Problem Set 04

Read

We’ll continue with Reeder’s notes.

• Pick up right after the proof of Lemma 10.10, and read through the statement of Theorem 10.12 (often called Burnside’s Normal $p$-complement Theorem). You can also read Example 3 or just save it for next time.

To Work On

1. Read the discussion in the paragraph following Lemma 10.10, and put together a proof of the following lemma. Then use the lemma to quickly prove the corollary that follows.

   Lemma (Transfer Evaluation). Let $G$ be a finite group, $H$ an abelian subgroup of $G$, and $T:G\rightarrow H$ the transfer map. Then for each $g\in G$, there exists $k_1,\ldots, k_r\in \mathbb{N}$ and $x_1,\ldots, x_r\in G$ such that \[T(g) = (x_1^{-1}g^{k_1}x_1)\cdots(x_r^{-1}g^{k_r}x_r)\] where $k_1 + \cdots + k_r = |G:H|$ and $x_i^{-1}g^{k_i}x_i \in H$ for all $1\le i\le r$.

   Corollary (Transfer to Central Subgroup). Let $G$ be a finite group, $H\le Z(G)$, and $T:G\rightarrow H$ the transfer map. Set $m= |G:H|$. Then $T(g) = g^m$ for all $g\in G$.

2. Let’s rewrite the previous Lemma and Corollary in the special case when $H$ is a Sylow $p$-subgroup of $G$ and the element $g$ is in $H$. Assume the setup is: $G$ is a finite group, $P$ is an abelian Sylow $p$-subgroup of $G$, and $T:G\rightarrow P$ is the transfer map. Assuming also that $g\in P$, what adjustment does Lemma 10.11 from Reeder’s notes imply you can make to the statement of the Transfer Evaluation lemma (written above)? (Hint: it’s about what group the $x_i$ can be chosen from.) Now use this adjusted lemma prove the following corollary.

   Corollary. Let $G$ be a finite group and $P\in \operatorname{Syl}_p(G)$. If $P\le Z(N_G(P))$ (or equivalently: $N_G(P) = C_G(P)$), then for $T:G\rightarrow P$ the transfer map and $m= |G:P|$, we have that $T(p) = p^m$ for all $p\in P$.

3. Write up—in your own words—a proof of Theorem 10.12. Start from the last corollary above. Then argue (as in Reeder’s notes) that the transfer map to $P$ is surjective, and conclude that the kernel of the transfer map is a normal $p$-complement.

4. Let’s see an easy example of using Burnside’s Normal $p$-complement Theorem. Let $G$ be a group of order $2^45^2$. Prove that $G$ is not simple following the steps below:
   • Quickly show that $n_5$ is either 1 or 16. If $n_5=1$, we’re done so…
   • Assume $n_5=16$ and $P\in \operatorname{Syl}_5(G)$. Explain why $P = N_G(P)$. Confirm that the hypotheses of Burnside’s Normal $p$-complement Theorem are met (noting that $P$ is abelian…do you see why?), and apply it to show $G$ is not simple in this case.
     
     
5. For this problem, let’s not use the transfer or Burnside’s theorem, but in the next problem set, we’ll see how Burnside’s theorem allows us to prove something much stronger. Using Sylow theory (and anything else from 210A), prove the following lemma about groups of order $p^2q$ (extending what you proved about groups of order $pq$ and $pqr$ in Problem Set 02).

   Lemma. Let $p,q$ be distinct primes. If $G$ has order $p^2q$, then $G$ is not simple.

   • There are two cases. When $p>q$, you should quickly find that $G$ is not simple. When $p< q$, focus on $n_q$: argue that $n_q = p^2$ and then combine this with the fact that $n_q \equiv 1$ mod $p$ (which some may prefer to work with as $n_q = 1 + pk$ for some $k\in \mathbb{Z}$) to show that $q=p+1$. The only possibility then is that $q=3$ and $p=2$, so now you can use Problem 4 from Problem Set 03.
     
     
6. A continuation from previous weeks: using the results we’ve built so far, try to determine all simple groups of order $n$ (if any) for as many values of $n$ between $1$ and $100$ as possible.
   • We have many general results to apply (groups or order $pq$, $pqr$, $p^2q$, $2\cdot odd$), but you can also try to analyze specific orders using all of our tools, like basic Sylow analysis and exploiting subgroups of small index (see Problem 2 of Problem Set 02). These two methods often get combined as follows: use Sylow analysis to understand the possibilities for $n_p$ for the various primes $p$; if $n_p$ is “small”, then remembering that $n_p = |G:N_G(P)|$ (for $p\in \operatorname{Syl}_p(G)$), you can try to exploit that $N_G(P)$ has small index to prove $G$ is not simple. Sometimes you can just simply exploit that $P$ has small index. Maybe take a look at groups of order 24, 36, 48, and 56.