Groups of order 16 are 2-groups, so how would Sylow's theorems even be used at all?

@Steve: Sylow theorems will give existence of subgroups of order 2,4,8; from this we can show that $G$ has normal subgroups of order 2,4,8; from this we can show $G$ has abelian normal subgroup of order 8; center of $G$ intersects each normal subgroup non-trivially...for each statement we need Sylow theorem; all these facts can be used when we face "isomorphism problem" for groups of order 16 constructed.

No, you do not need Sylow's theorems at all: again, when in a p-group, Sylow's theorems tell you nothing.

Then how can one prove existence of subgroups of $p$-groups?

You mean Cauchy's theorem - which is trivial for p-groups? Or the general fact that the center is non-trivial - which follows from letting the group act on itself via conjugation? What exactly are you taking as Sylow's theorems? The existence of a maximal p-subgroup (which would be your whole group in the p-group case)? That your whole group is conjugate to itself? Or that every subgroup of your group is contained in your group?

As stated in Alperin-Bell's book, he included "existance of p-subgroup, and every p-subgroup of G is in some Sylow-p subgroup". So in particular for p-groups, the existance of p-subgroups follows from Sylow theorem.

Continue!

I am looking at the Alperin-Bell book now: which statement do you mean? He shows the existence of a Sylow p-subgroup, but again, for a p-group, that is simply the entire group.

Sylow theorem in "Local structure."

Yes, he states four conclusions: (i) G has a Sylow subgroup; (ii) all Sylows are conjugate; (iii) every p-subgroup of G is contained in a Sylow; (iv) the number of Sylows is 1 mod p.

(iii)

That says nothing about the existence of p-subgroups, merely that if they exist, they are so contained.

ok

And again, in the p-group case, it is equivalent to saying "subgroups of your group are subgroups".

(iv) is still valid for p-groups also..

Again, in a p-group, the Sylow subgroup is just the whole group, so (iv) just says 1=1 mod p.

It is true that the number of subgroups of any given order is also 1 (mod p), but that does not follow from Sylow's theorems.

ok! I will look at all the things again..

ok, thanks for inviting me to chat! Never have done this before

same thing for me also..thanks!!