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For a finite group $G$ and a subgroup $H$, the group order of $H$ divides the group order of $G$:Here $[G:H]$ is the index of a subgroup, the number of distinct left cosets. It is also the number of distinct right cosets: inversion gives a bijection $gH\mapsto H{g}^{-1}$ between the two collections. No normal subgroup assumption is needed for Lagrange's theorem. The normal subgroup condition matters when constructing a quotient group, rather than when counting a partition of a set.

The beauty of Lagrange's theorem is that an arithmetic restriction follows from a structural picture. A group can be tiled by equal-sized cosets of any subgroup. To see the count, choose one representative from each left coset. Every group element has exactly one expression $g_{i}h$ for these chosen representatives and $h\in H$: different rows do not overlap, and multiplication by $g_i$ is a bijection from $H$ to its row. Counting rows times row size gives the formula. See MIT's symmetries notes, section 2.

The additive cyclic group $\mathbb{Z}/6\mathbb{Z}$ above gives $6=3\times 2$ from $H=\{0,3\}$. Another subgroup, $K=\{0,2,4\}$, has cosets $K$ and $1+K=\{1,3,5\}$, giving $6=2\times 3$. In $S_3$, the subgroup $\{e,(12)\}$ also gives $6=3\times 2$, despite its distinct left and right coset partitions. Thus equal row sizes persist in a nonabelian group. A proposed subgroup of size four in either group is impossible: four does not divide six. This filters possible subgroups before any search for group generators.

Apply Lagrange's theorem to the cyclic subgroup generated by a group element $g$. Its size is the order of a group element, so $\mathrm{ord}(g)\mid |G|$ and $g^{|G|}=e$. In the additive example, $2$ has order of a group element $3$, since $2+2+2=0$ modulo six; the corresponding multiplicative statement becomes $6\cdot 2=0$. A group of prime group order is consequently a cyclic group: any nonidentity group element generates a subgroup whose size is greater than one and divides that prime, forcing it to be the whole group. See MIT's cyclic-group notes. The same counting idea underlies the orbit-stabilizer theorem: the orbit of a group action is parametrized by cosets of its stabilizer subgroup.

Historically, Lagrange's theorem grew from investigations of permutations of roots of a polynomial; see St Andrews' account of Lagrange's early work. The modern statement about arbitrary finite groups should not be read back into the early work. Carmichael's 1921 review explicitly cautioned against attributing the full result to Lagrange; see Carmichael's historical note, reproduced by St Andrews. Finally, Lagrange's theorem supplies a necessary size condition; the failure of the converse to Lagrange's theorem below explains its limit.

Lagrange's theorem says that for a subgroup $H$ of a finite group $G$,To prove it, the left cosets $gH$ partition $G$. Indeed every element lies in its own coset; if $gH$ and $kH$ intersect, write $gh=k{h}^{\prime }$ and deduce $g=k{h}^{\prime }{h}^{-1}$, so $gH=kH$. Multiplication by $g$ is a bijection from $H$ to $gH$, and each coset therefore has $|H|$ elements. Summing over the $[G:H]$ distinct cosets proves the formula.

For the failure of the converse to Lagrange's theorem, take the alternating group $A_4$, of order $4!/2=12$, and the divisor $6$. Suppose it had a subgroup $H$ of order six. Its index would be two. An index-two subgroup is normal: for $g\notin H$, both $gH$ and $Hg$ are the complement of $H$, while for $g\in H$ they are $H$. Hence $A_4/H$ would be a group of order two.

Every three-cycle in $A_4$ has order three. Its image in the quotient has cube equal to the identity and, since the quotient has order two, must itself be the identity. Thus all three-cycles lie in $H$. There are eight of them: choose the omitted point in four ways and one of two cyclic orders on the other three points. Together with the identity this would give at least nine elements of $H$, contradicting $|H|=6$. Therefore $A_4$ has no subgroup of order six, although six divides its order.