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Lagrange's theorem for finite groups states that for a subgroup $H$ of a finite group $G$, $|G|=[G:H]|H|$. For $x\in G$, two powers among $1,x,\dots ,x^{|G|}$ coincide, giving $x^m=1$ for some positive $m$. The least such exponent $d$ is the order of a group element, and the cyclic subgroup generated by $x$ consists of the $d$ distinct powers $1,x,\dots ,x^{d-1}$. Apply Lagrange's theorem for finite groups to obtain $d\mid |G|$.

Cauchy's theorem for finite groups states that if a prime $p$ divides the order of a finite group, the group contains an element of order $p$.

The classification of a group of order eight isHere $D_8$ denotes the dihedral group of order eight, not a group of order sixteen. To prove exhaustiveness, every element order is $1,2,4$ or $8$. An element of order eight generates the whole group, giving $C_8$.

If no element has order four or eight, every nonidentity element is an involution. Since $(ab{)}^{-1}=ab$ while $a^{-1}=a$ and $b^{-1}=b$, we have $ab=ba$. Choose an involution $a$, another $b$ outside $⟨a⟩$, and then $c$ outside the four-element subgroup $⟨a,b⟩$. The factors commute and each added factor has trivial intersection with the preceding subgroup. The internal direct product theorem gives $G\cong C_2^3$, an elementary abelian group.

Otherwise choose $a$ of order four and set $A=⟨a⟩$. It has index two, so is normal. Choose $b\notin A$. The two cosets show that $a,b$ generate $G$ and that $b^2\in A$. There is no element of order eight, so $b^2$ is either $1$ or $a^2$.

If $G$ is abelian and $b^2=1$, the internal direct product theorem yields $G=A\times ⟨b⟩\cong C_4\times C_2$. If instead $b^2=a^2$, let $c=a^{-1}b$; commutativity gives $c^2=1$ and $c\notin A$, producing the same direct product.

If $G$ is nonabelian, conjugation by $b$ is a nonidentity group automorphism of $A$: otherwise $a,b$ would commute and hence so would all of $G$. Thus $ba{b}^{-1}=a^{-1}$. With $b^2=1$ these are the relations of $D_8$; with $b^2=a^2$ they are the relations of the quaternion group $Q_8$. In either presentation every word reduces to $a^i{b}^j$, $0\le i<4$, $0\le j<2$. The standard eight-element model maps onto $G$, and equal orders make that group homomorphism an isomorphism.

Finally group isomorphisms preserve element orders. The following numbers of elements distinguish all five possibilities:

Thus the list is both exhaustive and pairwise non-isomorphic.

The internal direct product theorem states that if $H,K⊴G$, $H\cap K=\{1\}$, and $HK=G$, then $G\cong H\times K$. Indeed, for $h\in H,k\in K$, the commutator $[h,k]$ lies in both $H$ and $K$, so it is trivial. The map $(h,k)\mapsto hk$ is consequently a homomorphism; the intersection condition makes it injective and $HK=G$ makes it surjective. Conversely, the two factors in a direct product satisfy these conditions.

In odd dimension, $-I$ is central and has determinant $-1$. Every $Q\in O(3)$ has a unique expression $Q=(-I{)}^{\epsilon }R$ with $R\in SO(3)$, soBy contrast, $SO(2)\times C_2$ is abelian, while reflections in $O(2)$ conjugate a rotation to its inverse. Thus $O(2)\ncong SO(2)\times C_2$.

The center of the unitary group consists exactly of scalar unitary matrices:The determinant gives a surjective homomorphismwhose kernel is the special unitary group $SU(2)$.

Nevertheless $U(2)$ is not isomorphic to $SU(2)\times \mathbb{T}$. The center of $U(2)$ has exactly one nonidentity element of order two, namely $-I$. The center of the proposed product is $\{\pm I\}\times \mathbb{T}$, which has three nonidentity elements of order two. Since an isomorphism preserves the center and element orders,