Exercise 2.4.2 ($A \subseteq B \Rightarrow \langle A \rangle \subseteq \langle B \rangle$)

Question

Prove that if $A$ is a subset of $B$, then $\langle A angle \leq \langle B angle$. Give an example where $A \subseteq B$ with $A 
e B$ but $\langle A angle = \langle B angle$.

Richard Ganaye · Accepted Answer

\begin{proof} By definition, 
$$\langle A angle = \bigcap _{K\in {\cal A}} K,\qquad 	ext{where}\quad {\cal A} = \{K \leq G \mid A \subseteq K\}.$$ 
Since $B \subseteq \langle B angle$(see Ex. 1) and $A \subseteq B$, then $A \subseteq  \langle B angle$, and $ \langle B angle \leq G$, thus $\langle B angle  \in {\cal A}$. Hence
$$\bigcap _{K\in {\cal A}} K \subseteq \langle B angle,$$
so
$$\langle A angle \subseteq \langle B angle.$$
(In other words $\langle A angle$ is the smallest subgroup of $G$ which contains $A$, and $\langle B angle$ is such a subgroup, so $\langle A angle \subseteq \langle B angle$.)

\bigskip
In conclusion, for all subsets $A, B $ of $G$,
$$A \subseteq B \Rightarrow \langle A angle \subseteq \langle B angle.$$

\bigskip
Consider a a counterexample the cyclic group $Z_3 = \langle x angle = \{1,x,x^2\}$, and $A = \{x\}$, $B = \{x,x^2\}$. Then $A \subseteq B,\ A 
e B$, but $\langle A angle = \langle B angle = Z_3$.
\end{proof}

Exercise 2.4.2 ($A \subseteq B \Rightarrow \langle A \rangle \subseteq \langle B \rangle$)

Answers

Comments

Exercise 2.4.2 ($A \subseteq B \Rightarrow \langle A \rangle \subseteq \langle B \rangle$)

Answers

Comments

Add answer