Problem Set 03

Read

We’re going to start reading about the so-called transfer homomorphism, and we’ll use notes by Mark Reeder as our main source, but the book by Machì (starting on page 225) is similar and worth looking at too (though note that Machì works with right cosets and right actions).

Download Reeder’s notes here.

Given a group $G$ and an abelian subgroup $H$, the transfer is a particular homomorphism from $G$ to $H$. Once we understand the definition of the transfer, our attention will turn to trying to answer the question: under what conditions can we determine the kernel?

• Read the beginning of Section 10.3 in the notes from the start of the section on page 68 through the end of the proof of Lemma 10.10 on page 70.
• Regarding notation:
  • Reeder uses $C_n$ for a cyclic group of order $n$ in the way Dummit and Foote use $Z_n$.
  • You’ll see semidirect product notation $M\rtimes P$. We will probably need little more than the “internal” definition given in Section 2.10 (pg 13).
• Some of the problems below are aimed at making sense of the reading and trying to add clarity, but also make sure to write down any questions that arise!

To Work On

1. Reeder’s notes mention that $\operatorname{Sym}(4)$ does not have a normal 2-complement or normal 3-complement. Give a brief explanation of this by applying the definition of normal p-complements to $\operatorname{Sym}(4)$. Also, use what you showed about groups of order $pq$, with $p< q$ primes, from Problem Set 02 to show that such groups always have a normal $p$-complement.

2. Lemma 10.10 from Reeder’s notes and the paragraph preceding it are the main things we want to grapple with this week.
   • Develop some of your own language, analogies, and/or pictures (…please draw pictures!) to better explain the equation $gx_i = x_{\sigma_g(i)}h_i(g)$ and all of the symbols involved.
   • Can you add clarity and/or pictures to the proof of Lemma 10.10? Or simply highlight key questions you have?
     
     
3. Revisit Problems 3-5 from Problem Set 02.
   • When thinking about a group $G$ of order $pqr$, consider trying a counting argument. To do this, assume $p<q<r$. Starting Sylow analysis, we ask the question: what happens if $n_p$, $n_q$, and $n_r$ are all larger than $1$? In this case $n_p\ge q$, $n_q\ge r$, and $n_r= pq$ (do you see why?). The “counting idea” is to try to count (or rather, get a lower bound on) the number of elements of order $p$, of order $q$, and of order $r$; in doing so, you get a lower bound on the number of elements in $G$, which may provide a contradiction. For example, each Sylow $p$-subgroup contains $p-1$ elements of order $p$ (do you see why?) and every two distinct Sylow $p$-subgroups have a trivial intersection (do you see why?); from this, we see that $G$ has $n_p(p-1) \ge q(p-1)$ elements of order $p$. Then continue with elements of order $q$ and $r$.
     
     
4. Our our hunt for nonabelian simple groups, we are currently stopped at groups of order $12$. To address this, try proving the lemma below. The starting point is Sylow: $n_3$ is 1 or 4. Now, if $n_3=4$, then this means a certain subgroup of $G$ has index $4$, so try acting on the left cosets of that subgroup and see what the kernel could be.

   Lemma. Let $G$ be a group of order 12. Then either $G$ has a normal Sylow $3$-subgroup or $G\cong \operatorname{Alt}(4)$ (in which case it has a normal Sylow $2$-subgroup).

5. A continuation/reformulation 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.