## Wednesday, April 16, 2014

### (Nil)Potent problems

Somehow, every year about this time, I start thinking about preliminary exams. (Also known at other institutions as subject exams, qualifying exams, etc.: the dreaded first hurdle of staying in grad school.) Specifically, I start thinking about questions that would be appropriate to ask on such exams, in difficulty, length, and subject matter.

Now, I'm going to avoid my usual rant about the content of most prelims in algebra -- this post isn't about prelims per se, and besides that, the problem I want to talk about is one that I emphatically wish algebra prelims wouldn't test! But it's a fun little argument nevertheless.

Recall the following definition:
A finite group $$G$$ is nilpotent if, for every normal subgroup $$N \triangleleft G$$, the center of $$G/N$$ is nontrivial.
(I wrote "finite" in the definition because this is not quite the correct definition once infinite groups are allowed. To get there, one has to define the ascending central series of $$G$$. For finite groups, however, the definitions are equivalent.) We say that an abelian group is nilpotent of class 0, and if $$G/Z(G)$$ is nilpotent of class $$k$$, then $$G$$ is nilpotent of class $$k+1$$. The following characterization of nilpotent groups is famous:
Theorem 1: A finite group $$G$$ is nilpotent iff $$G$$ is the direct product of its Sylow subgroups.
The question I asked myself today is: would this theorem be an appropriate problem to pose on an algebra prelim?

Of course, the answer to that question depends on a few things: on the level of the students coming into the program, plus the level that the committee setting the prelim want to pitch; whether they want to use the prelim to screen moderately good students out or in; and what the articulated expectations about students coming out of the first-year algebra course actually are. I can imagine programs where every student can write a proof of this theorem in their sleep; I personally don't remember if I'd actually ever seen this proved. (I'd skimmed over at least one proof in a textbook, that much I remember, but nilpotent groups actually weren't ever a focus in any of my courses.)

So I set out to come up with a proof. Attacking it directly didn't get me very far; but I got a fat hint from Bertram Huppert's wonderful Grundlehren volume (which I've been using for bedtime reading -- puts you right to sleep). I think the hint itself is actually the right problem to put on a prelim:
Lemma 2: A finite group is the direct product of its Sylow subgroups iff all its Sylow subgroups are normal.
In fact, it is this latter property that Huppert uses as the definition of "nilpotent".

The reason I think this is the right problem to use is that the proof is not entirely obvious, but a student who has studied their algebra will try the right things fairly quickly, whereas the student who just knows the definitions will flounder around and get nowhere. Also, the proof I found just by fumbling around takes about a page of notepaper, which is the right length (assuming that one is supposed to solve five such problems in a three-hour timespan, which I think is pretty standard). Now, I haven't gone back into any of my algebra textbooks to see if there is a substantially easier proof of Theorem 1 than the one I'll present here; but as I mentioned, I tried direct assault and got basically nowhere, whereas once I knew to try Lemma 2 first, the proof basically wrote itself. However, the overall proof of Theorem 1 that I found is a bit long for a prelim problem, in my opinion.

Anyway, I'll prove both of these statements. If you know of a proof short enough (and elementary enough) to work on a prelim exam, leave it in comments!

Proof of Lemma 2: Recall that a group $$G$$ is the direct product of its subgroups $$N_1, \ldots, N_\ell$$ if the $$N_i$$ are all normal, intersect pairwise trivially, and for every $$g \in G$$ there exists exactly one sequence $$h_1 \in N_1, \ldots, h_\ell \in N_\ell$$ such that $$g = h_1 h_2 \cdots h_\ell$$.

We see from the definition that the direction ($$\Rightarrow$$) is trivial.

In the other direction: we know that if $$p,q$$ are distinct primes dividing $$|G|$$, any Sylow-$$p$$-subgroup $$S_p$$ and Sylow-$$q$$-subgroup $$S_q$$ intersect trivially. Now, if $$a \in S_p$$ and $$b \in S_q$$,
$a^{-1} b^{-1} a b = (b^{-1})^a b \in S_q$
$a^{-1} b^{-1} a b = a^{-1} a^b \in S_p$so this commutator is the identity. It follows that any two distinct Sylow subgroups commute elementwise.

From this, we see that $$S_p S_q$$ is a normal subgroup of $$G$$, of size $$|S_p| \cdot |S_q|$$, and likewise for any product of distinct Sylow subgroups. It follows that $$G = S_{p_1} S_{p_2} \cdots S_{p_\ell}$$, where the $$p_i$$ run through all primes dividing $$|G|$$. Every element $$g \in G$$, in other words, can be written $$g = a_1 a_2 \cdots a_\ell$$, where $$a_i \in S_{p_i}$$.

We must show that this representation is unique. Suppose we have
$a_1 a_2 \cdots a_\ell = b_1 b_2 \cdots b_\ell$Then $$b_1^{-1} a_1 \in S_{p_2} \cdots S_{p_\ell} \cap S_{p_1} = \{ 1 \}$$. Proceeding down the line, each $$a_i = b_i$$. $$\square$$

Now we can prove the equivalence we really wanted:

Lemma 3: The equivalent conditions of Lemma 2 are also equivalent to $$G$$ being nilpotent.

Proof ($$\Rightarrow$$): Assume $$G$$ is the direct product of its Sylow subgroups. Each Sylow subgroup is a $$p$$-group for some prime $$p$$, and since every $$p$$-group has nontrivial center (and has only $$p$$-groups as homomorphic images), each Sylow subgroup is nilpotent. Since there are only finitely many primes we're working with, there is some $$k$$ such that all $$S_p$$ are nilpotent of class $$\leq k$$. This is expressible by an equation, which is preserved under direct products; hence $$G$$ is nilpotent of class $$\leq k$$ too.

($$\Leftarrow$$): We proceed by induction: we assume that the desired equivalence holds in all proper subgroups and homomorphic images of $$G$$. As base cases, if $$G$$ is a $$p$$-group for some prime $$p$$, there is nothing to show.

Otherwise, for some prime $$p$$, $$p \mid |Z(G)|$$. Hence there exists a Sylow-$$p$$-subgroup $$P$$ which intersects the center nontrivially. Say $$1 \lneq P \cap Z(G) = P_0 \triangleleft G$$. Our inductive assumption is that all the Sylow subgroups of $$G/P_0$$ are normal. We denote the quotient map from $$G$$ to $$G/P_0$$ by $$\pi$$.

Now, for every $$H \leq G/P_0$$, $$\pi^{-1}(H)$$ is a subgroup of $$G$$ of size $$|H| \cdot |P_0|$$. If $$H$$ was a Sylow-$$q$$-subgroup, where $$p \neq q$$ and $$q^\alpha$$ was the largest power of $$q$$ dividing $$|G|$$, then $$|H| = q^\alpha$$ and $$\pi^{-1}(H)$$ is a normal subgroup of $$G$$ of size $$q^\alpha \cdot |P_0|$$; by induction, $$\pi^{-1}(H)$$ is the direct product of a Sylow-$$q$$-subgroup $$Q$$ and $$P_0$$. Since $$\pi^{-1}(H)$$ is normal in $$G$$ and $$Q$$ is characteristic in $$\pi^{-1}(H)$$, $$Q$$ is normal in $$G$$, and is a Sylow-$$q$$-subgroup of $$G$$.

The only way the previous paragraph could fail, is if $$\pi^{-1}(H) = G$$, for then we could not apply the induction assumption. However, in that case, we have that $$G/P_0$$ is a $$q$$-group, so $$P = P \cap Z(G)$$ and these two primes are the only ones dividing $$|G|$$. In this case, $$P$$ is a subgroup of $$Z(G)$$, and hence normal. Let $$Q$$ be a Sylow-$$q$$-subgroup of $$G$$; then $$G = QP$$, and if $$g = qp$$ is an arbitrary element of $$G$$, then
$Q^g = Q^{qp} = Q^p = Q$so both $$P$$ and $$Q$$ are normal. $$\square$$