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The alternating conjugacy class splitting criterion gives seven conjugacy classes in $A_6$. Their cycle types and sizes are $1^6:1$, $2^2{1}^2:45$, $31^3:40$, $3^2:40$, $42:90$, and two $51$ classes of size 72. Representatives of the split classes are a 5-cycle and its square. Their sizes sum to 360. A normal subgroup must be a union of these classes containing the identity, so these explicit sizes can prove simplicity using Lagrange's theorem.

For the order-eight dihedral group generated by $r$ of order four and $s$ of order two with $srs=r^{-1}$, let $H=K=⟨s⟩$. There are exactly three double cosets: $H$, $H{r}^{2}H$, and $HrH$, of sizes two, two, and four. Their number does not divide eight. This differs from the number of ordinary cosets, which divides group order by Lagrange's theorem.

A double coset is generally not a subgroup, so Lagrange's theorem does not force its size to divide the ambient group order. In $S_3$, take $H=K=⟨(12)⟩$. Then $HeH$ has size two, while $H(13)H$ has size four, which does not divide six. The general size formula $|HgK|=|H||K|/|H\cap gK{g}^{-1}|$ explains why different double cosets can have different sizes.