= Lagrange's theorem
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{title2=$|G|=[G:H]|H|$}
{wiki=Lagrange's_theorem_(group_theory)}
= Lagrange theorem
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{synonym}
= Lagrange's theorem for finite groups
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{synonym}
For a <finite group> $G$ and a <subgroup> $H$, the <group order> of $H$ divides the <group order> of $G$:
$$
|G|=[G:H]|H|.
$$
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> $gH\mapsto Hg^{-1}$ 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 https://math.mit.edu/~djk/18.310/Lecture-Notes/symmetries_2007.html[MIT's symmetries notes, section 2].
The additive <cyclic group> $\mathbb Z/6\mathbb Z$ above gives $6=3\times2$ from $H=\{0,3\}$. Another <subgroup>, $K=\{0,2,4\}$, has <cosets> $K$ and $1+K=\{1,3,5\}$, giving $6=2\times3$. In $S_3$, the <subgroup> $\{e,(12)\}$ also gives $6=3\times2$, 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 $\operatorname{ord}(g)\mid|G|$ 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 $6\cdot2=0$. 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 https://ocw.mit.edu/courses/18-703-modern-algebra-spring-2013/50c134275caf32dbf4430ab097185157_MIT18_703S13_pra_l_4.pdf[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 https://mathshistory.st-andrews.ac.uk/Biographies/Lagrange/[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 https://mathshistory.st-andrews.ac.uk/Extras/Carmichael_Hardy_1/[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.
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