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Five examples arewhere $D_{12}$ denotes the symmetry group of a hexagon and ${\mathrm{Dic}}_3$ is the $3$-dicyclic group.

The first two are abelian, while the last three are not. Of the first two, only $C_{12}$ is cyclic. Among the nonabelian groups, $A_4$ has no element of order $6$; both of the others do. The dihedral group $D_{12}$ has seven involutions, whereas ${\mathrm{Dic}}_3$ has the unique involution $a^3$ and also has elements outside $⟨a⟩$ of order four. These invariants prove that all five groups are pairwise non-isomorphic.

From $ba{b}^{-1}=a^{-1}$ we obtain $ba=a^{-1}b$. Moving every occurrence of $b$ to the right and replacing $b^2$ by $a^n$ shows that every group element has the formThe element $b$ cannot lie in $⟨a⟩$: otherwise it would commute with $a$, forcing $a=a^{-1}$, contrary to $a$ having order $2n>2$. Thus the two cosets $⟨a⟩$ and $⟨a⟩b$ are disjoint and each has $2n$ elements. Hence every $n$-dicyclic group has order $4n$.

Existence is explicit. Put $\zeta =e^{\pi i/n}$ and takeThen $a$ has order $2n$,The $2n$ diagonal matrices $a^k$ and the $2n$ off-diagonal matrices $a^{k}b$ are distinct, so they form a dicyclic group of order $4n$.

Let $n=p^k$ and take $G=C_{p^k}$. If it embedded in $S_{n-1}$, that symmetric group would contain a permutation of order $p^k$. The order of a permutation is the least common multiple of its cycle lengths. For this least common multiple to be divisible by $p^k$, one cycle length must itself be divisible by $p^k$, requiring at least $p^k=n$ points. This is impossible in $S_{n-1}$. Thus $C_n$ is the required group.

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.

The orbit-stabilizer theorem states that for a finite group $G$ acting on a set $X$ and $x\in X$,Define $G/G_x\to Gx$ by $g{G}_x\mapsto gx$. This is well-defined because $g{G}_x=h{G}_x$ exactly when $h^{-1}g\in G_x$, equivalently $gx=hx$. The same equivalence proves injectivity, and the definition of the orbit proves surjectivity. Thus $|Gx|=[G:G_x]$, giving the formula.

The group acts transitively on the eight vertices. At a chosen vertex, an isometry fixing it may permute its three incident edges arbitrarily, and each of the six permutations is realized by an isometry. Thus the vertex stabilizer is isomorphic to $S_3$ and has order six. Orbit-stabilizer givesThis begins the three stabilizer descriptions in the symmetry group of a cube.

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

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 Smith normal form of an integer $m\times n$ matrix $A$ is a diagonal matrixfor whichwith $U\in G{L}_m(\mathbb{Z})$ and $V\in G{L}_n(\mathbb{Z})$. It exists and is unique.

The structure theorem for finitely generated modules over a principal ideal domain, specialized to $\mathbb{Z}$, says that every finitely generated abelian group has a unique invariant-factor decompositionEquivalently, its finite part is a direct sum of cyclic groups of prime-power order.