-free module over an integral domain need not be free; the ideal \(\gens{2,x}\subseteq \ZZ[x]\) is torsion-free and not free.
Projective modules. A projective module need not be free: \(\ZZ/2\) is a direct summand of \(\ZZ/6\cong \ZZ/2\times\ZZ/3\), hence projective over \(\ZZ/6\), and it is not free because it has \(2\) elements.
Tensor products and injectivity. Tensoring need not preserve injective maps: multiplication by \(2\) on \(\ZZ\) is injective, and tensored with \(\ZZ/2\) it is the zero map on \(\ZZ/2\).
Split exact sequences. An exact sequence need not split; \(0\to\ZZ/2\to\ZZ/4\to\ZZ/2\to 0\), since \(\ZZ/4\not\cong \ZZ/2\times\ZZ/2\).
Linear algebra
Further matrix counterexamples are on Matrix counterexamples.
Characteristic polynomials and similarity. Matrices with the same characteristic polynomial and the same minimal polynomial need not be similar; \(J_2(0)\oplus J_2(0)\) and \(J_2(0)\oplus J_1(0) \oplus J_1(0)\) both have characteristic polynomial \(x^4\) and minimal polynomial \(x^2\).
Real eigenvalues. A real matrix need not have a real eigenvalue: the rotation \(\begin{pmatrix}0&-1\\1&0\end{pmatrix}\) has characteristic polynomial \(x^2+1\).
Diagonalizability over \(\RR\) and \(\CC\). The same rotation is diagonalizable over \(\CC\), with eigenvalues \(\pm i\), and not over \(\RR\).
Fields
Normal extensions. A finite extension need not be normal; \(\QQ(2^{1/3})/\QQ\), since \(x^3-2\) has one root in \(\QQ(2^{1/3})\subseteq\RR\).
Transitivity of normality. \(\QQ(2^{1/4})/\QQ(\sqrt2)\) and \(\QQ(\sqrt2)/\QQ\) are normal, being of degree \(2\), but \(\QQ(2^{1/4})/\QQ\) is not.
Separability. An irreducible polynomial need not be separable; \(x^p - t\) over \(\FF_p(t)\) is irreducible and equals \((x-t^{1/p})^p\) over a splitting field. Every irreducible polynomial over a perfect field, in particular over a field of characteristic zero or a finite field, is separable.
Simple extensions. A finite extension need not be simple; \(\FF_p(s,t)/\FF_p(s^p,t^p)\) has degree \(p^2\), and every element \(\alpha\) of \(\FF_p(s,t)\) satisfies \(\alpha^p\in\FF_p(s^p,t^p)\), so every simple subextension has degree at most \(p\). By the primitive element theorem, every finite separable extension is simple.