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In additive notation every element $(a,b)\in {\mathbb{Z}}_2\times {\mathbb{Z}}_4$ obeys $4(a,b)=(0,0)$. Its order of a group element therefore divides four, whereas $1\in {\mathbb{Z}}_8$ has order eight. A group isomorphism preserves element orders, soFor the second product, define the group homomorphismIt is well-defined because multiples of six are divisible by both two and three. Its kernel contains only zero: simultaneous divisibility by two and three implies divisibility by six. Thus $\Psi$ is injective, and equal finite group orders make it surjective. Explicitly the six images areThereforeFor general positive integers $n,m$, the order of a direct-product element is the least common multiple of its component orders. Hence every element order divides $\mathrm{lcm}(n,m)=nm/\gcd (n,m)$, while $(1,1)$ attains this order. It generates the whole product precisely when $n,m$ are coprime:This is the cyclicity of a product of two finite cyclic groups, with the explicit isomorphism furnished by the Chinese remainder theorem.

A group homomorphism $f:G\to H$ preserves the operation: $f(ab)=f(a)f(b)$ for all $a,b\in G$. A group isomorphism is a bijective group homomorphism; its inverse also preserves the operation. The kernel of a group homomorphism is $\ker f=\{g\in G:f(g)=e_H\}$, a normal subgroup of $G$.

In the Klein four-group $V=\{1,a,b,ab\}$, the chainsare distinct composition series. All their composition factors are $C_2$.

The Jordan–Hölder theorem states that any two composition series of a group have the same length and the same composition factors up to permutation and group isomorphism. Thus the subgroup chains may differ, while the multiset of simple quotients is intrinsic.

Supposeare two factorizations into primes. Repeatedly apply the cyclic-subgroup construction of part (c), first removing $p_r$, then $p_{r-1}$, and so on. This constructs a composition series for $C_n$ whose composition factors are $C_{p_1},\dots ,C_{p_r}$. The second factorization constructs another composition series with composition factors $C_{q_1},\dots ,C_{q_s}$.

By the Jordan–Hölder theorem, $r=s$ and these composition factors agree up to group isomorphism and permutation. Since isomorphic finite groups have the same order, the corresponding primes agree. This proves uniqueness of prime factorization up to permutation, using existence of prime factorizations but not their uniqueness.

Let $1◃N◃G$ be the given length-two composition series. Thus $N$ and $G/N$ are simple groups. Let $M$ be any proper nontrivial normal subgroup of $G$, such as the penultimate term of another composition series.

If $M=N$, its two factors are already the given ones. Suppose $M\ne N$. Since $M\cap N$ is normal in the simple group $N$, it is either $1$ or $N$. The latter would mean $N\subseteq M$, but then $M/N$ would be a proper nontrivial normal subgroup of $G/N$, impossible. Thus $M\cap N=1$.

Also $M$ cannot be contained in $N$, since it would then be a nontrivial normal subgroup of $N$ distinct from $N$. The product $MN$ is a normal subgroup of $G$, strictly containing $N$. Simplicity of $G/N$ therefore forces $MN=G$. The second isomorphism theorem for groups givesIn particular $M$ is simple. Every other composition series has such an $M$ as its last proper term, and no nontrivial refinement inside a simple $M$ is possible. Hence every composition series has length two.

The general Jordan–Hölder theorem states that any two composition series of a finite group have the same length and the same composition factors up to group isomorphism and permutation. The argument above proves both assertions in this length-two case: another series either uses $N$ itself or has factors isomorphic to the original pair in the opposite order. No general uniqueness theorem was used to prove this special case.

Let $|G|=p^2$. The allowed center of a group theorem gives $|Z(G)|>1$. By Lagrange's theorem its order is $p$ or $p^2$. In the first case $G/Z(G)$ has order $p$ and is a cyclic group. A cyclic quotient by the center forces $G$ to be Abelian: if $G/Z(G)=⟨gZ(G)⟩$, write arbitrary elements as $g^{a}z,g^{b}w$ with $z,w\in Z(G)$, and commute them directly. Hence in every caseFor two examples at prime-cube order, take the dihedral group of a square, $⟨r,s\mid r^4=s^2=1,srs=r^{-1}⟩$, and the quaternion group $Q_8=\{\pm 1,\pm i,\pm j,\pm k\}$. Both have order $8=2^3$. They are nonabelian because $sr\ne rs$ and $ij=k\ne -k=ji$, respectively. They are distinct examples even up to group isomorphism: the dihedral group has five elements of order two and the quaternion group has only one. The request asks for two examples; it does not require the prime to be arbitrary.

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.

The universal property of a pushout of groups says: for every group $H$ and group homomorphisms $u_j:G_j\to H$ satisfying $u_1{\varphi }_1=u_2{\varphi }_2$, there is exactly one group homomorphism $u:P\to H$ such thatTo prove existence, the universal property of a free product supplies a unique group homomorphism $v:G_1\ast G_2\to H$ with $v{k}_j=u_j$. Every defining relator of $N$ maps toSince $\ker v$ is a normal subgroup, it contains the entire normal closure $N$. The universal property of a quotient group therefore gives a group homomorphism $u:P\to H$ through which $v$ factors. Its restrictions are the required $u_j$.

For uniqueness, $j_1(G_1)$ and $j_2(G_2)$ together generate $P$: their preimages already generate the free product. Consequently a group homomorphism out of $P$ is determined by its restrictions to these two subgroups. Any two candidates for $u$ therefore coincide.

This also proves that the pushout of groups is unique up to a unique compatible group isomorphism. Given another such object $P^{\prime }$, apply the universal property in both directions. Both composites restrict to the original maps from $G_1,G_2$, so uniqueness makes them the respective identity maps.

A group presentation of $G$ consists of a set $S$ of symbols and a subset $R\subseteq F(S)$ of relators, together with a group isomorphismHere $F(S)$ is the free group on $S$, and $⟨⟨R⟩⟩$ is the normal closure of $R$. The images of $S$ form a generating set of a group; the relators prescribe which words become the identity element.

For the canonical presentation of a group, use one symbol $s_g$ for every group element $g\in G$. The universal property of a free group gives a surjective group homomorphismLet $R(G)=\ker \epsilon$, namely all words whose evaluation in $G$ is the identity element. ThenIndeed $R(G)$ is already a normal subgroup, so its normal closure is itself, and the first isomorphism theorem identifies the resulting quotient group with $G$. This is the course's canonical presentation of a group.

An equivalent, smaller collection of relators is $s_g{s}_h{s}_{gh}^{-1}$ for all $g,h\in G$. These encode the multiplication table. They imply $s_1=1$ and $s_g^{-1}=s_{g^{-1}}$, so every word reduces to the single symbol recording its product. This proves directly that they normally generate $R(G)$, although the canonical collection above consists of every word in that group homomorphism kernel.

Put $A=T^2\setminus B^2$ and let $E=\partial A=S^1$ denote the attaching circle. The version of the Seifert-van Kampen theorem used here states that if $X=U\cup V$ is an open cover with $U,V,U\cap V$ path-connected and a common base point in $U\cap V$, thenThe expression on the right means the pushout of groups of the two inclusion-induced group homomorphisms; they need not be injective.

Choose a thin collar neighbourhood $C_A$ of $E$ inside $A$, and a thin equatorial band $C_S$ inside $S^2$. For example, before the square's side identifications take $C_A=\{v:1\le |v|<1+ϵ\}$ with $0<ϵ<1/2$, and take $C_S=\{(x,y,z)\in S^2:|z|<ϵ\}$. In the glued topological space setThese are open subsets: their preimages in the two pieces are open and respect the gluing. They cover $X$, and the three deformation retractionsfix $E$. Thus all three spaces are path-connected, and a base point $b\in E$ is preserved.

For completeness, the fundamental group of a once-punctured torus is $F(a,b)$, and its boundary loop of a once-punctured torus represents $[a,b]=ab{a}^{-1}{b}^{-1}$, up to orientation and a change of base point. Here is the geometry behind both facts. In the square model, move each point of $A$ radially outwards to the square boundary:The denominator never vanishes on $A$. The homotopy stays outside the missing open disk, stays inside the square, and fixes the square boundary, so it descends through the side identifications. The resulting retract is a wedge of two circles. Its fundamental group is the free group on the two oriented side based loops $a,b$. The inner boundary loop travels around the square once under this deformation retraction; its successive sides read $a,b,a^{-1},b^{-1}$.

The sphere $S^2$ is simply connected. Hence the group homomorphism from ${\pi }_1(E)\cong \mathbb{Z}$ to ${\pi }_1(V)$ is trivial, while its map to ${\pi }_1(U)$ sends a group generator to $[a,b]$. Applying the Seifert-van Kampen theorem givesThe final group isomorphism follows because the single relator makes $a,b$ commute, leaving the free abelian group on two group generators. The two hemispheres kill the same boundary loop; they do not create another group generator.

Use a topological graph consisting of the ray $[0,\infty )$ with vertices $v_n=n$ and one oriented graph loop ${\ell }_n$ attached at each $v_n$, for $n=0,1,2,\dots$. Set the base point to $b=v_0$. This is a connected, locally finite graph; it is deliberately infinite.

Define continuous maps $f,g:X\to X$ byHere each graph loop is parametrized by $s\in [0,1]$ with its endpoints at the attaching vertex. The formulas agree at every vertex, and their restrictions to every closed edge are continuous. The weak topology of a CW complex therefore makes $f,g$ continuous maps. Directly on the ray and on each graph loop,

We check the two fundamental group maps at exactly the specified base points. The ray is a spanning tree. At $b=v_0$ a free basis of a group consists of $a_n$, obtained by going along the ray to $v_n$, following ${\ell }_n$, and returning. At $f(b)=v_1$, use the corresponding free basis of a group $c_n$ with connecting paths starting at $v_1$; in particular $c_0$ travels backwards along the first edge before following ${\ell }_0$. The standard spanning tree description applies to this infinite topological graph as well: every loop and every homotopy between based loops has compact image, so lies in a finite subcomplex.

The induced group homomorphisms areThus $f_{\ast }$ misses $c_0$: words in $c_1,c_2,\dots$ cannot equal $c_0$ by the normal form for free groups and free product of groups. Consequently $f_{\ast }$ is not surjective. Meanwhile $c_0$ is a nontrivial free group element killed by $g_{\ast }$, so $g_{\ast }$ is not injective. Neither induced map is a group isomorphism, while functoriality of the fundamental group correctly gives $g_{\ast }{f}_{\ast }=\mathrm{id}$.

Any two finite subnormal series have refinements with the same successive factors up to ordering and group isomorphism. For descending chains $A_i$ and $B_j$, insert $A_{i+1}(A_i\cap B_j)$ between $A_i$ and $A_{i+1}$ and insert $B_{j+1}(B_j\cap A_i)$ in the other chain. The Zassenhaus butterfly lemma identifies the factor indexed by $(i,j)$ in the first refinement with the factor indexed by $(j,i)$ in the second. If both chains are composition series, their simple factors admit no nontrivial refinement, giving the Jordan–Hölder theorem.

For a subgroup $H\le G$ and a normal subgroup $N⊴G$, the map $h\mapsto hN$ from $H$ onto $HN/N$ is a group homomorphism with kernel $H\cap N$. The first isomorphism theorem gives the displayed group isomorphism. The product $HN$ is a subgroup because $N$ is normal. In particular, if $HN=G$ and $H\cap N=1$, then $H\cong G/N$.

The odd integers form an abelian group under the displayed operation. The identity element is $-1$ and the inverse element of $a$ is $-a-2$. The bijection $a\mapsto (a+1)/2$ is a group isomorphism to $(\mathbb{Z},+)$, since it converts the shifted operation into addition. This illustrates how an unfamiliar identity and inverse can arise from ordinary addition by a change of coordinates.

Every real orthogonal two-by-two matrix of determinant one has the unique formIndeed the first column is a unit vector, the perpendicular second column has one of two signs, and positive determinant selects the displayed one. The map $z\mapsto R(z)$ from the circle group is bijective and satisfies $R(zw)=R(z)R(w)$ by direct multiplication, proving a group isomorphism. Writing $z=e^{i\theta }$ parametrizes these matrices as planar rotations, with the angle defined modulo $2\pi$.

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.

Take the four-point setthe vertices of a polygon of a rectangle whose side lengths are four and two. Three choices of structure give three different symmetry groups on these same points.

As an unstructured set, $R$ has all $4!=24$ permutations as symmetries, forming the symmetric group $S_4$. If Euclidean distance is part of the data, the allowed maps are only the identity, a half-turn rotation, and the reflections across the two coordinate axes. They form a Klein four-group of group order four. To see that nothing is missing, fix the image of one vertex of a polygon: its three other distances, $2$, $4$, and $\sqrt{20}$, are distinct, so they uniquely determine the images of the remaining vertices.

Preserving affine structure permits more. The affine mapcycles all four points of $R$. It sends a horizontal side of length four to a vertical side of length two, so it is not an isometry. Yet it preserves the rectangle's vertex set and all affine combinations among those points.

Let $A=\mathrm{diag}(2,1)$ and let $r(x,y)=(-y,x)$ be the square's quarter-turn rotation. ThenConjugating every square symmetry by $A$ gives eight affine maps preserving $R$. This group isomorphism makes the affine symmetry group a dihedral group, although some of its maps do not preserve the original Euclidean distances. There are no further affine symmetries: an affine map preserves the common midpoint of the diagonals, so it must preserve opposite pairs. The image of one vertex has four choices and the image of an adjacent vertex has two; those choices determine the map through three noncollinear points.

ThusAn arbitrary permutation need not be an affine map. For example, exchanging the two top vertices while fixing the two bottom vertices sends the common diagonal midpoint to two different points, which no single affine map can do. The phrase "the symmetry of these points" becomes unambiguous only after stating which structure matters.

A topological invariant is a property or mathematical object preserved by homeomorphisms of topological spaces. Connectedness and compactness are examples. The fundamental group is an invariant up to group isomorphism; it is also preserved up to isomorphism by homotopy equivalence, a stronger invariance property which does not hold for every topological invariant.