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The Sylow theorems also give $n_5\mid 14$ and $n_5\equiv 1(\mathrm{mod}5)$. Of $1,2,7,14$, only $1$ is congruent to one modulo five. Thus $H$ is a normal subgroup of $G$ as well.

By Lagrange's theorem, $H\cap K=\{1\}$, since its group order divides both $5$ and $7$. For $h\in H$ and $k\in K$, the group commutator $hk{h}^{-1}{k}^{-1}$ lies in $H$ by the normal subgroup property of $H$, and in $K$ by the normal subgroup property of $K$. It is therefore the identity element, so all elements of $H$ commute with all elements of $K$. This is the fact that normal subgroups of coprime order commute.

The subgroup $HK$ consequently has $|H||K|=35$ elements: the expression $hk$ is unique because $H\cap K=\{1\}$. Since $H$ and $K$ have prime orders, they are cyclic groups. Write $H=⟨h⟩$ and $K=⟨k⟩$. The product of commuting elements of coprime order has order $35$, so $hk$ generates $HK$. Finally, $HK$ is a normal subgroup because both its factors are normal subgroups; equivalently, it has subgroup index two. Therefore

The Sylow theorems give $n_7\mid 10$ and $n_7\equiv 1(\mathrm{mod}7)$. Among the divisors $1,2,5,10$, only $1$ satisfies the modular congruence. HenceUniqueness and the conjugacy assertion of the Sylow theorems make $K$ a normal subgroup.

Write $|G|=pq$ with $p<q$. The Sylow theorems give $n_q\mid p$ and $n_q\equiv 1(\mathrm{mod}q)$. Since $p<q$, necessarily $n_q=1$. Its unique Sylow subgroup $N$ is a normal subgroup and has prime order $q$, soA Sylow subgroup $H\cong C_p$ has trivial intersection with $N$, and $|NH|=|N||H|=pq$, so $NH=G$. ThusThe semidirect product action is a group homomorphism $C_p\to \mathrm{Aut}(C_q)\cong C_{q-1}$. If the action is trivial, $G\cong C_{pq}$ by the Chinese remainder theorem; if it is nontrivial, $p\mid q-1$. In particular $G$ has both a cyclic normal subgroup and a cyclic quotient, making it a metacyclic group and hence a solvable group.

The Abelian group criterion for a product of two prime squares applies because $11<19$ and $11\nmid 19^2-1=360$. Here its Sylow theorems proof is especially short:The candidates for $n_{19}$ are $1,11,121$, of which only $1$ has the required remainder. The candidates for $n_{11}$ are $1,19,361$, again leaving only $1$. Thus both Sylow subgroups $P,Q$ are normal subgroups. Their group commutators lie in $P\cap Q=\{1\}$, and $PQ=G$, so $G\cong P\times Q$. Both factors are abelian groups of prime-square order by (a).

The Fundamental theorem of finitely generated abelian groups gives preciselyThese four groups are pairwise nonisomorphic: their characteristic Sylow subgroups differ in whether they contain an element of order $11^2$ or $19^2$. The first is also the cyclic group $C_{43681}$.

Suppose $N$ were a normal subgroup of index $m$ and put $Q=SL(2,{\mathbb{F}}_p)/N$. The supplied reduction map and part (c) show that $Q$ is generated by the images $s,t$ of $S,ST$. Their orders divide $4$ and $6$, respectively, because $S^4=I$ and $(ST{)}^6=I$. By Lagrange's theorem, their orders also divide $|Q|=m$. Since $\gcd (m,6)=1$, both $\gcd (m,4)$ and $\gcd (m,6)$ equal one. Thus $s=t=1$, forcing $Q$ to be trivial and $m=1$, a contradiction. No such normal subgroup exists for $m>1$.

The normal closure $⟨⟨B⟩{⟩}_G$ is the smallest normal subgroup of $G$ containing $B$, equivalently the intersection of all normal subgroups containing $B$. Any such normal subgroup contains every $g{b}^{\pm 1}{g}^{-1}$ and hence every finite product of these elements. Thus $N\subseteq ⟨⟨B⟩{⟩}_G$.

Include the empty product so that $1\in N$, in particular when $B=\varnothing$. Products of elements of $N$ are obtained by concatenating their factor lists. Inversion reverses the list and replaces ${ϵ}_i$ by $-{ϵ}_i$, so $N$ is closed under inverses. For any $k\in G$,Therefore $N$ is a normal subgroup and it contains $B$. Minimality of the normal closure gives the reverse inclusion, provingThe notation $n\in \mathbb{N}$ in a finite-product description is understood to allow the empty product; otherwise the empty generating subset would require a separate identity convention.

In a group presentation, $X$ is a set of formal generators and $R$ is a subset of the free group $F(X)$, consisting of relators which are required to become the identity. The presented group isThe universal property of a group presentation says that a function $X\to H$ extends to a group homomorphism from the presented group precisely when every relator evaluates to $1$ in $H$: first extend freely to $F(X)$, then factor through the quotient because its kernel contains the normal closure of $R$.

A rotation $\alpha$ by $2\pi /n$ and a reflection $\beta$ of a regular $n$-gon satisfyThe last identity holds because reflection reverses the sense of rotation, and it implies $\alpha \beta \alpha \beta =1$. Hence all defining relators vanish, giving a homomorphism $\theta :D_{2n}\to M_n$ with $\theta (\sigma )=\alpha$ and $\theta (\tau )=\beta$.

To prove it is an isomorphism without presupposing the order of the presented group, use $\sigma \tau =\tau {\sigma }^{-1}$. Move every occurrence of $\tau$ to the left, cancel pairs by ${\tau }^2=1$, and reduce exponents modulo $n$. Every element has one of the formsThus $|D_{2n}|\le 2n$. For the usual regular polygon with $n\ge 3$, the $n$ rotations ${\alpha }^j$ are distinct and the $n$ reflections $\beta {\alpha }^j$ are distinct; no rotation is a reflection, since their planar orientation behavior differs. The supplied description says these exhaust $M_n$, so $|M_n|=2n$, and every one lies in the image of $\theta$. Surjectivity and the order bound force $|D_{2n}|=2n$ and trivial kernel. Thus $\theta$ is an isomorphism.

For the requested Cayley graph, use right multiplication by the generators. The $\sigma$ edges are the two directed 3-cyclesThe $\tau$ edges are the undirected pairsThese pairs use ${\sigma }^j\tau =\tau {\sigma }^{-j}$. Thus the underlying graph is a triangular prism, with the two rotation cycles oriented oppositely when reflection pairs are drawn as radial edges.

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.

A normal subgroup $N$ is powerfully embedded in a finite p-group or pro-p group $G$ if $[N,G]\le N^p$ for odd $p$, or $[N,G]\le N^4$ for $p=2$; use closed generated subgroups in the pro-p group case. The condition implies that $N$ is a powerful p-group. For odd $p$, both $[N,G]$ and $N^p$ remain powerfully embedded.

The p-residual is the smallest normal subgroup with p-group quotient, equivalently the subgroup generated by all elements of order prime to $p$. It is characteristic and satisfies $O^p(O^p(G))=O^p(G)$: an extension of two finite p-groups is a p-group. A finite group has a normal p-complement exactly when its p-residual has order prime to $p$.

A proper convex lattice subgroup $P$ is prime when $a,b\ge 1$ and $a\wedge b=1$ imply $a\in P$ or $b\in P$. Equivalently the ordered right cosets of $P$ form a totally ordered set. Prime refers to this order property, and does not require a normal subgroup or prime index.

If $H$ is a subgroup and $K$ a normal subgroup of a group $G$, their product set $HK=\{hk:h\in H,k\in K\}$ is a subgroup. Normality gives $(h_1{k}_1)(h_2{k}_2)=h_1{h}_2(h_2^{-1}{k}_1{h}_2)k_2\in HK$ and $(hk{)}^{-1}=h^{-1}(h{k}^{-1}{h}^{-1})\in HK$. Together with the identity element this proves the subgroup test. For finite subgroups, $|HK|=|H||K|/|H\cap K|$: the pairs representing one product differ exactly by multiplying the first factor by an element of $H\cap K$ and its second factor by its inverse.

The inverse limit of the quotients of an abstract group by its normal subgroups of finite $p$-power index. Maps to pro-p groups factor uniquely through the canonical dense image in this completion. Unlike profinite completion, only finite p-groups are used.

For a normal subgroup $N$ with Prüfer rank $r$ and quotient of rank $s$, let $H=⟨h_1,\dots ,h_k⟩\le G$. Choose at most $s$ elements $x_j\in H$ generating the image of $H$. Write $h_i=w_i(x_1,\dots ,x_s)n_i$ with $n_i\in N\cap H$. The finitely generated subgroup $⟨n_1,\dots ,n_k⟩$ needs at most $r$ generators, so $H$ needs at most $r+s$. The full intersection $H\cap N$ need not be finitely generated.

In the quaternion group, $-1=i^2=j^2=k^2$ commutes with each generator because any element commutes with its powers. Hence $N=\{1,-1\}$ is central and is a normal subgroup. The reverse products satisfy $ji=-k$, $kj=-i$, $ik=-j$, so none of $\pm i,\pm j,\pm k$ is central. Therefore $Z(Q_8)=N$.

The four cosets are $N,iN,jN,kN$. Each nonidentity coset squares to $N$, and $(iN)(jN)=kN$. Thus the quotient group is the Klein four-group. Explicitly, mapping $\pm 1$ to its identity and mapping $\pm i,\pm j,\pm k$ to its three different nonidentity elements is a surjective group homomorphism with group kernel $N$. The first isomorphism theorem gives the title result.

For a group $G$ and a normal subgroup $N$, the quotient group $G/N$ is the set of cosets $gN$, with operationIts identity element is the coset $N$, and the inverse element of $gN$ is $g^{-1}N$. A quotient group treats all elements within one coset as the same input while retaining a consistent operation on the classes. The normal subgroup condition is what makes this operation independent of which representatives are chosen. ^[ref].

Take the additive group $G=\mathbb{Z}/6\mathbb{Z}$ and $N=\{0,3\}$. Since $G$ is an abelian group, $N$ is a normal subgroup. There are three elements in the quotient group:Adding two copies of $2+N$ gives $4+N=1+N$; adding $1+N$ and $2+N$ gives $3+N=N$. Hence the quotient group has the addition table of $\mathbb{Z}/3\mathbb{Z}$. It does not select one representative and discard the others: its elements are the entire cosets.

This is useful whenever distinctions within a normal subgroup should be ignored. For example, the sign of a permutation retains only even or odd parity of a permutation in $S_3$. Its group kernel is the alternating group $A_3$, and the quotient group $S_3/A_3$ has two elements. The first isomorphism theorem explains why a group homomorphism's image is a quotient group of its domain.

The quotient map of groups sends each group element to its coset modulo a normal subgroup. It is a surjective group homomorphism, with group kernel exactly $N$. In the additive six-element example, it sends $0,3$ to $N$, $1,4$ to $1+N$, and $2,5$ to $2+N$. The quotient map of groups records the identification that produces the new group.

The rank of a profinite group is $\mathrm{rk}(G)={\sup }_{K{\le }_{c}G}d(K)$, where $K$ runs through closed subgroups and $d(K)$ is the least number of topological generators. If an open subgroup $H$ has rank $r$, its normal core $C$ is open and has rank at most $r$. Every closed $K\le G$ has a normal subgroup $K\cap C$ generated by at most $r$ elements, while $K/(K\cap C)$ has order at most $[G:C]$. A finite group of order at most $M$ is generated by at most $\lfloor {\log }_{2}M\rfloor$ elements, since each necessary new generator at least doubles the generated subgroup. Lifting generators yields $d(K)\le r+\lfloor {\log }_2[G:C]\rfloor$. Thus finite rank is preserved in both directions under passage to an open subgroup.

A transitive group action is regular if every point stabilizer is trivial. Equivalently each ordered pair of points is connected by exactly one group element. Identifying one point with the identity identifies the action with left multiplication on the group. A regular normal subgroup inside a larger primitive group can supply translations or nonabelian left multiplication; the whole primitive group usually has a nontrivial stabilizer acting by automorphisms of this regular subgroup.

For a normal subgroup $K$ of $G$, each $c_h:k\mapsto h^{-1}kh$ is a group automorphism of $K$. Its inverse is $c_{h^{-1}}$, but the dependence on $h$ is a group antihomomorphism under ordinary function composition, because $(gh{)}^{-1}k(gh)=h^{-1}(g^{-1}kg)h$. The assignment $h\mapsto (k\mapsto hk{h}^{-1})$ instead is a group homomorphism to the automorphism group. Equivalently the first convention is a right group action, while the second is a left group action.