Hamathster
§ DS-02 Subgroup
2024-05-02
Algebra-I
00

Contents

§2 Subgroups
§2.1 Definition of a Subgroup
§2.2 Examples of Subgroups

§2 Subgroups

§2.1 Definition of a Subgroup

Definition 2.1 Let GG be a group and let H⊂GH \subset G. We say that HH is a subgroup of GG if the following conditions are satisfied:

(1) For all h1,h2∈Hh_1, h_2 \in H, we have h1h2∈Hh_1 h_2 \in H. (Closure under multiplication)

(2) 1G∈H1_G \in H.

(3) If h∈Hh \in H, then h−1∈Hh^{-1} \in H.

§2.2 Examples of Subgroups

Example 2.1 R>0⊂R×\mathbb{R}_{>0} \subset \mathbb{R}^{\times} is a subgroup.

Proof.

R>0\mathbb{R}_{>0} is a subgroup of R×\mathbb{R}^{\times} because

(1) For all h1,h2∈R∗>0h_1, h_2 \in \mathbb{R}*{>0}, we have h1h2∈R∗>0h_1 h_2 \in \mathbb{R}*{>0}.

(2) 1∈R>01 \in \mathbb{R}_{>0}.

(3) If h∈R∗>0h \in \mathbb{R}*{>0}, then h−1=1h∈R∗>0h^{-1}=\dfrac{1}{h} \in \mathbb{R}*{>0}.

 □~\tag*{$\square$}

Example 2.2 Z⩾0⊂(Z,+)\mathbb{Z}_{\geqslant 0} \subset(\mathbb{Z},+) is not a subgroup.

Proof.

Z∗⩾0\mathbb{Z}*{\geqslant 0} is not a subgroup of Z\mathbb{Z} because not every element has an inverse in Z∗⩾0\mathbb{Z}*{\geqslant 0}. For example, for z=1z=1, its inverse under addition should be −1-1, but −1∉Z⩾0-1\notin\mathbb{Z}_{\geqslant 0}.

 □~\tag*{$\square$}

Example 2.3 SLn(R):={M∈GLn(R)∣det⁡M=1}⊂GLn(R)SL_n(\mathbb{R}):=\left\{M \in GL_n(\mathbb{R}) \mid \det M=1\right\} \subset GL_n(\mathbb{R}) is a subgroup.

Proof.

SLn(R)SL_n(\mathbb{R}) is a subgroup of GLn(R)GL_n(\mathbb{R}) because

(1) For all M1,M2∈SLn(R)M_1, M_2 \in SL_n(\mathbb{R}), we have M1M2∈SLn(R)M_1 M_2 \in SL_n(\mathbb{R}), since

det⁡(M1M2)=det⁡(M1)det⁡(M2)=1×1=1.\det\left(M_1 M_2\right)=\det\left(M_1\right) \det\left(M_2\right)=1 \times 1=1 .

(2) The identity matrix 1GLn(R)∈SLn(R)1_{GL_n(\mathbb{R})} \in SL_n(\mathbb{R}), since det⁡(1GLn(R))=1\det\left(1_{GL_n(\mathbb{R})}\right)=1.

(3) For every M∈SLn(R)M \in SL_n(\mathbb{R}), we have M−1∈SLn(R)M^{-1} \in SL_n(\mathbb{R}), since det⁡(M−1)=1det⁡(M)=1\det\left(M^{-1}\right)=\frac{1}{\det(M)}=1.

 □~\tag*{$\square$}

Example 2.4 Let C×:=C∖{0}\mathbb{C}^{\times}:=\mathbb{C} \setminus\{0\}. Define mm to be complex multiplication: for all z1,z2∈C×z_1, z_2 \in \mathbb{C}^{\times},

m(z1,z2)=z1×z2.m\left(z_1, z_2\right)=z_1 \times z_2 .

Then:

(a) C×\mathbb{C}^{\times} is a group.

(b) Let

S1={z∣∣z∣=1}.S^1=\{z\mid | z |=1\}.

Then S1⊂C×S^1 \subset \mathbb{C}^{\times} is a subgroup.

Proof.

(a) C×\mathbb{C}^{\times} is a group because

(1) Complex multiplication is associative.

(2) For every z∈C×z \in \mathbb{C}^{\times}, we have 1×z=z×1=z1 \times z=z \times 1=z.

(3) For every z∈C×z \in \mathbb{C}^{\times}, let z−1=zˉ∣z∣2z^{-1}=\frac{\bar{z}}{|z|^2}.

(b) S1S^1 is a subgroup of C×\mathbb{C}^{\times} because

(1) ∣z1∣=∣z2∣=1⇒∣z1×z2∣=1\left|z_1\right|=\left|z_2\right|=1 \Rightarrow \left|z_1 \times z_2\right|=1, so S1S^1 is closed under multiplication.

(2) ∣1∣=1|1|=1, so the identity lies in S1S^1.

(3) If ∣z∣=1|z|=1, then ∣zˉ∣=1|\bar{z}|=1, and z−1=zˉz^{-1}=\bar{z}.

 □~\tag*{$\square$}