First pretend that a normal subgroup is trivial

Quick description

When dealing with a group theory problem involving a normal subgroup H of a group G, pretend first that H is trivial, and solve this simpler problem. To then handle the general case, quotient everything by H and apply what you have just done. In many cases, this solves the problem "modulo H"; to finish the job, one has to figure out how to deal with some residual objects lying in H.

Prerequisites

Basic group theory

Example 1

Theorem 1 Let H, K be normal solvable subgroups of a group G. Then the subgroup HK generated by H and K is also normal and solvable.

Proof. Normality is easy: every inner automorphism $h \mapsto ghg^{-1}$ of G preserves H and K, so will certainly preserve HK. The issue is to verify solvability.

When H is trivial, there is nothing to prove; HK is just K, and so the solvability of K implies the solvability of HK. To take advantage of this, let's quotient out by H (thus "forcing" H to become trivial). Write $G \to G/H$ for the quotient homomorphism. Observe that HK and K have the same image under this map: $\pi(HK)=\pi(K)$ . From this we soon conclude that the commutator groups [HK,HK] and [K,K] have the same image: $\pi([HK,HK]) = \pi([K,K])$ . We iterate this and see that the derived series $(HK)^{(i+1)} = [(HK)^{(i)}, (HK)^{(i)}], K^{(i+1)} = [K^{(i)}, K^{(i)}]$ of HK and K have exactly the same projection under $\pi$ . As K is solvable, this means that some member of the derived series of HK has trivial projection under $\pi$ , i.e. it lies in H. To finish off the job, we now work entirely inside H. Since H is solvable, every subgroup of H is also solvable; so if one continues the derived series of HK further, it will eventually terminate after finitely many steps, as required. $\Box$

Example 2

Theorem 2 Let H, K be normal nilpotent subgroups of a group G. Then the subgroup HK generated by H and K is also normal and nilpotent.

Proof This one is a bit trickier. Our task is to keep proceeding along the lower central series $(HK)_{i+1} = [HK,(HK)_i]$ and show that it is eventually trivial.

The exact same argument as before shows that some element $(HK)_i$ of this is contained in H. However, the nilpotency of H doesn't then kill the problem off immediately, because we still have to take commutators of $(HK)_i$ with elements of HK, which can include elements outside of H - in particular, we have to understand what happens if we take commutators between elements of $(HK)_i$ and elements of $K$ .

The trick now is to introduce a new normal subgroup, namely the intersection $H \cap K$ of H and K. (Again, this subgroup is preserved by all inner automorphisms and is hence normal.) Let's pretend first that this subgroup was trivial. Observe that the commutator $[h,k] = hkh^{-1}k^{-1}$ of two elements $h\in H, k \in K$ lie in both H and K (by normality) and is thus trivial. In other words, every element of H commutes with every element of K. In particular, every element of $(HK)_i \subset H$ commutes with every element of K; also, HK is now equivalent to the direct product $H \times K$ . So when taking commutators between an element of $(HK)_i$ and an element hk of HK, the K component k of hk has no impact on the commutator, and we are now just taking commutators between elements of $(HK)_i$ and elements of H. At this point the nilpotency of H kicks in and makes the lower central series trivial after finitely many steps.

OK, this handles the case when $H \cap K$ is trivial. What if it is not trivial? Well, we quotient it out, and argue as in the previous theorem. At the end of the day, what happens is that we obtain some subsequent element $HK_j$ of the lower central series that lies in $H \cap K$ .

Now what? Well, we still have the problem that we can't work completely inside H yet; we have to take commutators of elements in $(HK)_j \subset H \cap K$ with elements in HK, and in particular with elements of K. Now, there's no reason why an element of K has to commute with an element of $H \cap K$ . But suppose that $H \cap K_2 = H \cap [K,K]$ was trivial; then K would commute with everybody in $H \cap K$ , and we could use the nilpotency of H as before to conclude the argument.

What if $H \cap K_2$ is not trivial? Well, it is a normal subgroup of G, so we can quotient it out and apply the previous argument. The upshot is that we get another element $(HK)_k$ of the lower central series which is now contained in $H \cap K_2$ . If $H \cap K_3$ was trivial, then this group is centralized by K and we can argue as before; so, by moving even further down the lower central series, we eventually land inside $H \cap K_3$ , then $H \cap K_4$ , etc. Since K is itself nilpotent, we eventually obtain triviality. $\Box$

Example 3

Theorem 3 Let G be a group, with lower central series $G_1 = G, G_2 = [G,G_1], G_3 = [G,G_2]$ , etc. Then $[G_i,G_j] \subset G_{i+j}$ .

Proof It's quite natural to use induction here, so let's induct on k := i+j, assuming the claim is proven for smaller values of i+j. Fixing this k, we see that the claim is obvious for i=1, so we can perform a second induction, assuming that i>1 and the claim is already proven for i-1 (and fixed k), thus $[G_{i-1},G_{j+1}] \subset G_{i+j}$ .

It suffices to verify the claim when $G_{i+j}$ is trivial, since otherwise we can quotient out by $G_{i+j}$ (which is normal, being preserved by all inner (or outer!) automorphisms) and reduce to this case. By the second induction hypothesis, this means that $G_{i-1}$ commutes with $G_{j+1}$ . Our task is to show that every element of $G_i$ commutes with $G_j$ .

By construction, $G_i$ is generated by commutators $[g,g_{i-1}]$ , where $g \in G$ and $g_{i-1} \in G_{i-1}$ . So it will suffice to show that all the commutators $[g,g_{i-1}]$ commute with an element $g_j$ of $G_j$ . Equivalently, we want to show that the inner automorphism $\rho_{g_j}$ of conjugation by $g_j$ leaves $[g,g_{i-1}]$ invariant.

By the first induction hypothesis, $[g_{i-1},g_j]\in G_{i+j-1}$ , thus $\rho_{g_j}$ moves $g_{i-1}$ by an element $h$ in $G_{i+j-1}$ . But since $[G,G_{i+j-1}]=G_{i+j}$ is trivial, $G_{i+j-1}$ is central and so $h$ eventually has no impact on $\rho_{g_j}( [g_{i-1},g_j] )$ . Also, $\rho_{g_j}$ moves $g$ by an element in $[G,G_j] = G_{j+1}$ , which by hypothesis commutes with $g_{i-1}$ . Putting all this together we see that $\rho_{g_j}$ leaves $[g_{i-1},g_j]$ invariant, as required. $\Box$

General discussion

In fancier language, all one is doing here is exploiting a short exact sequence $0 \to H \to G \to G/H \to 0$ to factor a group theory problem into two smaller pieces.