ok, for a finite group G, and any subgroup H:

[G:H] = |G|/|H| (we know this is an integer from Lagrange's theorem).

if H is a normal subgroup, then G/H is a group, and since |G/H| is the number of cosets of H (these are the elements of G/H), we have:

|G/H| = [G:H] = |G|/|H|.

so, for any group G with a normal subgroup H:

|G| = |H|*|G/H| (and even if H is not normal, we STILL have |G| = |H|*[G:H]).

now if p is a prime number, with p|ab, then either p|a or p|b (this is a defining property of prime numbers).

so if p divides the order of G, it has to divide one of |H| or |G/H| (maybe both, it could happen).

now if p divides |H|, we're good, since |H| is assumed less than |G|, and we can apply our induction hypothesis.

but maybe it doesn't. in that case p HAS to divide |G/H|. since |H| < |G|, and |G| = |H|*|G/H|, it follows that |G/H| < |G| as well, so we can apply our induction hypothesis to G/H (a smaller group than G).

of course, the elements of G/H are cosets of H, so an element of G/H of order p looks like this:

xH in G/H with: (xH)^{p}= H (H is the identity element of G/H).

but (xH)^{p}= (x^{p})H, so we have x^{p}is an element of H (since (xH)^{p}= (x^{p})H = H).

of course, that doesn't appear to be much help...we want an element g of G for which g^{p}is in the subgroup {e}, not H.

so let's look at H for a second. let's take a typical element h of H. what can we say about it?

well, we know that h DOESN'T have order p. this means that gcd(|h|,p) = 1. this means that h^{p}generates <h> just like h does.

that is, for EVERY h in H, h^{p}generates <h>.

now ANY element h of H lies in the subgroup <h> of H. and the union of all these cyclic subgroups of H is H itself.

that is: if a is in H, a lies in <h> for some h in H (we might have to take h = a, but that's OK).

but <h> = <h^{p}> so a = (h^{p})^{k}, for some h in H (which again, might be a, might not).

thus a = (h^{k})^{p}, a is a p-the power of h^{k}. clearly if h is in H, so is h^{k},

so ANY element of H is the p-th power of some element of H (we can do this because p DOESN'T divide |H|, not even "partially", because p is PRIME).

ok, so we know we have this x in G with x^{p}in H.

so x^{p}= h, for some h in H. thus h^{-1}x^{p}= e.

now h^{-1}is in H, so there is some element (let's call it h') with:

h'^{p}= h^{-1}.

here is where the abelian part comes in:

consider h'x.

(h'x)^{p}= (h'x)(h'x)....(h'x) (p times).

= (h')(h')....(h')(x)(x)....(x) (each factor p times)....because G is ABELIAN.

= (h')^{p}x^{p}= h^{-1}x^{p}= e, so h'x has order p, as desired.

**************

to answer some of your questions:

yes, [G:H], the INDEX of H in G, is the number of left (OR right) cosets of H in G. even if H is not normal, the number of left cosets and right cosets are the same (even if they aren't the same cosets). of course, in an ABELIAN group they always WILL be the same (all subgroups of an abelian group are normal, one of the nice things about abelian groups).

i agree that the phrase: "it is easily checked..." is a cover-up. you need some more explanation, there. i hope you can follow my argument above.