NoteNextra-origin/content/Math4302/Math4302_L8.md

Math4302 Modern Algebra (Lecture 8)

Subgroups

Cyclic group

Subgroup of cyclic group is cyclic

Every subgroup of a cyclic group is cyclic.

Order of subgroup of cyclic group

If a\in G and |\langle a\rangle| be the smallest positive n such that a^n=e, then \langle a\rangle=\{e,a,a^2,\cdots,a^{n-1}\} and a^{m_1}=a^{m_2}\iff m_1=m_2\mod n. (n divides m_1-m_2)

Size of subgroup of cyclic group

Let G=\langle a\rangle and H=\langle a^m\rangle. Then |H|=\frac{|G|}{d} where d=\operatorname{gcd}(|G|,|H|). In particular, \langle a^m\rangle=G\iff \operatorname{gcd}(n,m)=1.

GCD decides the size of subgroup

Suppose G=\langle a\rangle, |G|=n.

Then \langle a^{m_1}\rangle=\langle a^{m_2}\rangle\iff \operatorname{gcd}(n,m_2)=\operatorname{gcd}(n,m_1).

Proof

\implies:

\langle a^{m_1}\rangle=\langle a^{m_2}\rangle\implies \operatorname{gcd}(n,m_1)=\operatorname{gcd}(n,m_2)

\impliedby:

Suppose d=\operatorname{gcd}(n,m_1)=\operatorname{gcd}(n,m_2).

Enough to show a^{m_1}\in \langle a^{m_2}\rangle. (then we conclude \langle a^{m_1}\rangle=\langle a^{m_2}\rangle and by symmetry \langle a^{m_2}\rangle=\langle a^{m_1}\rangle.)

Equivalent to show that a^{m_1}=(a^{m_2})^k for some integer k. That is n divides m_1-km_2 for some k\in \mathbb{Z}.

From last lecture, we know that d can be written as d=nr+m_2 s for some r,s\in \mathbb{Z}.

Multiply by \frac{m_1}{d}, (since d divides m_1, this is an integer).

So m_1=nr\frac{m_1}{d}+m_2s\frac{m_1}{d}.

Therefore n divides m_1-(\frac{m_1}{d}s)m_2, so k=\frac{m_1}{d}s. works.

Corollaries for subgroup of cyclic group

Let G=\langle a\rangle be a cyclic group of finite order.

1. If H\leq G, then |H| is a divisor of |G|. (More generally true for finite groups.)
2. For any d divides |G|, there is exactly one subgroup of G of order d. \langle a^m\rangle where m=\frac{|G|}{d}.

Examples

(\mathbb{Z}_18,+).

The subgroup with size 6 is \langle 3\rangle=\{0,3,6,9,12,15\}=\langle 15\rangle.

Note that \operatorname{gcd}(18,3)=3=\operatorname{gcd}(18,15).

\langle 6\rangle=\{0,6,12\}.

\langle 9\rangle=\{0,9\}.

\langle 2\rangle=\{0,2,4,6,8,10,12,14,16\} (generators are 2,4,8,10,14,16 since they have gcd 2 with 18).

Non-cyclic groups

Let G be a group and a,b\in G, then we use \langle a,b\rangle to mean the subgroup of G generated by combination of a and b.

\langle a,b\rangle\coloneqq \{e,a,b,ab,ba,a^{-1},b^{-1},(ab)^{-1},(ba)^{-1},\ldots\}

This is a subgroup of G since it is closed and e=a^0.

Klein 4 group

Klein 4 group is abelian but not cyclic.

*	e	a	b	c
e	e	a	b	c
a	a	e	c	b
b	b	c	e	a
c	c	b	a	e

The subgroups are

\langle e\rangle=\{e\}

\langle a\rangle=\{e,a\}

\langle b\rangle=\{e,b\}

\langle c\rangle=\{e,c\}

Therefore G is not cyclic and not isomorphic to \mathbb{Z}_4.

Here G=\langle a,b\rangle=\{e,a,b,ab=c\}.

More generally, if we have a_i\in G, where i\in I, then \langle a_i,i\in I\rangle= all possible combinations of a_i with their inverses. Is a subgroup of G.

Another way to describe is that \langle a_i,i\in I\rangle=\bigcap_{H\leq G, a_i\in H,i\in I}H.

Definition of finitely generated group

If G is a group and if there is a finite set a_1,\ldots, a_n\in G such that G=\langle a_1,\ldots, a_n\rangle, then G is called finitely generated.

Examples

Any finite group is finitely generated.

---

(\mathbb{Q},+) is not finitely generated.

Suppose for the contrary, there is a finite set \frac{a_1}{b_1},\ldots,\frac{a_n}{b_n}\in \mathbb{Q} such that

\mathbb{Q}=\langle \frac{a_1}{b_1},\ldots,\frac{a_n}{b_n}\rangle=\{t_1\frac{a_1}{b_1},\ldots,t_n\frac{a_n}{b_n}|t_1,t_2,\ldots,t_n\in \mathbb{Z}\}
$$.

Pick prime $p$ such that $p>|b_1|,\ldots,|b_n|$. Then $\frac{1}{p}\in \mathbb{Q}$.

\frac{1}{p}=t_1\frac{a_1}{b_1}+t_2\frac{a_2}{b_2}+\cdots+t_n\frac{a_n}{b_n}=\frac{A}{b_1b_2\cdots b_n}

This implies that $pA=b_1b_2\cdots b_n$.

Since $p$ is prime, $p|b_i$ for some $i$.

However, by our construction, $p>|b_i|$ and cannot divide $b_i$.

Contradiction.

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