Math 402, Monday 7/12/04

DIRECT PRODUCTS OF GROUPS

Definition: The direct product of two groups G₁ and G₂ is the group G₁x G₂ whose underlying set is G₁x G₂={(a,b) : a Є G₁ and b Є G₂ }, and whose operation is component-wise multiplication:

(a,b) (a’,b’)= (aa’,bb’)

(Note: sometimes Artin calls this just the product of the two groups. I don’t like this because there are other kinds of products, for example the semi-direct product which we will not study.)

Example:

1) Consider the direct product Z/2Z x Z/3Z={0,1}x{0,1,2}.
The operation on is +, i.e. (+mod2, +mod3) instead of multiplication here.
Write an element of order 2 in this group: (1,0) is the only one

What order does x=(0,1) have? 3

Because (0,1)+(0,1)+(0,1)=(0,2)+(0,1)=(0,3)=(0,0)
Compute (1,0)+(1, 2)= (0,2)

We have not yet proved that G₁x G₂ is a group.

-since the operation is defined component-wise, closure and associativity follow from those of G₁ and G₂. Make a guess and verify the other two properties:

-the identity is: 1_G1xG2 =( 1_G1 ,1_G2 )

-the inverse of (a,b) is ( a^-1, b^-1)

Question: If the order of a in G₁ is n and the order of b in G₂ is m, what is the order of (a,b) in G₁x G₂?

Answer: least common multiple of m and n

Proof:

Let L be any positive integer.

Then (a, b)^L=(a^L, b^L) =1_G1xG2 iff a^L =1_G1 and b^L =1, i.e. if and only if (iff) n divides L and m divides L.

L is precisely the order of (a,b) (rather than a multiple of the order) iff no smaller integer works. In other words, L has to the the smallest positive integer divisible by both m and n, which means that L is the lcm of m and n.

Question: When is G₁x G₂ abelian?

Answer:Precisely when both groups are abelian.
Proof: G₁x G₂ is abelian iff (a,b)(c,d)=(c,d)(a,b) for any elements a, c in G1 and b, d in G2. This is equivalent to (ac, bd)=(ca, db), i.e. to ac=ca & bd=db for any elements a, c in G1 and b, d in G2.
Describe the maps, complete the statements and prove the following

Proposition:
a) There exist injective group homomorphisms:

i₁:G₁→ G₁x G₂ and i₂ :G₂→ G₁x G₂,

whose images are Im i₁ = G₁x{1} and Im i₂= {1)xG₂

b) There exist surjective group homomorphisms (projections):

p₁: G₁x G₂ → G₁ and p₂ : G₁x G₂→ G₂,

whose kernels are Ker p₁= {1}xG2 and Ker p₂= G1x{1}

PROOF:

a) Define i₁:G₁→ G₁x G₂ as i₁(a)=(a,1) and i₂:G₂→ G₁x G₂ as i₂(b)=(1,b). Then i1 satisfies:

• group hom: i₁(aa’)=(aa’,1)=(a,1)(a’,1)= i₁(a) i₁(a’) for all a, a’ in G₁
• injective: Ker i₁={a in G₁: (a,1)=1=(1,1)}={a in G₁: a=1}={1} so the map is injective

Similarly for i2.

b) Define p₁: G₁x G₂→ G₁ as p₁ ((a,b))=a and p₂: G₁x G₂→ G₂ as p₂((a,b))=b, the projection maps in the first, respectively second, coordinate.

For the map p₁ show:

• group hom: p₁((a,b)(a’,b’))= p₁ ((aa’,bb’))=aa’= p₁ ((a,b)) p₁ ((a’,b’)) for all a, a’ in G₁ and b, b’ in G_2.
• Surjective: clearly, by definition
• Ker p₁={(a, b) in G₁x G₂ : p1((a,b))=a=1_G1}={1}x G2.

Similarly for the other map.

Let H and K be two subgroups of a group G.

When is G a direct product of H and K? ( i.e. G isomorphic to HxK?)

Define the map

f: H x K → G ,

f((h,k))=hk

Let’s see what conditions on H and K ensure that f is an isomorphism (inj., surj. & hom.)

0) Prove first that H and K must both be normal in G, because Hx{1}&{1}xK are normal in HxK.

1) We need: H ∩ K={1}. Why? What does this mean in terms of f?

2) We need Im f = G. Write Im f in terms of H and K to get that G=?=

3) Finally we need f to be a group homomorphism.

a) Prove first that f is homomorphism <=> hk=kh for any h ЄH and any k ЄK.

b) Prove that 0) and 1) ensure hk=kh for any h ЄH and any k ЄK.

Write a proposition summarizing all of above.

Do problem 2.8.4 as examples (after finishing the rest of the handout).

FINITE CYCLIC GROUPS

Question #1: Consider Z/6Z. Is it isomorphic to Z/2Z x Z/3Z? If yes, provide an isomorphism. If no, explain why not.

Yes! Take the map Z/6Zà Z/2Z x Z/3Z, which maps 1 to (1,1). (map a generator to another)

Extend by the homomorphism property to a map on entire Z/6Z.

(i.e. 2=1+1 must map to (1,1)+(1,1)=(0,2), 3 maps to (1,0), 4 maps to (0,1), 5 maps to (1,2), 6=0 maps to (0, 0). Notice that the order of each element matches that of its image, as it should be!).

Then the map you get is a group homomorphism, by definition. It is injective because only 0 maps to (0,0). It is surjective because you get all elements in Z/2Z x Z/3Z.

Yay, it’s an isomorphism!

Alternatively, prove that Z/2ZxZ/3Z is generated by (1,1) so it must be cyclic of order 6, so it must be isomorphic to Z/6Z.

Question #2: Same question but about Z/12Z and Z/2Z x Z/6Z.

Z/12Z={0, 1, 2, …, 11} is cyclic of order 12

Choices of possible generators: 1, 5, 7, 11; each of these has order 12.

The other elements are the

identity 0 (order 1), 3 and 9 (order 4).

6 (order 2) 2, 4, 8, 10 (order 6),

Z/2Z x Z/6Z is also of order 12, but it’s not cyclic! It has no element of order 12, so it requires at least 2 generators. It has elements:

0=(0,0) of order 1, (0,2), (0,4) of order 3

(1,0), (1,3), (0,3) of order 2, (0,1), (0,5), (1,1), (1, 5), (1,2), (1,4) of order 6

The groups could not be isomorphic because isomorphisms preserve the orders of elements.

Make a conjecture and prove it:

When is a cyclic group of order mn equal to a direct product of cyclic groups of order m and n?

Proposition: Z/mnZ ≈ Z/mZ x Z/nZ if and only if (iff) m and n are relatively prime integers.

Proof:

The Structure Theorem for Finitely Generated Abelian Groups:

If G is an abelian group with finitely many generators, then G is isomorphic to a direct product of infinite cyclic groups and finite cyclic groups whose orders are powers of prime numbers.

G ≈ (Zx Zx…xZ) x Z/(p₁)^a1Z x Z/(p₂)^a2Z x … x Z/(p_n)^anZ

Proof: Skipped (Chapter 12, page 472 if you want to take a look)

Application:

1) List all three abelian groups of order 8 (up to isomorphism).

2) Classify all abelian groups of order less than or equal to 15, up to isomorphism