The union of two normal subgroups may not even be a subgroup. When multiples of 3 and 5 are combined, they don't contain the number 8, and are not a subgroup of Z. However, any set of elements, including the union of preexisting subgroups, can be used to generate a new subgroup.
If Q is a collection of group elements, let R contain Q, the inverses of Q, and all finite products taken from Q and its inverses. If a and b are in R, then a/b is in R, hence it is a subgroup, in fact the smallest subgroup that contains Q.
Next expand R, so that it includes xw/x for every w in R and x in G. This is still a subgroup, possibly larger. Note that y(xw/x)/y = (yx)w/(yx) is in R, so R satisfies the criterion for a normal subgroup. In fact R is the smallest normal subgroup containing Q.
Let K be normal in G and let H be any subgroup of G.
Let R be the subgroup generated by K and H,
also called K join H,
and let J be the intersection of K and H.
For every x in G, x*K/x winds up in K.
This still holds if x belongs to R, hence K is normal in R.
Except for J itself, each left coset of J is completely separate from J. If part of a coset of J is in H, the entire coset is in H, since all of J is in H - and similarly for K. A coset of J might belong to K, or H, but not both, as J is their intersection.
Let X be the left cosets of J in H, and Y the left cosets of J in K. Formally, X is a collection of cosreps, somewhat arbitrary, and so is Y. Let 1, the group identity, represent J in both X and Y. Since J is the complete intersection of H and K, 1 is the only element common to X and Y.
If an element e1 is x1y1j1, and e2 is x2y2j2, what can we say about e3, the product of e1 and e2? It begins life as x1y1j1x2y2j2. Since K is normal in R, the left and right cosets of x2 are the same. We can move y1j1 to the right of x2, replacing it with a new value from K. This gives x1x2y3j3y2j2. Now everything to the right of x2 is an element of K, and x1x2 is another coset of J in H, so the element e3 is the product of elements taken from x, y, and J, in that order. Since R consists of finite products from H and K, all of R is represented by these triples.
If two triples lead to the same element, i.e. x1y1j1 = x2y2j2, multiply by x2 inverse on the left. Now the product x2 inverse times x1, call it x3, is multiplied by something in K, and produces something in K. Clearly x3 is in K, and in H, hence in J, so the original cosreps x1 and x2 are the same. Divide these out, leaving elements in K.
Now y1 = y2 and j1 = j2, and the triples were in fact the same. Elements of R correspond 1-1 with triples drawn from x, y, and J.
The xyj representation is convenient, but don't assume you can just multiply components together. After all, R is not a direct product. When we pull elements past each other, y1 is replaced with y3, and so on. However, the projection onto H/J is a proper group homomorphism. This is not surprising, since K is normal in R. If e1 begins with x1, and e2 starts with x2, then e1e2 starts with x1x2. The cosets of J in H faithfully represent the cosets of K in R.
If R is finite, the order of R is |H|*|K|/|J|. Divide through by |J|, and the index of J in R is the index of J in H times the index of J in K. Cosets of J in H become cosets of K in R, and cosets of J in K become cosets of H in R. Therefore the index of J in R is the index of K in R times the index of H in R. Multiply each factor by the index of R in G, and the index of J in G ≤ the index of K in G times the index of H in G.
Of course there is an easier way to prove this index inequality, that does not require H or K to be normal. Let G act on the left cosets of H and K, whence it acts on the cartesian product of these two sets. The stabilizer of H cross K is precisely the intersection of H and K. The orbit, or index of the stabilizer, is at most the product of the two indexes. Therefore the index of H∩K is at most the index of H times the index of K.
If H and K are disjoint, intersecting only in the identity element, then we don't have to worry about J at all. Thus R becomes the direct product of H and K.