Problem Set 05

Read

1. Review automorphism groups of groups. Take a look back in your 210A textbook about automorphisms of groups and the relationship of $N_G(H)$ and $\operatorname{Aut}(H)$ when $H\le G$. In Dummit & Foote, this would be Section 4.4. There doesn’t seem to be much review in Reeder’s notes, but the book by Machì has some good stuff in Section 1.3.
2. Continue with Reeder’s notes: start from Example 3 (right after Theorem 10.12) and continue through the end of Corollary 10.15.

To Work On

1. After reviewing automorphism groups, put together brief explanations/proofs of the following in your own words.

   Lemma. Let $H$ be a group.

   1. If $H$ is cyclic of order $n$, then $\operatorname{Aut}(H)\cong (\mathbb{Z}/n\mathbb{Z})^\times$.¹
   2. If $H$ is an elementary abelian $p$-group of order $p^n$, then $\operatorname{Aut}(H)\cong \operatorname{GL}_n(\mathbb{Z}/p\mathbb{Z})$.²

   Lemma. Let $G$ be a group and $H\le G$. Then $N_G(H)/C_G(H)$ is isomorphic to a subgroup of $\operatorname{Aut}(H)$. In particular,

   1. if $H$ is cyclic of order $n$, then $|N_G(H)/C_G(H)|$ divides $\phi(n)$;³
   2. if $H$ is an elementary abelian $p$-group of order $p^n$, then $|N_G(H)/C_G(H)|$ divides $\prod_{k=0}^{n-1}(p^n - p^{k})$.

2. Write up a proof of Corollary 10.13 from Reeder’s notes in your own words (using the above lemmas).

3. Prove the following variant of Corollary 10.15 from Reeder’s notes (using the above lemmas). Please don’t cite Corollary 10.15, but you can use the ideas you see in the proof. (Notice how this result generalizes our earlier results about groups of order $pqr$ and $p^2q$.)

   Lemma. Let $G$ be a finite group, and let $p$ be the smallest prime divisor of $|G|$. If $G$ is simple, then one of the following occur:

   • $|G|=p$,
   • $|G|$ is divisible by $12$, or
   • $|G|$ is divisible by $p^3$.

4. Prove the following lemma using the approach in Problem #4 in Problem Set 04. Using results we developed (like the “Small Index Lemma”, see #2 in Problem Set 02), work to show that $n_q$ must be as large as possible (i.e. $n_q = 2^4$); then Burnside’s Theorem applies.

   Lemma. Let $q$ be an odd prime. If $|G| = 2^4q^2$, then $G$ is not simple.

   Once you finish the lemma, tie it together with our previous work (and fill in any gaps) to prove the following corollary.

   Corollary. Let $q$ be an odd prime. If $|G|$ divides $2^4q^2$, then $G$ is not simple unless $|G|$ equals $2$ or $q$.

5. Prove the following classification of simple groups of order 60. To do this, you’ll want show that such a group acts on a set of size 5, and for that, aim to show that $n_2 = 5$ (using all the theory we developed, including Burnside’s Theorem).

   Lemma. If $G$ is a simple group of order $60$, then $G\cong\operatorname{Alt}(5)$.

6. Let’s work to prove that $\operatorname{Alt}(n)$ is simple for $n\ge 5$. This is typically done by induction, so let’s start by proving this for $n=5$.

   Lemma. $\operatorname{Alt}(5)$ is simple.

   One approach to the previous lemma is as follows. Assume $N$ is a normal subgroup of $\operatorname{Alt}(5)$. Then $N$ is a union of conjugacy classes of elements of $\operatorname{Alt}(5)$ (do you see why?). Now prove that the sizes of the conjugacy classes in $\operatorname{Alt}(5)$ are 1,15,20,12,12, and then show that the only way some of these numbers could add up $|N|$ (using that $|N|$ is a divisor of 60) is if $|N|$ is 1 or 60. Now, in 210, you determined conjugacy classes in $\operatorname{Sym}(5)$, and the difference for $\operatorname{Alt}(5)$ is this: suppose $\alpha \in \operatorname{Alt}(5)$, $\mathcal{C}_A(\alpha)$ is the conjugacy class of $\alpha$ in $\operatorname{Alt}(5)$, and $\mathcal{C}_S(\alpha)$ is the conjugacy class in $\operatorname{Sym}(5)$; then either $\mathcal{C}_A(\alpha) = \mathcal{C}_S(\alpha)$ (which happens when $C_{\operatorname{Sym}(5)}(\alpha) \not\le \operatorname{Alt}(5)$) or $|\mathcal{C}_A(\alpha)| = \frac{1}{2}|\mathcal{C}_S(\alpha)|$ (which happens when $C_{\operatorname{Sym}(5)}(\alpha) \le \operatorname{Alt}(5)$).

   Lemma. $\operatorname{Alt}(n)$ is simple for $n\ge 5$.

   Try working by induction: so assume $\operatorname{Alt}(n-1)$ is simple (with $n\ge 6$) and consider $G = \operatorname{Alt}(n)$. Suppose $N$ is a normal subgroup. Notice that for each $x\in \{1,\ldots,n\}$ the stbilizer $G_x$ is isomorphic to $\operatorname{Alt}(n-1)$. Since $G_x$ is simple and $G_x\cap N\trianglelefteq G_x$, either $G_x \le N$ or $G_x\cap N = 1$. In the first case, try to prove $N=G$. In the second case, argue that $|N| \le n$, so each element of $N$ has a conjugacy class of size at most $n$; then prove that the only element of $G$ with such a small conjugacy class is the identity, implying $N=1$ in this case.

7. Continuation from previous weeks: using everything we’ve developed so far, try to determine all simple groups of order $n$ (if any) for as many values of $n$ between $1$ and $200$ as possible.
   • Use all of our general results and then start looking at the orders left over. Maybe try investigating groups of order 120 or 132. Likely $120$ will put a little fight (mainly because $\operatorname{Sym}(5)$ has order $120$ and is close to being simple); consider trying this: show that a simple group $G$ of order $120$ must have $n_5 = 6$, so you get an injective homomorphism from $G$ to $\operatorname{Alt}(6)$ (do you remember why?)…but this means that $\operatorname{Alt}(6)$ has a subgroup of order $120$, which you can argue is not possible since it would have index $3$ in $\operatorname{Alt}(6)$.

Footnotes