Basics in Group theory
UW-Madison
Groups
I’ll start recalling some definitions and results of the theory of groups.
Let G be a group and X is a subset of G. I’ll denote by hXi the group generated by X. One can think of
X as the smallest subgroup of G containing X.
Definition. Suppose that H, K ⊆ G are subgroups. Then,
HK = {hk : h ∈ H, k ∈ K}.
Normality of H or K in G implies that HK is a subgroup. On the other hand, HK is not necessary a
subgroup of G. However we see from the definitions that HK ⊆ hH, Ki. Where we have equality if and
only if the left hand side is a group.
Theorem (Isaacs, 2.18, 2.31). The set HK ⊆ G is a subgroup if and only if HK = KH. In particular,
if either H or K is normal in G, then HK = KH.
Definition. If G is a group and X is some subset of G, we define
CG (X) = {g ∈ G : gx = xg for all x ∈ X},
the centralizer of X in G, the set of all elements that commute with all elements of X.
It is not hard to show that this is in fact a subgroup of G. One key example is when we take X = G. This
group, CG (G) is called the center of G and is denoted Z(G). The following is an easy, but very useful
result.
Lemma. Let G be a group and N be a central subgroup ( N ⊆ Z(G)). If G/N is cyclic, then G is abelian.
Proof. Exercise.
Definition. We define
NG (X) = {g ∈ G : g −1 Xg = X},
the ‘normalizer of X in G.
Again, it is not hard to show that NG (X) is a subgroup. If H is a subgroup of G, then NG (H) is the
biggest subgroup of G containing H as a normal subgroup. In particular H E G ⇐⇒ G = NG (H).
Normal subgroups are of fundamental importance in group theory, and many of the techniques we will see
will help us construct normal subgroups or exploit their existence.
Correspondence Theorem
The primary result in group theory allowing for the exploitation of normal subgroups is the so-called
Correspondence Theorem.
Theorem (Isaacs, 3.7). Let φ : G → H be a surjective homomorphism and let N = ker φ. Define the
following sets of subgroups
S = {U |N ⊆ U ⊆ G}
and
T = {V |V ⊆ H}.
Then φ and φ−1 are inverse bijections between S and T . Furthermore, these maps respect containment,
indices, normality and factor groups.
The last sentence means that if U1 , U2 ∈ S and V1 and V2 are the corresponding elements of T , that is
Vi = φ(Ui ),
Ui = φ−1 (Vi ) = {x ∈ G : φ(x) ∈ Vi },
then U1 ⊆ U2 if and only if V1 ⊆ V2 . In this case, |U2 : U1 | = |V2 : V1 | and U1 E U2 if and only if V1 E V2 .
If this latter statement holds then U2 /U1 ∼
= V2 /V1 .
From correspondence follows the classic isomorphism theorems:
Let K, M, N be subgroups of G such that N E G, M E G and M ⊆ N . Then:
• K ∩ N E K and K/(K ∩ N ) ∼
= KN/N .
• G/N ∼
= (G/M )/(N/M ).
Note that if in particular K and N are finite groups we have that |KN ||K ∩ N | = |K||N |. This is in fact
true, see [Isaacs, 4.17], for arbitrary finite subgroups K, N of a group G. i.e we don’t need N to be normal
or G to be finite. This fact is quite useful, and one way to remember -at least for me- is the diamond
diagram.
KN G
GG
ww
GGn
w
GG
ww
w
GG
w
w
w
N
K GG
GG
ww
w
GG
ww
n GGG
ww m
ww
K ∩N
m
Characteristic Subgroups
A subgroup H of G is characteristic if for every automorphism φ of G, φ(H) = H. [Recall that an
automorphism is a bijective homomorphism from G to itself]. We write HcharG if H is characteristic in
G. Here are basic properties of characteristic subgroups. We suppose that H ⊆ K are subgroups of G.
1. If HcharG, then H E G.
2. If HcharK and KcharG, then HcharG. [The corresponding statement with normal in place of characteristic is not true].
3. If HcharK and K E G, then H E G.
4. Marty Isaacs’ “the” test. The idea is that any subgroup H of G that has a precise, group theoretic
definition must necessarily be characteristic. For example, “the” center is the set of all elements that
commute with all elements of G. Applying an automorphism to G won’t change this property and hence
“the” center of G is characteristic.
Theorem (the “N/C theorem”). : If H is a subgroup, then CG (H) E NG (H) and NG (H)/CG (H) is
isomorphic to a subgroup of Aut(H) [the group of automorphisms of H].
Proof. Let g ∈ NG (H). Then the map φg : H → H sending x ∈ H to xg := g −1 xg is an automorphism of
H. Now Φ : NG (H) → Aut(G) sending g to φg is an homomorphism with kernel CG (H).
When H = G the image of Φ, Inn(G) , is called the group of inner automorphisms of G.
Direct Products
If H and K are groups, then the group H × K = {(h, k) : h ∈ H, k ∈ K}, and the group operation on
H × K is done componentwise, namely (h1 , k1 ) · (h2 , k2 ) = (h1 h2 , k1 k2 ). From this definition, it is not hard
to see that H̃ = {(h, 1) : h ∈ H} ⊆ H × K and K̃ = {(1, k) : k ∈ K} ⊆ H × K are normal subgroups,
where 1 denotes the identity of H or K. Further, H̃ ∩ K̃ = 1 and H̃ K̃ = {ab : a ∈ H̃, b ∈ K̃} = H × K.
The surprising fact is that the converse is true.
Theorem (Isaacs, 7.2). Suppose that G is a group with normal subgroups H and K such that HK = G
and H ∩ K = 1. Then, G ∼
= H × K.
Direct products are one of the easiest ways to construct new groups from old ones. In fact, there are many
important groups that are direct products of smaller groups.
Theorem (Isaacs, 7.10, 7.15). Suppose that G is a finite abelian group. Then G has a unique representation as a direct product of finite cyclic groups of prime-power order.
In order to prove Theorem 7.2, Isaacs uses the following lemma, which has independent interest.
Lemma (Isaacs, 7.1). Suppose that M and N are normal in G and M ∩ N = 1. Then mn = nm for all
m ∈ M and n ∈ N .
Now, using this lemma I will indicate an idea that appears very commonly on the qualifying exam.
Lemma. Suppose that G is a group, and A, B and C are normal subgroups of G. Suppose also that
G = AB = AC = BC and that A ∩ B = A ∩ C = B ∩ C = 1. Then, G is abelian.
Proof. First, we consider CG (A). Since A ∩ B = 1 and A ∩ C = 1, it follows from Lemma 7.1 that
B, C ⊆ CG (A). Thus, BC ⊆ CG (A). However, G = BC so G = CG (A) and hence A ⊆ Z(G). Since the
statement is completely symmetric in A,B,C we have that B ⊆ Z(G). Thus, AB = G ⊆ Z(G) and hence
G = Z(G), so G is abelian.
Problems
1. Show that in the situation of the last lemma A ∼
=B∼
= C.
2. If Zn is a cyclic group of order n, prove that Aut(Zn ) is an abelian group of order φ(n), where
φ(n) = |{1 ≤ r ≤ n : gcd(r, n) = 1}|.
3. Prove that if p is prime then Aut(Znp ) ∼
= GLn (Fp ), where Fp is the field of order p. Use this to show
that |Aut(Znp )| = (pn − 1)(pn − p)(pn − p2 )....(pn − pn−1 ).
4. Suppose that G = H × K. Prove that Aut(G) has a subgroup isomorphic to Aut(H) × Aut(K).
Furthermore, show that if H and K are finite of relatively prime order Aut(H) × Aut(K) ∼
= Aut(G).
5. Suppose that G is a group such that every non trivial element has order 2. Show that G is abelian.
Does the same result hold if we replace 2 by 3?
6. Let G be the subgroup of GL3 (F3 ) consisting of upper triangular matrices with 1 on the diagonal.
Show that if x ∈ G, x3 = I, moreover show that every element of order 3 in GL3 (F3 ) is conjugated
to an element of G. Is G abelian?
7. Suppose that G is a finite group with gcd(|G|, |Aut(G)|) = 1. What can you say about G?
• Use the NC theorem and deduce that G is abelian.
• Use that φ(pn ) = pn−1 (p − 1), Isaacs 7.10-7.15 and problem 4 to deduce that G has only cyclic
factors of the form Zp . In other words G ∼
= Znp11 × ... × Znpkk , for some set of different primes pi .
• Use problems 3 and 4 to deduce that each ni = 1.
• Finally, show that G is a cyclic group of order n a square free integer, such that if p and q are
prime factors of n we have p 6≡ 1 (mod q) and q 6≡ 1 (mod p).
8. Let G be a group such that Aut(G) is a cyclic group. Show that Aut(G) has finite even order, unless
G∼
= Z2 .
9. Find and example of groups H E K E G, but H E G.
10. (Dedekind’s lemma) Let H ⊆ K ⊆ G and L ⊆ G. Show that K ∩ HL = H(K ∩ L).
11. Let H,K and L subgroups of a group G. Show that if H ⊆ K, H ∩ L = K ∩ L and HL = KL, then
H = K.
12. Show that a group can’t be the union of two proper subgroups. Give an example of a group which
is the union of three proper subgroups.
13. Let G be a finite group such that it is the union of 3 proper subgroups. Show that Z2 × Z2 is a
homomorphic image of G.
14. In an abelian group every subgroup is a normal subgroup. If G is a group such that every subgroup
is normal, is G abelian?
0 i
0 1
. Show that:
and b =
15. Let Q8 be the subgroup of GL2 (C) generated by a =
i 0
−1 0
i) Q8 is a nonabelian group of order 8.
ii) Every subgroup of Q8 is normal.
iii) Q8 is the union of three proper subgroups.
Q8 is called the quaternion group. For those of you who have seen the more standard presentation
of Q8 , show that a → i and b → j defines an isomorphism between Q8 -notes and usual Q8 .
16. Show that up to isomorphism there are only two subgroups of order 6. Conclude that Aut(Z2 ×Z2 ) ∼
=
S3 .
17. Jan-93, Aug-95, Jan-96, Aug-96, Aug-98, Aug-99, Aug-00, Jan-02, Jan-04, Aug-05, Jan-06, Aug-06.