# Group Theory Qual Review

Robert Won

Prof. Rogalski

## 1 (Some) qual problems and (some) techniques

ˆ (Spring 2008, 1) Let G be a finite group and H a proper subgroup. Show that G is not the set-theoretic union of the conjugates of H.

Consider the intersection and count.

ˆ (Spring 2008, 2) Classify all groups with 99 elements.

These types of problems are very common, so do a lot of these as practice. Your tools include Sylow, semidirect products, etc.

ˆ (Spring 2008, 3) Let p be prime. ${ \mathrm { I f ~ } } | G | = p ^ { n }$ and N is a normal subgroup, show that N intersects the center of G nontrivially.

A normal subgroup is a union of conjugacy classes. Count.

ˆ (Spring 2007, 1) Let p be a prime and G a group of order $p ^ { 3 }$

(a) Prove that G has a normal subgroup of order $p ^ { 2 }$

(b) Assume that G has a cyclic normal subgroup N of order $p ^ { 2 }$ generated by some element n. Let g be an element not in N .

i. If the order |g| of g is $p ^ { 3 }$ , classify the possible G up to isomorphism.

ii. If the order |g| of g is $p ,$ classify the possible G up to isomorphism

Use Sylow, semidirect products.

ˆ (Fall 2007, 1) Let G be a group of order $2 4 0 = 2 ^ { 4 } \cdot 3 \cdot 5$

(a) How many p-Sylow subgroups might G have, for $p = 2 , 3 , 5 ?$

(b) If G has a subgroup of order 15, show that it has an element of order 15.

(c) Say G does not have a subgroup of order 15. Show that the number of 3-Sylows is 10 or 40.

Use Sylow, use Sylow again on the subgroup of order 15, semidirect products.

ˆ (Fall 2006, 2.1) Let p be a prime number. $( \mathbb { Z } / p ^ { 2 } \mathbb { Z } ) ^ { \times }$ denotes the multiplicative group consisting of all congruence classes $\hat { x } \in \mathbb { Z } / p ^ { 2 } \mathbb { Z }$ such that $\operatorname* { g c d } ( x , p ) = 1$

(a) Show that the order of 1 +ˆ p in $( \mathbb { Z } / p ^ { 2 } \mathbb { Z } ) ^ { \times }$ is equal to $p .$

(b) Use (a) to construct a non-abelian group of order $p ^ { 3 }$

(c) Describe the non-abelian group in (b) via generators and relations.

Semidirect products, etc.

ˆ (Fall 2006, 2.2) Let G be a group. Let $r \geq 2$ be an integer. Assume that G contains a non-trivial subgroup H of index $[ G : H ] = r$ . Prove the following.

(a) If G is simple, then G is finite and |G| divides r!.

(b) If $r \in \{ 2 , 3 , 4 \}$ , then G cannot be simple.

(c) For all integers $r \geq 5$ , there exist simple groups G which contain non-trivial subgroups H of index $[ G : H ] = r$

If G is simple, act on cosets of H by multiplication to give an injection $G  S _ { n }$ . This is a common technique when you are dealing with simple groups. Also see Dummit and Foote pp. 201-213.

## 2 (Some) group things to know

ˆ Basic facts and definitions. (homomorphisms, isomorphism theorems, subgroups, normal subgroups, normalizers, centralizers, quotient groups, cyclic groups, dihedral groups, symmetric groups, etc.)

$H \leq G$ . Given $a , b \in G$ , either $a H = b H \Leftrightarrow a ^ { - 1 } b \in H$ or $a H \cap b H = \emptyset$ . So cosets partition G and $\vert a H \vert = \vert H \vert$

$H \leq G .$ . Then $| G / H | = | G | / | H | = [ G : H ]$

$K \leq H , H \leq G$ . Then $[ G : K ] = [ G : H ] [ H : K ]$

ˆ The kernel of a group homomorphism is a normal subgroup.

ˆ G act on A, then for each $g \in G$ , we get $\sigma _ { g } : A  A$ . This $\sigma _ { g }$ is a permutation of A and the map $G \to S _ { a } , g \mapsto \sigma _ { g }$ is a homomorphism.

ˆ (Orbit-stabilizer) $| \mathcal { O } _ { x } | = [ G : G _ { x } ] = | G | / | G _ { x } |$

ˆ Automorphisms

If $H \leq G$ , then G acts by conjugation on H as automorphisms of H. Also $G / C _ { G } ( H ) \cong \mathrm { a }$ subgroup of $\operatorname { A u t } ( H )$

For any $H \leq G , N _ { G } ( H ) / C _ { G } ( H ) \cong \mathrm { a }$ subgroup of $\operatorname { A u t } ( H )$

$G / Z ( G ) \cong$ subgroup of Aut(G).

p a prime =⇒ Aut(Zp) ∼= Zp−1.

## ˆ Isomorphism Theorems

First Isomorphism Theorem: If $\varphi : G \to H$ is a homomorphism, then ker $\varphi \ \leq \ G$ and $G / \ker \varphi \cong \varphi ( G )$

ϕ injective ⇐⇒ ker $\varphi = 1$

Second Isomorphism Theorem: $A \leq G , B \leq G$ and $A \leq N _ { G } ( B )$ (or $B \leq G )$ . Then $A B \leq G$ and $B \triangleleft A B , A \cap B \triangleleft A$ and $A B / B \cong A / A \cap B .$ •

|AB| = |A||B|/|A ∩ B|.

Third Isomorphism Theorem: $H \leq G$ and $K \leq G$ with $H \leq K$ . Then $K / H \triangleleft G / H$ and ${ \overline { { G } } } / { \overline { { K } } } \cong G / K$

ˆ Characteristic subgroups

Characteristic subgroups are normal.

If $H \leq G$ is the unique subgroup of a given order, then H char G.

K char H and $H \triangleleft G \Longrightarrow K \triangleleft G .$

ˆ (Lagrange’s Theorem) G a finite group, $H \leq G$ , then $| H | \mid G |$

ˆ (Cauchy’s Theorem) G a finite group and p a prime such that $p \mid \mid G \mid$ then G has an element of order p.

ˆ (Sylow’s Theorem)

Sylow p-subgroups of G exist.

If $P \in S y l _ { p } ( G )$ and Q any p-subgroup of G, then $Q \leq g P g ^ { - 1 }$

$n _ { p } \equiv 1 { \pmod { p } }$ and $n _ { p } = [ G : N _ { G } ( P ) ]$

$n _ { p } = 1 \Longleftrightarrow P \triangleleft G \Longleftrightarrow P$ char G ⇐⇒ All subgroups generated by elements of p-power order are p-groups.

ˆ (Fundamental Theorem of Finitely Generated Abelian Groups)

$$
G \cong \mathbb { Z } ^ { r } \times \mathbb { Z } _ { n _ { 1 } } \times \cdot \cdot \cdot \times \mathbb { Z } _ { n _ { s } }
$$

Invariant factors: $n _ { i } \mid n _ { i + 1 } f o r 1 \leq i \leq s - 1$

Elementary divisors

If n is the product of distinct primes, the only abelian group of order n is the cyclic group of order $n , \mathbb { Z } _ { n }$

$$
\mathbb { Z } _ { m } \times \mathbb { Z } _ { n } \cong \mathbb { Z } _ { m n } \Longleftrightarrow ( m , n ) = 1 .
$$

ˆ (Class equation)

$\begin{array} { r } { | G | = | Z ( G ) | + \sum _ { i } [ G : C _ { G } ( x _ { i } ) ] } \end{array}$ (one $x _ { i }$ from each conjugacy class).

ˆ Commutators

$[ x , y ] = x ^ { - 1 } y ^ { - 1 } x y$ is called the commutator (= 1 iff x and y commute).

$G ^ { \prime } = \langle [ x , y ] \mid x , y \in G \rangle$ is the commutator subgroup (= 1 iff G abelian).

$$
x y = y x [ x , y ] .
$$

$$
H \triangleleft G { \mathrm { i f f ~ } } [ H , G ] \leq H .
$$

G0 char G and $G / G ^ { \prime }$ is abelian (the largest abelian quotient).

If $G ^ { \prime } \leq H , H \leq G$ , then G/H is abelian

ˆ Direct products

If $H , K \leq G$ and $H \cap K = 1$ , then $H K \cong H \times K$

ˆ Semidirect products

Let K, H be groups $\varphi : K \to \operatorname { A u t } ( H )$ a homomorphism. If $\sigma : K  K$ is an automorphism of K then

$$
H \rtimes _ { \varphi } K \cong H \rtimes _ { \varphi \circ \sigma } K .
$$

p-groups

$| P | = p ^ { a } , p \mathrm { ~ a ~ }$ prime, then:

1. The center of p is non-trivial:

2. H E P then $H \cap Z ( P ) \neq 1$ . So every normal subgroup of order p is contained in the center.

3. $H < P$ then $H < N _ { P } ( H )$

4. Every maximal subgroup of P is of index p and is normal in P .

ˆ Upper central series

$Z _ { 0 } ( G ) = 1 , Z _ { i + 1 } ( G ) / Z _ { i } ( G ) = Z ( G / Z _ { i } ( G ) )$ (so $Z _ { i + 1 } ( G )$ is the preimage in G of the center of $G / Z _ { i } ( G )$ under the natural projection).

Zi(G) char G.

ˆ Nilpotent groups

G is nilpotent if $Z _ { n } ( G ) = G$ for some n. (So abelian groups are nilpotent).

${ \mathrm { I f ~ } } | P | = p ^ { a }$ for prime a, then P is nilpotent. (p-groups have non-trivial center).

$| G | = p _ { 1 } ^ { a _ { 1 } } \cdot \cdot \cdot p _ { s } ^ { a _ { s } }$ , and $P _ { i } \in S y l _ { p _ { i } } ( G )$ . TFAE:

1. G nilpotent;

2. $H < G$ then $H < N _ { G } ( H )$ (normalizers grow);

3. $P _ { i } \triangleleft G ;$

4. $G \cong P _ { 1 } \times . . . \times P _ { s }$

Finite abelian group is direct product of its Sylow subgroups.

Finite group is nilpotent iff every maximal subgroup is normal

Subgroups and factor groups of nilpotent groups are nilpotent

ˆ Lower central series

$G ^ { 0 } = G , G ^ { i } = [ G , G ^ { i - 1 } ]$ . Then $G ^ { 0 } \geq G ^ { 1 } \geq \cdots$

A group is nilpotent iff $G ^ { n } = 1$ for some n.

ˆ Derived series (Commutator series)

$$
G ^ { ( 0 ) } = G , G ^ { ( i + 1 ) } = [ G ^ { ( i ) } , G ^ { ( i ) } ] .
$$

$G ^ { ( i ) }$ char G.

G is solvable iff $G ^ { ( n ) } = 1$ for some n.

Nilpotent groups and subgroups of solvable groups are solvable

If $G / N$ and N are solvable, then G is solvable.