It is a little hard to know what to answer here since the difficulty (and the method) of finding the prime power decomposition of a given abelian (finite) group, depends on how the group is given to you in the first place. The theiorem says of course that a finite abelian group is isomorphic to a product of cyclic groups, each of prime power order. OK. There is also another theorem that says a finite abelian group is isomorphic to a product of cuyclic groups, where the irder of each cyclic factor divides the order of the next one in the product.
So one way of giving the group in the first place is to give a product of the second type, and ask you to express it as a product of the first type. E.g. try expressing Z/(14Z) x Z/(42Z) x Z/(252Z) as a product of cyclic factors each of prime power order. You can make up lots of these examples and solve them. This transition is one basic fact and skill you should have in this subject. Similar examples can be constructed just by starting from any product fo cyclic groups whatever, without regard to how the order of one relates to the order of the next, e.g. Z/(16) x Z/(30Z) x Z/(54Z). Try some of these, reducing them to either a product ofmprime power factors, of a product of factors whose orders do divide each other.
Another way to express a finite abelian group is as a quotient of a free abelian group by a subgroup of maximal rank. All such subgroups are themselves free abelian, so the quotient can always be expressed as the "cokernel" of a homomorphism of free abelian groups, and this homomorphism can always be expressed as a matrix of integers. This is called giving a "presentation matrix" for the quotient group. An m by n matrix of integers defines a homomorphism from Z^n to Z^m, and the image is spanned by the columns. Thus the "cokernel" is the quotient of the target, i.e. of Z^m, modded out by the subgroup spanned by the n columns. In order for this cokernel to be finite, i.e. for the matrix to have maximal rank, there must then be at least m columns, i.e. we must have n ≥ m. Giving a presentation matrix for a group requires finding a finite set of m generators and a finite set of n "relations" among those generators, i.e. linear equations they satisfy. A finite group has a finite set of generators, since just taking all the elements as generators would do, and it is a theorem that any finite set of generators has a finite set of generators for its relations, so every finite abelian group has a presentation matrix.
The easy case for expressing the cokernel as a quotient of cyclic groups is when the matrix is "diagonal", i.e. n ≥ m, the first m columns form a square diagonal m by m matrix, and the rest of the columns are zero. In this case if the jth column has a single non zero entry equal to r in the jth place, this gives rise to a cyclic factor Z/(rZ). So the problem of expressing any group given by any presentation matrix, reduces to the problem of diagonalizing an integer matrix. Thus one proof, and one of the proofs I give in my notes, is to show how to use just the Euclidean algorithm to diagonalize any integer matrix. This expresses your group as a product of cyclic groups, and then you can proceed to change this product into one of the two standard ones mentioned above. Or, one can show that the diagonalizing porocess can be continued in such a way as to give factors whose orders divide each iother consecutively, i.e. one can do the diagonalization so as to give the "second" standard decomposition mentioned above. Then one can change it into a prime power decomposition. Indeed this is one proof I give in my notes.
I did write my notes so that this topic was treated last, after I had covered group theory including the Sylow theorems, but you can perhaps tell from this description here that only the concept of the Euclidean algorithm is needed to carry out the decomposition in practice, at least if you assume as given a presentation matrix. All this is discussed on pages 11-16 of my notes 845-1 on my webpage above. You can see there that the method also applies to finitely generated abelian groups that may not be finite, and gives a decomposition with possibly some factors of Z. This is the case when the matrix may not have maximal rank m. An earlier discussion of cyclicmproducts occurs on pages 47-50 of the notes for 844-2.
And don't be afraid of the words "Sylow p -subgroup". For an abelian finite group, if the group has order np^k where p does not divide n, the sylow p subgroup it is the unique subgroup of order p^k. So in the decomposition, it is just the product of all cyclic factors whose order is a power of p. So all that is happening in these discussions of looking at "Sylow subgroups" is that all factors involving the same prime are being bunched together. E.g. an abelian group G of order 36 = 2^2.3^2, has two Sylow subgroups, one of order 4 and one of order 9, and G is the product of those subgroups, which follows from the "Chinese remainder theorem". I.e. if these subgroups are H and K, the map HxK-->G taking (x,y) to xy, is an isomorphism [exercise].
