Groups, Joining and Intersecting Subgroups

Joining and Intersecting Subgroups

Using the a/b criterion, the intersection of arbitrarily many subgroups is a subgroup. Using the x*H/x criterion, the arbitrary intersection of normal subgroups is normal. We don't have to worry about an empty intersection; every subgroup contains the identity element.

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.

Joining a Subgroup and a Normal Subgroup

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.

An Abnormal Join

Watch what happens when H and K are not normal. Let G = A₅, the alternating group on 5 letters. Let H be a 5 cycle, and let K be a 3 cycle. The intersection is 1. Thus H join K looks like it should be a group of size 15; yet H and J generate G.

Joining Disjoint Normal Subgroups

Let H and K be normal in G, and let R be H join K, as described above. Now the projection onto H/J and the projection onto K/J are both group homomorphisms. In other words, we can simply multiply x₁x₂ and y₁y₂.

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.