= Solution
<Lagrange's theorem> says that for a <subgroup> $H$ of a <finite group> $G$,
$$
\boxed{|G|=[G:H]|H|,\qquad |H|\mid|G|.}
$$
To prove it, the left <cosets> $gH$ partition $G$. Indeed every element lies in its own <coset>; if $gH$ and $kH$ intersect, write $gh=kh'$ and deduce $g=k h'h^{-1}$, 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 \b[$A_4$ has no <subgroup> of order six, although six divides its order].
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