NoteNextra-origin/content/Math4302/Math4303_L12.md

Math4303 Modern Algebra (Lecture 12)

Groups

Direct products

\mathbb{Z}_m\times \mathbb{Z}_n is cyclic if and only if m and n have greatest common divisor 1.

More generally, for \mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times \cdots \times \mathbb{Z}_{n_k}, if n_1,n_2,\cdots,n_k are pairwise coprime, then the direct product is cyclic.

Proof

For the forward direction, use \mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}=\mathbb{Z}_{n_1n_2}. if n_1, n_2 are coprime.

For the backward, suppose to the contrary that for example \gcd(n_1,n_2)=d>1, then G=\mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2}\times H, where any element in H has order \leq |H| and any element in \mathbb{Z}_{n_1}\times \mathbb{Z}_{n_2} has order <\frac{n_1n_2}{d}, therefore, all the elements in G will have order strictly less than the size n_1n_2\ldots n_k of the group.

Corollary for composition of cyclic groups

If n=p_1^{m_1}\ldots p_k^{m_k}, where p_i are distinct primes, then the group

G=\mathbb{Z}_n=\mathbb{Z}_{p_1^{m_1}}\times \mathbb{Z}_{p_2^{m_2}}\times \cdots \times \mathbb{Z}_{p_k^{m_k}}

is cyclic.

Example for product of cyclic groups and order of element

\mathbb{Z}_{8}\times\mathbb{Z}_8\times \mathbb{Z}_12

the order for (1,1,1) is 24.

What is the maximum order of an element in this group?

Guess:

8*3=24

Structure of finitely generated abelian groups

Theorem for finitely generated abelian groups

Every finitely generated abelian group G is isomorphic to

Z_{p_1}^{n_1}\times Z_{p_2}^{n_2}\times \cdots \times Z_{p_k}^{n_k}\times\underbrace{\mathbb{Z}\times \ldots \times \mathbb{Z}}_{m\text{ times}}

Example

If G is abelian of size 8, then G is isomorphic to one of the following:

• \mathbb{Z}_2\times \mathbb{Z}_2\times \mathbb{Z}_2 (non cyclic)
• \mathbb{Z}_2\times \mathbb{Z}_4 (non cyclic)
• \mathbb{Z}_2 (cyclic)

And any two of them are not isomorphic

---

Find all abelian group of order 72.

Since 72=2^3*3^2, There are 3 possibilities for the 2^3 part, and there are 2 possibilities for the 3^2 part.

Note that \mathbb{Z}_8\times\mathbb{Z}_9, where 8,9 are coprime, \mathbb{Z}_8\times\mathbb{Z}_9=\mathbb{Z}_{72}, is cyclic.

There are 6 possibilities in total.

Corollary for divisor size of abelian subgroup

If g is abelian and |G|=n, then for every divisor m of n, G has a subgroup of order m.

Warning

This is not true if G is not abelian.

Consider A_4 (alternating group for S_4) does not have a subgroup of order 6.

Proof for the corollary

Write G=\mathbb{Z}_{p_1}^{n_1}\times \mathbb{Z}_{p_2}^{n_2}\times \cdots \times \mathbb{Z}_{p_k}^{n_k} where p_i are distinct primes.

Therefore n=p_1^{m_1}\ldots p_k^{m_k}.

For any divisor d of n, we can write d=p_1^{m_1}\ldots p_k^{m_k}, where m_i\leq n_i.

Now for each p_i, we choose the subgroup H_i of size p_i^{m_i} in \mathbb{Z}_{p_i}^{n_i}. (recall that every cyclic group of size r and any divisor s of r, there is a subgroup of order s. If the group is generated by a, then use a^{\frac{r}{s}} to generate the subgroup.)

We can construct the subgroup H=H_1\times H_2\times \cdots \times H_k is the subgroup of G of order d.

Cosets

Definition of Cosets

Let G be a group and H its subgroup.

Define a relation on G and a\sim b if a^{-1}b\in H.

This is an equivalence relation.

• Reflexive: a\sim a: a^{-1}a=e\in H
• Symmetric: a\sim b\Rightarrow b\sim a: a^{-1}b\in H, (a^{-1}b)^{-1}=b^{-1}a\in H
• Transitive: a\sim b and b\sim c\Rightarrow a\sim c : a^{-1}b\in H, b^{-1}c\in H, therefore their product is also in H, (a^{-1}b)(b^{-1}c)=a^{-1}c\in H

So we get a partition of G to equivalence classes.

Let a\in G, the equivalence class containing a

aH=\{x\in G| a\sim x\}=\{x\in G| a^{-1}x\in H\}=\{x|x=ah\text{ for some }h\in H\}

This is called the coset of a in H.

Example

Consider G=S_3