The proof is by induction on the order of G. If |G| = 1, then $p^0$ is the only prime power that divides |G|, and G itself is a subgroup of order $p^0$. Suppose |G| > 1 and assume inductively that the theorem is true for all groups of order less than |G|. By class equation we have |G| = |Z(G)| + [G: C($a_1$)] + .... + [G: C($a_r$)], where for each i, [G: C($a_i$)] > 1. Furthermore , |Z(G)| $\geq$ 1 and |C($a_i$)| < |G|.

Suppose there is an index j such p doesn't divide [G:($a_j$)]. Then by theorem $p^k$ must divide |C($a_j$)| because $p^k$ !!!**divides |G| by hypothesis*!!!