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Define $(xN)(yN)=xyN$. If representatives are changed to $x{n}_1,y{n}_2$, normality moves $n_1$ past $y$ and leaves the product in $xyN$, so multiplication is well-defined.

Let $G^{\prime }$ be the set of finite products of commutators. It is closed under products, and $[x,y{]}^{-1}=[y,x]$, so it is a subgroup. Moreoverso conjugation preserves products of commutators and $G^{\prime }◃G$. In $G/G^{\prime }$, one has $xy{G}^{\prime }=yx{G}^{\prime }$ because $(yx{)}^{-1}xy=[x,y]$, hence the quotient is abelian. Conversely, if $G/N$ is abelian, every commutator maps to the identity, so $G^{\prime }\le N$. This is the universal property of the commutator subgroup.

Write a matrix as $(r,b)$, where $r=a^2$ is a nonzero quadratic residue modulo $11$. The square subgroup isMatrix multiplication becomesThus $H$ is the affine semidirect product of cyclic groups of orders eleven and fiveThere are five choices for $r$ and eleven for $b$, so $|H|=55$. It is nonabelian, since

LetConjugation by the complement givesHence the ten nonidentity translations split into the two orbitseach of size five. For $r\ne 1$,Since $1-r\ne 0$, varying $t$ gives every $b\in {\mathbb{F}}_{11}$. Conjugation cannot change $r$ in the abelian quotient $H/N$. Therefore, withthe seven conjugacy classes areof sizes $1,5,5,11,11,11,11$, agreeing with the conjugacy classes in the affine semidirect product of orders eleven and five.

The commutators with $(4,0)$ generate every translation because multiplication by $4-1=3$ is invertible in ${\mathbb{F}}_{11}$. Hence the commutator subgroup is $N$, and the abelianization isPut $\zeta =e^{2\pi i/5}$. The one-dimensional characters factor through the abelianization, giving five characters

For the remaining characters, put $\omega =e^{2\pi i/11}$ and define a character of $N$ byThe complement has two free orbits on the nontrivial ${\theta }_m$, indexed by $m\in Q$ and $m\in 2Q$. By induction from an abelian normal subgroup with a free character orbit, the charactersare irreducible of degree five and vanish outside $N$.

SetThe quadratic periods modulo eleven give the nonsquare sum $\overline{\eta }=(-1-i\sqrt{11})/2$. Multiplication by a square preserves $Q$ and multiplication by a nonsquare exchanges the two square classes, so the complete character table isFinally,and there are seven rows for the seven conjugacy classes. Thus these are all irreducible characters, as described by the irreducible characters of the affine semidirect product of orders eleven and five.

Schur lemma says that an intertwining linear map between irreducible finite-dimensional complex representations is either zero or an isomorphism. In particular, every endomorphism of an irreducible complex representation is a scalar multiple of the identity.

A continuous representation of a topological group is a continuous homomorphismfor a finite-dimensional complex vector space $V$. It is a unitary representation if $V$ has a positive-definite Hermitian inner product for whichfor every $g\in G$ and $v,w\in V$.

For $G=S^1$, start with any positive-definite Hermitian form and average it using normalized Haar measure:Translation invariance makes this form $S^1$-invariant, and positivity is preserved, proving unitarity. Since $S^1$ is abelian, the operators $\rho (z)$ commute; since they are unitary, they are normal. Simultaneous diagonalization therefore decomposes $V$ into common one-dimensional eigenspaces. Thus every representation of the circle group is a direct sum of one-dimensional representations.

WriteThe group law and inverse areA calculation givesEvery element of the centre $Z$ occurs by taking, for example, $x^{\prime }=y=0$ and $y^{\prime }=1$. Hence the commutator subgroup is exactly $Z$. The image of a one-dimensional representation is abelian, so the one-dimensional representation kills the commutator subgroup and its kernel contains $Z$.

Now let $(\rho ,V)$ be a complex representation of $G/Z_0$. Its restriction to the central subgroupis a representation of the circle group. Decompose it into its distinct weight spaces:Because $Z/Z_0$ is central, every $\rho (g)$ commutes with its action and preserves every common eigenspace. Thus the $V_i$ are $G/Z_0$-subrepresentations, as asserted by the central circle weight-space decomposition.

Let $n_i=\dim V_i$. The mapis a one-dimensional representation of $G/Z_0$; after composition with $G\to G/Z_0$, its kernel contains $Z$. On $z\in Z/Z_0$ its value isso ${\theta }_i^{n_i}=1$. A continuous character of $S^1$ has the form $z\mapsto z^m$; the displayed identity forces $m=0$. Therefore every ${\theta }_i$ is trivial. Since the ${\theta }_i$ were distinct, $d=1$ and ${\theta }_1=1$.

It follows that every finite-dimensional complex representation of $G/Z_0$ kills the entire nontrivial central circle $Z/Z_0$. Its kernel is therefore nontrivial, so no such representation is faithful. This is the real Heisenberg quotient has no faithful finite-dimensional representation.

The rotations form a cyclic subgroup $R\cong C_n$ of index two, so it is normal. The quotient $D_{2n}/R$ has order two and is cyclic. Hence $1⊴R⊴D_{2n}$ has abelian factors, proving $\boxed{D_{2n}\text{ is solvable}}$. Its commutator subgroup is contained in the abelian rotation subgroup, so its second derived subgroup is already trivial.

The cyclic subgroup $⟨x^3⟩$ is normal: conjugation by $y$ sends $x^3$ to $x^{12}=x^3$, and conjugation by $x$ fixes it. Modulo this subgroup, the relation $yx{y}^{-1}=x^4$ becomes $yx{y}^{-1}=x$, so the quotient group is Abelian. Thus the commutator subgroup $G^{\prime }$ is contained in $⟨x^3⟩$.

Conversely the displayed relation gives the commutator $yx{y}^{-1}{x}^{-1}=x^3$, so $⟨x^3⟩\subseteq G^{\prime }$. HenceSince $x$ has element order $9$, $x^3\ne e$, so the first commutator subgroup is nontrivial. ThereforeHere derived length counts the number of steps needed for the derived series to reach the trivial group.

For dimension at least three, $[I+a{E}_{ik},I+E_{kj}]=I+a{E}_{ij}$ for distinct $i,j,k$, with $[x,y]=x^{-1}{y}^{-1}xy$. In dimension two over a field with more than three elements, choose $s$ with $s^2\ne 1$ and commute an elementary matrix with $\mathrm{diag}(s,s^{-1})$. In both cases all generating transvections lie in the commutator subgroup, so the special linear group over a finite field is perfect outside ${\mathrm{SL}}_2({\mathbb{F}}_2)$ and ${\mathrm{SL}}_2({\mathbb{F}}_3)$.

// Target: algebra.bigb

A finite p-group is powerful if $[G,G]\le G^p$ for odd $p$, or $[G,G]\le G^4$ for $p=2$. Here powers denote generated subgroups. The same definition applies to pro-p groups using closed power and commutator subgroups. Its Frattini subgroup is $G^p$. For odd $p$, successive power layers admit linear correction of roots, which makes every element of $G^p$ an actual $p$th power.

A pro-p group whose closed commutator subgroup lies in its closed power subgroup $G^{p^{ϵ}}$, with $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. This is the profinite analogue of a powerful p-group. Finitely generated torsion-free examples are exactly the uniform pro-p groups. The additive group of a finite product of p-adic integers is an abelian example.

The intersection of the maximal proper open subgroups of a profinite group. A subset topologically generates the profinite group exactly when its image topologically generates its quotient by this intersection: a proper closed generated subgroup is contained in a maximal open subgroup, obtained from a suitable finite quotient. For a pro-p group, $\Phi (G)=\overline{G^p[G,G]}$, where the bar denotes closure of the algebraically generated power and commutator subgroups. Closedness cannot simply be discarded in infinite rank.

For a nontrivial simple soluble group, the commutator subgroup is either the identity or the whole group. It cannot be the whole group because the derived series would never terminate. The group is therefore Abelian. Every nonidentity element then generates a normal cyclic subgroup, which must be the whole group. An infinite cyclic group has proper nontrivial subgroups; a finite cyclic group is simple exactly when its order is prime. Thus the soluble simple groups are precisely the cyclic groups of prime order.

The subgroup generated by elementary transvections in finite matrix sizes, with identity stabilization. A third distinct index expresses each transvection as a commutator; stable block identities show that this subgroup is normal and that its quotient in the stable general linear group is abelian. Thus it is precisely the commutator subgroup.

Every abstract subgroup of finite index in a finitely generated pro-p group is open. Fixed-generator commutator collection in nilpotent groups makes the abstract commutator subgroup closed. The resulting abelian quotient is a quotient of a finite power of the p-adic integers, so its subgroup of $p$th powers is closed and of finite index. Thus the algebraic $G^p[G,G]$ is open and every abstract index-$p$ subgroup is open. An arbitrary finite quotient is a finite p-group; induction on its order, using a normal subgroup of index $p$ and the Schreier lemma, gives the assertion. The normal core reduces nonnormal finite-index subgroups to this case.

A group $G$ is two-step nilpotent whenEquivalently, its commutator subgroup is contained in the center of a group of $G$. This allows every word to be collected into powers of generators followed by powers of their pairwise commutators.