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The Schur multiplier of a group $G$ isthe second group homology group with trivial integral coefficients. If $G\cong F/R$ is a free presentation, Hopf's formula states that

Write $I_F=\ker (\mathbb{Z}F\to \mathbb{Z})$ for the augmentation ideal. The presentation relation sequence iswhere $r[R,R]\mapsto 1\otimes (r-1)$. If $F$ is free on a set $S$, then $I_F$ is free as a left $\mathbb{Z}F$-module on the elements $s-1$, so the two modules immediately preceding $\mathbb{Z}$ are free $\mathbb{Z}G$-modules. Resolving the relation module $R/[R,R]$ by free modules and splicing produces a free resolution of $\mathbb{Z}$.

Apply the right-exact functor $\mathbb{Z}{\otimes }_{\mathbb{Z}G}-$ to this partial resolution. Its degree-two homology is the kernel ofThe coinvariant module on the left is $R/[F,R]$. On the right, the map $f-1\mapsto f[F,F]$ identifies the coinvariants with the abelianization $F/[F,F]$. The displayed map is induced by the inclusion $R\hookrightarrow F$, so its kernel isThis proves Hopf's formula.

For an abelian group $A$, the Schur multiplier of an abelian group is $M(A)\cong {\bigwedge }^{2}A$. One way to see the direct-sum rule is the degree-two Künneth theorem:A cyclic group has zero second integral group homology, whileConsequently

The hidden subgroup problem for a group $G$ is specified by an oracle $f:G\to S$ that hides an unknown subgroup $H\le G$: it is constant on each left coset of $H$ and takes distinct values on distinct cosets. The task is to determine $H$, usually by finding a generating set for it.

A complex representation of $G$ is a complex vector space $V$ together with a homomorphismIt is faithful when $\ker \rho =\{e\}$, equivalently when $\rho$ is injective. Every finite group has a faithful complex representation: in the regular representation, $G$ permutes the basis $\{e_h:h\in G\}$ byand a group element fixing every basis vector must be the identity.

If $g$ is conjugate to $g^k$, then $\rho (g)$ is similar to $\rho (g^k)=\rho (g{)}^k$. Thus the two matrices have the same spectrum, while an eigenvalue $\lambda$ of $\rho (g)$ gives the eigenvalue ${\lambda }^k$ of $\rho (g{)}^k$. Hence

Now take a $p$-cycle $g\in S_p$. It is conjugate to $g^k$ for every $1\le k<p$. Faithfulness makes $\rho (g)$ have order $p$, so its spectrum contains a nontrivial $p$th root of unity $\lambda$. The preceding implication puts all the $p-1$ distinct valuesin the spectrum. Therefore $\dim V\ge p-1$. This is the spectrum orbit bound for a faithful symmetric-group representation.

Finally let $G$ be the group of all permutations of $\mathbb{N}$. It has no faithful finite-dimensional complex representation. For every $n$, $G$ contains a subgroup $(C_2{)}^n$ generated by $n$ disjoint transpositions. In a $d$-dimensional complex representation, commuting involutions are simultaneously diagonalizable and hence map into the group of diagonal sign matrices, which has order $2^d$. A faithful restriction would require $2^n\le 2^d$ for every $n$, an impossibility.

A covering map is locally a disjoint union of homeomorphisms onto the base. A space is simply connected when it is path-connected and every loop is null-homotopic. By the uniformization theorem, the simply connected Riemann surfaces areup to analytic isomorphism.

A lattice is a discrete subgroupwith ${\omega }_1/{\omega }_2\notin \mathbb{R}$. The Weierstrass elliptic functionconverges normally on compact subsets of $\mathbb{C}\setminus L$, defines a meromorphic map $\mathbb{C}\to {\mathbb{C}}_{\infty }$, and is nonconstant because it has a double pole at every lattice point. Reindexing the normally convergent derivative series shows that ${\wp }_L^{\prime }$ is $L$-periodic; evenness fixes the integration constants, so ${\wp }_L(z+\omega )={\wp }_L(z)$.

The invariance makeswell-defined and unique on $\mathbb{C}/L$. It is analytic because the quotient projection is locally biholomorphic.

Neither map is a covering. For any nonzero $\omega \in L$, periodicity and oddness of ${\wp }_L^{\prime }$ giveso ${\wp }_L^{\prime }(\omega /2)=0$. The map is not a local homeomorphism there; since the quotient projection is locally biholomorphic, the descended map fails for the same reason.

There is no covering $\mathbb{C}/L\to {\mathbb{C}}_{\infty }$. The torus is connected, while every connected covering of the simply connected sphere is a homeomorphism. A torus is not homeomorphic to a sphere, for example because their fundamental groups are ${\mathbb{Z}}^2$ and $0$.

The character of $(\rho ,V)$ isThe stated action is a representation because successive action by $(g_2,h_2)$ and $(g_1,h_1)$ gives

Identifying ${\mathrm{Hom}}_{\mathbb{C}}(V,W)$ with $V^{\ast }\otimes W$ gives the character of a Hom representationHere the final equality follows after unitarizing the finite-group representation.

The permutation character of $\mathbb{C}G$ counts fixed points:Thus

On the other hand, the character of
${⨁}_i\mathrm{Hom}(V_i,V_i)$ isThe column form of character orthogonality says that this has exactly the same two values above. Complex representations of a finite group are semisimple, and semisimple representations with the same character are isomorphic. Hence the two-sided regular representation decomposition isas a $G\times G$ representation.

The Dirichlet unit theorem states that for a number field of signature $(r_1,r_2)$,

For $K=\mathbb{Q}(\sqrt{5})$,The field is real quadratic, so the unit rank is one and its only roots of unity are $\pm 1$. To see that $\varphi$ is fundamental, suppose a positive unit satisfies $1<u<\varphi$. If its norm is $1$, its integral trace $u+u^{-1}$ lies strictly between $2$ and $\sqrt{5}$; if its norm is $-1$, its trace $u-u^{-1}$ lies strictly between $0$ and $1$. Neither interval contains an integer. Reducing any positive unit by a suitable power of $\varphi$ now proves the units of the quadratic field Q square root of five formula

If ${\mathcal{O}}_L^{\times }/{\mathcal{O}}_K^{\times }$ is finite, the two unit groups have the same rank, namely one. Put $d=[L:K]$ and let $(r_1,r_2)$ be the signature of $L$. ThenTherefore $r_1=4-2d$ and $r_2=2d-2$. For a proper extension, nonnegativity forceswith signature $(0,2)$.

This degree occurs: take $L=K(i)$. It is a totally imaginary quadratic extension of $K$, so the unit ranks agree and the quotient is finite. It is nontrivial because $i\in {\mathcal{O}}_L^{\times }$ but $i\notin {\mathcal{O}}_K^{\times }$ (and its coset has order two). This is the finite relative unit quotient over a real quadratic field example.

The Seifert-van Kampen theorem says that if $X=U\cup V$, where $U,V,U\cap V$ are path connected open sets containing $x_0$, then

The cell attachment is the quotientwith the image of $x_0=f(\ast )$ as base point. For $n=2$, take one open set that deformation retracts onto $X$ together with a boundary collar, and another that consists of the interior of the disc with a collar and is contractible. Their intersection deformation retracts onto $S^1$. Its generator maps to $[f]$ in the first set and to the identity in the second. Van Kampen therefore proves the fundamental group after attaching a 2-cell formula

Applying van Kampen to the two circles givesAttach three discs along loops representing $a^2$, $b^3$, and $(ab{)}^2$. The resulting presentation complex for the symmetric group on three letters has groupSending $a$ to $(12)$ and $b$ to $(123)$ gives a surjection onto $S_3$. The relations imply $aba=b^{-1}$, so every word reduces to one ofThe presented group has at most six elements and surjects onto the six-element group $S_3$, so this map is an isomorphism.

Sincethe phase and group velocities are

The primitive root ${\zeta }_n$ is a zero of the cyclotomic polynomial ${\Phi }_n$, of degree $\phi (n)$. Its minimal polynomial over $K$ divides ${\Phi }_n$, sowhere divisibility also follows from the embedding of the Galois group of the cyclotomic field into $(\mathbb{Z}/n\mathbb{Z}{)}^{\times }$. Over $\mathbb{Q}$, Gauss's irreducibility theorem for cyclotomic polynomials says that ${\Phi }_n$ is irreducible. Therefore

LetBoth have splitting field $\mathbb{Q}(\sqrt{2},\sqrt{3})$. On the roots of $f$, the Galois group has two orbits of size two and embeds asThe roots of $g$, namely $\pm \sqrt{2}\pm \sqrt{3}$, form one orbit and give the regular Klein-four subgroupThe subgroups are not conjugate because their orbit decompositions differ.

Let $F$ be the splitting field over ${\mathbb{F}}_p$ of $X^q-X$. Its derivative is $-1$, so it has $q$ distinct roots. The roots are closed under addition, subtraction, multiplication, and inversion, using $(x+y{)}^q=x^q+y^q$ and $(xy{)}^q=x^q{y}^q$. They therefore form a field with $q$ elements.

Every field with $q$ elements has multiplicative group of order $q-1$, so every element satisfies $x^q=x$; it is therefore a splitting field of the same polynomial and is unique up to isomorphism. If an irreducible factor over ${\mathbb{F}}_p$ has a root $\alpha \in {\mathbb{F}}_q$, its degree is $[{\mathbb{F}}_p(\alpha ):{\mathbb{F}}_p]$, which divides $d$ by the tower law. In particular it is at most $d$.

Dedekind's factorization/Frobenius theorem says that for a prime not dividing the discriminant, the degrees of the distinct irreducible factors modulo $p$ give the cycle lengths of an element of the Galois group.

The polynomial is irreducible over $\mathbb{Q}$ by the rational-root test followed by a comparison of possible monic quadratic factors. Its discriminant is $547^2$, a square, so its transitive Galois group lies in $A_4$. Modulo five,and the cubic has no root in ${\mathbb{F}}_5$, giving a three-cycle. A transitive subgroup of $A_4$ containing a three-cycle cannot be the Klein four group; hence

The topological group $S^1=\{z\in \mathbb{C}:|z|=1\}$ has complex multiplication and the subspace topology. It is compact and abelian, so every finite-dimensional irreducible complex representation is one-dimensional. A continuous character lifts along $t\mapsto e^{it}$ to a continuous homomorphism $\mathbb{R}\to {\mathbb{C}}^{\times }$, hence has the form $t\mapsto e^{i\lambda t}$. Periodicity by $2\pi$ forces $\lambda =n\in \mathbb{Z}$, giving precisely $z\mapsto z^n$.