Math4302 Modern Algebra (Lecture 7)

Subgroups

Last time, let G be a group and a\in G. |\langle a\rangle|= smallest positive n such that a^n=e.

\langle a\rangle=\{a^0,a^1,a^2,\cdots,a^{n-1}\}.

Every subgroup of a cyclic group is cyclic.

G=\langle a\rangle.

Proof

Let H\leq G be a subgroup.

If H=\{e\}, we are done.

Otherwise, let m be the smallest positive integer such that a^m\in H. We claim H=\langle a^m\rangle.

\langle a^m\rangle\subseteq H. trivial since a^m\in H and H is a subgroup.
H\subseteq\langle a^m\rangle. Suppose a^k\in H, need to show a^k\in \langle a^m\rangle Divide k by m: k=qm+r, 0\leq r\leq m-1, Then a^k\in H\implies a^{qm+r}\in H. Also a^m\in H, then (a^m)^q\in H, so a^mq\in H, a^-mq\in H, so a^{k}a^{-mq}\in H, so a^r\in H, so r has to be zero.
By our choice of m, k=mq, so a^k=a^mq\in \langle a^m\rangle.

Example

Every subgroup of (\mathbb{Z},+) is of the form

like the multiples of n: n\mathbb{Z}=\langle n\rangle for some n\geq 0.

In particular, if n,m\geq 1 are in \mathbb{Z}, then the subgroup \{nr+ms|r,s\in \mathbb{Z}\}\leq \mathbb{Z}.

is equal to d\mathbb{Z} where d=\operatorname{gcd}(n,m).

Skip \operatorname{gcd} part, check for Math 4111 notes in this site.

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

Proof

Recall |H| is the smallest power of a^m which is equal to e.

Let d=\operatorname{gcd}(m,n), so m=m_1d, n=n_1d. and \frac{n}{\operatorname{gcd}(m,n)}=n_1,

(a^m)^{n_1}=a^{mn_1}=a^{m_1dn_1}=a^{m_1n}=(a^n)^{m_1}=e.
If (a^m)^k=e, the a^{mk}=e\implies mk is a multiple of n,
- If a^\ell=e, divide \ell by n, \ell=nq+r, 0\leq r\leq n-1, then e=a^\ell=a^{nq+r}=a^r, r has to be zero, so a^\ell=a^r=e. n|\ell.
n_1d|m_1dk, but by the definition of smallest common divisor, m_1,n_1 should not have common divisor other than 1. So n_1|m_1k, n_1|k\implies k\geq n_1.

Example Applying the lemma

Let G=\langle a \rangle, |G|=6, H=\langle a^4\rangle. Then |H|=\frac{6}{d}=3 where d=\operatorname{gcd}(6,4)=2.

To check this we do enumeration \langle a^4\rangle=\{e,a^4,a^2\}.

Find generator of \mathbb{Z}_9:

Using the coprime, we have g=\{1,2,4,5,7,8\}.

Corollary: \langle a^m\rangle=G\iff |H|=n\iff \frac{n}{d}=n\iff \operatorname{gcd}(m,n)=1 m,n are coprime.