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A group homomorphism satisfies $\phi (g_1{g}_2)=\phi (g_1)\phi (g_2)$; a bijective one is a group isomorphism. Its kernel of a group homomorphism is $K=\{g:\phi (g)=1_H\}$. It is a subgroup, and for $k\in K$, $g\in G$,Thus $gK{g}^{-1}\subseteq K$, and applying the same argument to $g^{-1}$ gives equality. Therefore $K$ is a normal subgroup.

Write $B_n=\left(\begin{array}{cc}1& n\\ 0& 1\end{array}\right)$. Direct matrix multiplication gives $B_n{B}_m=B_{n+m}$, $B_0=I$ and $B_n^{-1}=B_{-n}$. Consequently these matrices form the integer shear group, a subgroup of $GL(2,\mathbb{R})$. It is an infinite cyclic group, generated by $B_1$, since $B_1^n=B_n$ and no nonzero integer power is the identity.

The map $n\mapsto B_n$ is a bijective group homomorphism from the additive group of integers, proving the required group isomorphism. There are proper subgroups: for example $\{B_{2n}:n\in \mathbb{Z}\}$ omits $B_1$. In fact every subgroup is $\{B_{mn}:n\in \mathbb{Z}\}$ for an integer $m\ge 0$, by the classification of subgroups of an infinite cyclic group; $m=0$ gives the trivial subgroup and $m=1$ the whole group.

Yes. Take $X=S^1$, the circle, and $A=\{1\}\subset S^1$. The constant function $r(z)=1$ is a continuous map and satisfies $ri={\mathrm{id}}_A$, so $A$ is a retract of $X$.

However, their fundamental groups areA homotopy equivalence between path-connected spaces induces a group isomorphism of their fundamental groups, allowing the usual change of base point. The infinite cyclic group is not the trivial group. Therefore a retract need not have the homotopy type of its ambient space.

Let $n=(0,0,1)$ and use the open coverThe cap $V$ is open in $Y$: it equals $Y\cap \{z>1/2\}$ because the attached chord has $z=0$. The cap is homeomorphic to an open disk, and $U\cap V=V\setminus \{n\}$ is a punctured open disk, hence homotopy equivalent to a circle. Both sets and their intersection are path-connected. Choose the base point $b$ in that intersection.

We first calculate ${\pi }_1(U,b)$. A stereographic projection identifies $S^2\setminus \{n\}$ with ${\mathbb{R}}^2$ and sends the two chord endpoints to $(-1,0),(1,0)$. Thus $U$ becomes the Euclidean plane with an extra closed interval attached at those two points. This description is a homeomorphism: gluing the two closed pieces along their endpoints gives exactly the original subspace topology.

The Euclidean plane has a deformation retraction onto $L=[-1,1]\times \{0\}$ while fixing its endpoints. An explicit homotopy isExtend it by the identity map on the extra chord. The pasting lemma for closed subspaces gives a deformation retraction of $U$ onto two distinct closed intervals sharing just their endpoints, a circle. Consequently ${\pi }_1(U,b)\cong \mathbb{Z}$, with the routine base point transport if the retraction moves $b$.

The inclusion-induced group homomorphism ${\pi }_1(U\cap V,b)\to {\pi }_1(U,b)$ is trivial: it factors through $S^2\setminus \{n\}\cong {\mathbb{R}}^2$, which is contractible. Also ${\pi }_1(V,b)=1$. All hypotheses of the Seifert-van Kampen theorem are now checked, and it gives the fundamental group of a sphere with an attached chord:where both maps from the amalgamating infinite cyclic group are trivial. A group generator follows the chord from one endpoint to the other and returns along any path on the sphere, with a connecting path from the chosen base point.

A poly-infinite cyclic group has a finite subnormal series all of whose nontrivial factors are infinite cyclic groups. Intersecting its series with a subgroup again gives trivial or infinite cyclic factors; repetitions may be deleted. Splicing a kernel series with the preimage of a quotient series proves closure under extensions. Such groups are finitely presented groups: induct on the series, split the last extension by an infinite cyclic quotient, and use finite presentation of a semidirect product.

For $G=⟨g⟩$ infinite, every subgroup is $⟨g^m⟩$ for a unique integer $m\ge 0$. This is the classification of subgroups of the additive group of integers transported through a group isomorphism. For $m>0$ the subgroup index is $m$ and the subgroup is again an infinite cyclic group; for $m=0$ it is trivial and has infinite subgroup index.

The trefoil is the torus knot $T_{2,3}$. It is the simplest nontrivial knot by crossing number of a knot: $c(T_{2,3})=3$. A standard three-crossing alternating knot diagram is reduced, so the Tait crossing-number theorem establishes minimality. The four-crossing knot diagram shown below also represents the trefoil knot, illustrating that a particular projection need not achieve that minimum. Its Seifert genus is one.

Here is a small algebraic certificate that a trefoil knot cannot be deformed into an unknot. One Seifert matrix for a trefoil knot isTaking the determinant gives the Alexander polynomial of a knotwhere $\doteq$ allows multiplication by a Laurent unit $\pm t^n$ in the Laurent polynomial ring. No such multiplication turns this three-term Alexander polynomial of a knot into the unknot's value $1$. Since the Alexander polynomial of a knot is a knot invariant, this computation rules out every possible ambient isotopy between the two knots; it does not merely rule out the deformations we happened to try.

The knot group supplies another certificate. The presentation $⟨a,b\mid a^2=b^3⟩$ maps onto the symmetric group $S_3$ by $a\mapsto (12)$ and $b\mapsto (123)$, since both sides of the relation map to the identity. These images generate a non-abelian group, so the knot group cannot be the unknot's infinite cyclic group. However, a trefoil knot and its mirror of a link have isomorphic knot groups: the abstract knot group alone does not distinguish all knots.

The unknot $U$ is the knot type represented by the planar circleEquivalently, an unknot is a knot that bounds a smoothly embedded disk in the three-sphere. The disk must itself be embedded; merely filling a loop with a map that is allowed to intersect itself does not establish that the knot is an unknot. Flattening a neighborhood of an embedded spanning disk explains why this criterion recovers the planar circle.

A complicated drawing can still represent an unknot. Starting with a crossing-free knot diagram of $U$, insert a small kink by a type I Reidemeister move. The resulting knot diagram has one crossing, but removing the kink exhibits the same unknot. Repeating in separate small regions gives knot diagrams of the unknot with arbitrarily many crossings, although the crossing number of a knot is $c(U)=0$. A drawing's apparent complexity is therefore not a certificate of knotting.

The normalized Alexander polynomial of a knot and Jones polynomial of the unknot are both $1$, and its knot group is the infinite cyclic group $\mathbb{Z}$, because the knot exterior is a solid torus. These are useful comparison values. In particular, a nonconstant Alexander polynomial of a knot certifies that a knot is not an unknot, but the converse fails: nontrivial untwisted Whitehead doubles also have Alexander polynomial of a knot $1$.