Cosets and Lagrange's theorem

For a subgroup $H$ and an element $g$ of $G$, the left coset $gH$ consists of the products $gh$ as $h$ ranges over $H$. The word left describes the position of the representative, not a direction in a picture. A right coset instead consists of products $hg$. These need not agree. The map sending $h$ to $gh$ is a bijection from $H$ onto $gH$, with inverse given by multiplication on the left by $g^{-1}$.

Two left cosets are equal exactly when $g^{-1}k$ belongs to $H$. To see why overlap is decisive, suppose $gh_1=kh_2$. Rearranging gives $k=gh_1h_2^{-1}$, so every element of $kH$ lies in $gH$; reversing the argument gives equality. Equivalently, belonging to the same coset is an equivalence relation on the group. The cosets are its equivalence classes, which explains their partition property without relying on a diagram.

The index $[G:H]$ is the number of distinct left cosets. If $G$ is finite, its partition into that many pieces, each of size $|H|$, gives the group-order formula. In particular, the order of a subgroup divides the order of the group. Finiteness matters when writing this as an equation of ordinary integers and drawing divisibility conclusions. The partition argument itself still makes sense for infinite groups, but cardinal arithmetic requires separate care.

Apply the theorem to the cyclic subgroup generated by an element. Its size is that element's order, so every element order in a finite group divides the group order. A group with prime order is therefore cyclic: any nonidentity element has an order greater than one dividing that prime, hence has full order. This is a genuine classification deduction from counting, even though no assumption of commutativity was included at the start.

If a proposed subgroup has an order that does not divide the finite group order, it cannot exist. If the order does divide, the theorem alone is silent. Extra structure may construct such a subgroup, and later Sylow theorems guarantee particular prime-power orders, but a necessary condition is not automatically sufficient. Likewise, knowing that an element order divides the group order does not mean every divisor occurs as an element order.

For a concrete warning, the group with four elements obtained by adding pairs of residues modulo two has identity $(0,0)$ and three nonidentity elements. Doubling any pair gives the identity, so every nonidentity element has order two. Four divides the group order, yet no element has order four. Compare this with the cyclic group modulo four, which does contain such elements. Group size restricts possible behavior without determining the full multiplication structure.