Is Every Group of Order p^2 Abelian?

[Kummer]

I still can't figure out the last question I posted, which was:

Let G be a finite group. For every prime divisor p of |G| show that there exists an element x in G of order p (i.e. o(x) = p) such that [G: C_G(x)] and p are coprime.
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(Accept strong induction.)
You can use the conjugacy class equation,
|G| = |Z(G)|+\sum [G:C(x)]
Where x is the sum of all non-trivial conjugacy classes.
Now p divides |G|. We have two cases: Case#1 is that there is an index [G:C(x)] is not divisible by p. Case#2 is that all indexes [G:C(x)] are divisible by p. If Case#1 then p does not divide [G:C(x)] where x is some element forming a non-trivial conjugacy class. Since [G:C(x)]|C(x)| = |G| it means p must divide C(x). If Case#2 then since the sum is divisible by p it implies Z(G) must be divisible by p if G is non-abelian i.e. Z(G)<G then by strong induction there is an element of order p in Z(G). So the problem reduces to working with abelian G. Now let a (different from identity) be an element and form the subgroup H=\left< a \right> if H=G then the group is cyclic and the proof is complete. Otherwise H<G. Now if p divides |H| then proof is complete by strong induction so it is safe to say H is not divisible by p. Now form the factor group G/H (note H is normal since G is abelian) it is divisible by p since |G/H||H|=|G| and p does not divide H. So G/H must contain an element b which has order p by strong induction, i.e. (bH)^p \in H. We will not claim that b^m has order p (where m=|H|). First (b^m)^p = (b^p)^m = e since b^p\in H. Second we show this is the smallest exponent because if it was not then b^m = e but then (bH)^m = H \in G/H so p would be a divisor of m a contradiction. So b^m has order p.