Subgroup inclusion reverses index size.Suggest a slogan
Let \(G\) be a group and \(H\leq K \leq G\) subgroups. Then \[ \abs{G} = [G:1] \geq [G:H] \geq [G:K] \geq [G:G] = 1 \] as cardinals. In particular, for arbitrary subgroups \(H, K\leq G\), applying this to \(H \cap K \leq H\leq G\) and \(H\cap K\leq K\leq G\) gives \([G: H \cap K] \geq [G:H]\) and \([G:H\cap K]\geq[G:K]\).
By the tower law, \([G:H]=[G:K]\,[K:H]\geq[G:K]\) since \([K:H]\geq1\); the outer inequalities are the cases \(H=1\) and \(K=G\).