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Choose an angle $2\pi \alpha$ with $\alpha \notin \mathbb{Q}$. Rotation through that angle has infinite order, so the subgroup it generates is isomorphic to $\mathbb{Z}$.

The group also contains a copy of ${\mathbb{Z}}^2$. Choose real numbers $\alpha ,\beta$ such that $1,\alpha ,\beta$ are linearly independent over $\mathbb{Q}$, for example $\alpha =\sqrt{2}$ and $\beta =\sqrt{3}$. DefineThis is a homomorphism. If $\varphi (m,n)=I$, then $m\alpha +n\beta \in \mathbb{Z}$, and the stated rational independence forces $m=n=0$. Thus $\varphi$ is injective and its image is isomorphic to ${\mathbb{Z}}^2$.

If a homomorphism $A_5\to \{\pm 1\}$ were surjective, its kernel would be a normal subgroup of index two and hence order $30$. This contradicts the simplicity of $A_5$. Therefore no such surjective homomorphism exists.

The conjugacy classes in $A_5$ consist of the identity, the $15$ double transpositions, the $20$ three-cycles, and two classes of $12$ five-cycles. Thus their sizes areA normal subgroup is a union of conjugacy classes containing the identity, and its order must divide $60$ by Lagrange's theorem. Checking sums of a subset of $15,20,12,12$ shows that no numberdivides $60$. Hence the only possible normal-subgroup orders are $1$ and $60$. This proves the Simplicity of the alternating group A5.

Let $\varphi :A_5\times A_5\to \{\pm 1\}$ be a homomorphism. Its restrictions to $A_5\times \{1\}$ and $\{1\}\times A_5$ cannot be nontrivial, because any nontrivial subgroup of $\{\pm 1\}$ is the whole group and part (iii) rules out such a surjection. Both restrictions are therefore trivial. Since$\varphi$ itself is trivial and in particular is not surjective.

Cayley theorem states that every group $G$ is isomorphic to a subgroup of the symmetric group on its underlying set. For $g\in G$, defineEach $L_g$ is a permutation, and $L_g{L}_h=L_{gh}$, so $g\mapsto L_g$ is a homomorphism. If $L_g$ is the identity permutation, then $g=L_g(1)=1$; hence the homomorphism is injective. For finite $G$ of order $n$, this embeds $G$ in $S_n$.

Because $n$ is not a prime power, distinct primes $p<q$ divide $n$. By Cauchy's theorem, $G$ has subgroups $H,K$ of orders $p,q$. Combine the two coset actions to obtainThe kernel of the first action is contained in $H$, and that of the second is contained in $K$. Their intersection is trivial because $H\cap K=\{1\}$, so the combined action is faithful.

Since $p\ge 2$, $q\ge 3$, and $n\ge 6$,Adding fixed points embeds this symmetric group into $S_{n-1}$. Thus every group of non-prime-power order $n$ is a subgroup of $S_{n-1}$, as summarized by symmetric-group embedding at one less than the group order.

A Sylow $2$-subgroup of $A_5$ has order four. Since $A_5$ has no element of order four, each is a Klein four-group. For every $i\in \{1,\dots ,5\}$, take the identity and the three double transpositions fixing $i$. For instance, the subgroup fixing $5$ isThese five point stabilizers are precisely the Sylow $2$-subgroups. Indeed, the $15$ double transpositions occur three to each such subgroup, so there areof them. Together with the previous parts, this gives the Sylow subgroups of S3, S4 and A5.

A Sylow $2$-subgroup of $S_4$ has order $8$. For example,It is the normalizer of $⟨(1234)⟩$. The six $4$-cycles form three inverse pairs, and a dihedral group of order eight has a unique cyclic subgroup of order four. Hence the three Sylow $2$-subgroups are

Let $|G|=p^{a}m$ with $p\nmid m$. The Sylow theorems state:

If $(x,y)\sim (x+a,y+b)$ in the torus, thenThis is Klein-equivalent to $(2x,y)$ by taking $(c,d)=(2a,b)$. Hence $\tilde{\pi }$ descends to a well-defined continuous map

The deck group of the Klein bottle has an index-two subgroup consisting of transformations with even $c$; it is generated by translations $(x,y)\mapsto (x+2,y)$ and $(x,y)\mapsto (x,y+1)$. This is exactly the image under $\tilde{\pi }$ of the torus deck lattice. Therefore the orientation double cover of the Klein bottle has two sheets:

By the finite case of the structure theorem, a finite abelian group is a direct sum of cyclic groups of prime-power order. If it is indecomposable, this decomposition can have only one nonzero summand, so the group is cyclic of order $p^n$ for some prime $p$.

Conversely, every nontrivial subgroup of the cyclic group $C_{p^n}$ contains its unique subgroup of order $p$. Hence two nontrivial subgroups cannot have trivial intersection. They therefore cannot be the two summands of an internal direct sum. Thus the indecomposable finite abelian groups are precisely

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 rational-root test shows that $f=X^3-X-1$ is irreducible. Its discriminant iswhich is not a square in $\mathbb{Q}$. The Galois group of an irreducible cubic therefore gives

Quadratic subfields correspond to order-three subgroups. The unique such subgroup is $A_3$, and its fixed field is generated by the alternating product of root differences, whose square is $\Delta$. Thus the unique quadratic subfield is

Let the roots be $\alpha ,\beta ,\gamma$ and choose
$\delta =(\alpha -\beta )(\alpha -\gamma )(\beta -\gamma )$, so ${\delta }^2=-23$. Put $\theta =\alpha +\delta$. A nonidentity even permutation moves $\alpha$ and fixes $\delta$, so it cannot fix $\theta$. An odd permutation negates $\delta$. If it fixes $\alpha$, it plainly does not fix $\theta$. If it sends $\alpha$ to $\beta$ and fixed $\theta$, then $\beta -\alpha =2\delta$. Cancelling the factor $\beta -\alpha$ in the expression for $\delta$ would giveBut $f^{\prime }(\gamma )=3{\gamma }^2-1$, so this would make ${\gamma }^2$ equal to $1/6$ or $1/2$, contradicting the irreducibility of its cubic minimal polynomial. The other transpositions are identical. Hence $\theta$ has trivial stabilizer, so