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A field $k$ is a splitting field for a finite group $G$ if every simple module over its group algebra $kG$ is absolutely simple: it remains simple over every extension field. Equivalently, the quotient by the Jacobson radical is a product of full matrix algebras over $k$. In characteristic $p$, containing all $m$th roots of unity, where $|G|=p^{a}m$ and $p\nmid m$, is sufficient. This is the modular splitting-field theorem; it is substantially stronger than merely having eigenvalues for one chosen group element.

The stabilizer of $x$ is the subgroup $\{g:gx=x\}$.

A subgroup is a subset of a group that is itself a group under the restricted operation.

For finite sets $A,B\subseteq G$ in an abelian group $G$, their sumset isEach distinct value occurs only once in this set, even if many ordered pairs produce it. When $B=A$, repetitions are allowed: $a+a$ belongs to $A+A$. For the two integer sets introduced above,For $A$, the value $3$ comes from $(0,3),(1,2),(2,1),(3,0)$. For $B$, two ordered pairs give the same sum only when one reverses the other, apart from a pair being identical to itself. Thus its four pairs $(b,b)$ and six pairs with distinct members, counted without regard to order, produce ten distinct sums.

Every nonempty finite set of integers $S\subseteq \mathbb{Z}$ satisfies $|S+S|\ge 2|S|-1$. To see this, list its members as $s_1<\cdots <s_n$. The $2n-1$ distinct sumsall belong to $S+S$. A nonconstant arithmetic progression attains this bound. The ordered list is essential: in the cyclic group $\mathbb{Z}/6\mathbb{Z}$, the subgroup $H=\{0,2,4\}$ instead satisfies $H+H=H$, of cardinality three. Addition wraps around, for example $4+4=2$ in this group.

Suppose a finite group $G$ acts by Riemannian isometries on a compact Riemannian manifold $M$, and Gassmann equivalent subgroups $H_1,H_2$ act freely. The quotient manifolds $H_i\backslash M$ have equal Laplace-Beltrami operator spectra. For an eigenfunction space $E$, the quotient multiplicity is $|H_i{|}^{-1}{\sum }_{h\in H_i}\mathrm{tr}(h|E)$, and the equality follows by summing its character of a representation over conjugacy classes. Nonconjugacy inside $G$ alone does not guarantee that the quotients are nonisometric.

A Sylow $p$-subgroup of a finite group has the largest power of $p$ dividing the group order.

A subset $A$ of a multiplicatively written group is symmetric when it contains the identity element and $A^{-1}=A$. Consequently $⟨A⟩={\bigcup }_{m\ge 1}{A}^m$.

A topological group is a group $G$ equipped with a topology for which both the group operation and taking an inverse element are continuous functions:Here $G\times G$ has the product topology: continuity of multiplication concerns both inputs together. Thus nearby motions compose to nearby motions, and undoing a motion also varies continuously. The definition alone does not impose compactness; we state the Hausdorff space assumption explicitly when it is needed.

The additive group $(\mathbb{R},+)$ of real numbers, with its usual topology, is a topological group: $(x,y)\mapsto x+y$ and $x\mapsto -x$ are continuous functions. Its identity element is $0$. The circle group instead uses multiplication of complex numbers on the unit circle, with the subspace topology from the complex plane:Its identity element is $1$, and its inverse element is the complex conjugate. Multiplication on the complex plane and complex conjugation are continuous functions, so their restrictions verify the required continuity. Angles add, but points multiply: a quarter-turn followed by another quarter-turn gives $i\cdot i=-1$, and the next two quarter-turns return to $1$. Ordinary addition would fail to stay on the unit circle, since $1+1=2$.

For any fixed $a\in G$, the translations $x\mapsto ax$ and $x\mapsto xa$ are homeomorphisms, with inverses given by translation by $a^{-1}$. Consequently the neighbourhoods of the identity element determine the entire topology. This is why local information near one motion can control a whole family of symmetries. See Tornier, Haar Measures, section 1.1 for these basic properties of topological groups.

A torsion-free group is a group in which the identity element is the only element of finite order of a group element.

If an infinite group has a single nonidentity conjugacy class, it is a torsion-free group. Otherwise every nonidentity element has the same finite order $n$. Powers show $n$ is prime. For $n\ge 3$, choose $xg{x}^{-1}=g^2$; then $x^n=1$ gives $g=g^{2^n}$, impossible by Fermat's little theorem. For $n=2$, the group has exponent two and is abelian, so every conjugacy class is a singleton and the hypothesis would force the group to have only two elements. Infinitude excludes this final case.

A group is a torsion group when every element is a torsion element, meaning that each element has finite order. The orders need not have a common finite bound. A torsion group has no nontrivial torsion-free group as a subgroup; in particular it cannot contain a nonabelian free group. Infinite finitely generated groups of this kind can be constructed using a torsion group construction by p-power relators.

Enumerate all nonidentity words $w_i$ in the rank-two free group and impose relations $w_i^{p^{i+2}}=1$. The resulting two-generated group hasIt is infinite by p-deficiency at least one implies infinitude. Each element has order a power of $p$, so it is a torsion group. The infinitely many relators are essential to this particular construction; finite generation does not imply a finite group presentation.

For a $p$-subgroup $D$ of a finite group $G$, set $I_D={\mathrm{Tr}}_D^G((kG{)}^D)$, a transfer ideal of conjugation-fixed elements in $Z(kG)$. A block of a group algebra idempotent $b$ lies in $I_D$ if and only if a defect group of a block of $b$ is conjugate to a subgroup of $D$. In particular, defect groups can equivalently be defined as minimal subgroups $D$ with $b\in I_D$.

Here is a proof using the Brauer morphism. A subgroup of smallest order with $b\in I_D$ exists, since a Sylow subgroup $P$ gives $b={\mathrm{Tr}}_P^G(b/[G:P])$. Write $b={\mathrm{Tr}}_D^G(a)$ and replace $a$ by $ba$. If ${\mathrm{Br}}_D(b)=0$, then $ba$ belongs to the kernel of the Brauer morphism, which is the sum of proper-subgroup relative traces. Hence $b\in {\sum }_{Q<D}{I}_Q$. The algebra $bZ(kG)$ is a commutative Artinian ring and a local ring, so multiplying this equality by $b$ forces some ideal $b{I}_Q$ to contain its identity $b$. This contradicts minimality. On the other hand, ${\mathrm{Br}}_E(I_D)=0$ unless $E$ is conjugate into $D$: the $E$-action on $G/D$ has no fixed coset otherwise, and nonfixed orbit sizes vanish in characteristic $p$. Thus these minimal subgroups are precisely the maximal subgroups with nonzero Brauer image. They are all conjugate. Trace transitivity proves the stated criterion for larger $D$.

A group action on a set $X$ is transitive when every two points lie in the same orbit of a group action, equivalently when for all $x,y\in X$ there is a group element $g$ with $gx=y$.