NoteNextra-origin/content/Math4302/Math4302_L5.md

Math4302 Modern Algebra (Lecture 5)

Groups

Subgroups

A subset H\subseteq G is a subgroup of G if

• e\in H
• \forall a,b\in H, a b\in H
• a\in H\implies a^{-1}\in H

H with * is a group

We denote as H\leq G.

Example

For an arbitrary group (G,*),

(\{e\},*) and (G,*) are always subgroups.

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(\mathbb{Z},+) is a subgroup of (\mathbb{R},+).

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Non-example:

(\mathbb{Z}_+,+) is not a subgroup of (\mathbb{Z},+).

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Subgroup of \mathbb{Z}_4:

(\{0,1,2,3\},+) (if 1\in H, 3\in H)

(\{0,2\},+)

(\{0\},+)

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Subgroup of \mathbb{Z}_5:

(\{0,1,2,3,4\},+)

(\{0\},+)

Cyclic group with prime order has only two subgroups

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Let D_n denote the group of symmetries of a regular $n$-gon. (keep adjacent points pairs).

D_n=\{\sigma\in S_n\mid i,j\text{ are adjacent } \iff \sigma(i),\sigma(j)\text{ are adjacent }\}

\begin{pmatrix}
1&2&3&4\\
2&3&1&4
\end{pmatrix}\notin D_4

D_4 has order 8 and S_4 has order 24.

|D_n|=2n. (n option to rotation, n option to reflection. For \sigma(1) we have n option, \sigma(2) has 2 option where the remaining only has 1 option.)

Since 1-4 is not adjacent in such permutation.

D_n\leq S_n (S_n is the symmetric group of n elements).

Lemma of subgroups

If H\subseteq G is a non-empty subset of a group G.

then (H is a subgroup of G) if and only if (a,b\in H\implies ab^-1\in H).

Proof

If H is subgroup, then e\in H, so H is non-empty and if a,b\in H, then b^{-1}\in H, so ab^{-1}\in H.

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If H has the given property, then H is non-empty and if a,b\in H, then ab^-1\in H, so

• There is some a,a\in H, aa^{-1}\in H, so e\in H.
• If b\in H, then e\in H, so eb^{-1}\in H, so b^{-1}\in H.
• If b,c\in H, then c^{-1}, so bc^{-1}^{-1}\in H, so bc\in H.

Cyclic group

G is cyclic if G is a subgroup generated by a\in G. (may be infinite)

\mathbb{Z}_n\leq D_n\leq S_n.

Cyclic group is always abelian.