Exercise 3.1.16 ($G = \langle S \rangle \Rightarrow \overline{G} = \langle \overline{S} \rangle$)

Question

Let $G$ be a group, let $N$ be a normal subgroup of $G$ and let $\overline{G} = G/N$. Prove that if $G = \langle x, y \rangle$ then $\overline{G} = \langle \overline{x}, \overline{y} \rangle$. Prove more generally that if $G = \langle S \rangle$ for any subset $S$ of $G$, then $\overline{G} = \langle \overline{S} \rangle$.

Richard Ganaye · Accepted Answer

\begin{proof} 
I give two distinct proofs for these two questions (even if the first part is a consequence of the last).
\begin{enumerate}
\item[{\bf(a)}] (``bottom up'' approach.)

Suppose that $G = \langle x, y angle$.  If $g \in G$, let $\overline{g}$ denote the coset $gN \in G/N$.
By Proposition 9, every element $g \in G$ is of the form $g = a_1^{\varepsilon_1}a_2^{\varepsilon_2}\cdots a_n^{\varepsilon_n}$, where $a_i  \in \{x ,y \}$ and $\varepsilon_i = \pm 1$ for each $i$.

Then $\overline{g} = \overline{a_1}^{\varepsilon_1} \overline{a_2}^{\varepsilon_2}\cdots  \overline{a_n}^{\varepsilon_n}$, where $\overline{a_i} = a_i N \in \{\overline{x}, \overline{y}\}$ for each $i$. Therefore $\overline{g} \in \langle \overline{x}, \overline{y}angle$. Since this is true for every  $\overline{g} \in G/N$, this shows that
$$\overline{G} = \langle \overline{x}, \overline{y} angle.$$

\item[{\bf(b)}] ( ``top down'' approach.)

Now suppose that $G = \langle S angle$, where $S$ is any subset of $G$. Put $\overline{S} = \{\overline{x} \mid x \in S\}$. Consider any subgroup $\overline{H} \leq \overline{G}$ such that $\overline{S}\subseteq \overline{H}$, so
$$\overline{S} \subseteq \overline{H} \leq \overline{G}.$$

Let $\pi : G 	o G/N$ be the natural projection defined by $\pi(g) = gN = \overline{g}$, and define $H = \pi^{-1}(\overline{H})$. Then $H$ is a subgroup of $G$ which contains $N$ (see Exercise 1), and since $\pi$ is surjective, $\pi(H) = \overline{H}$.

If $x \in S$, then $\overline{x} \in \overline{S} \subseteq \overline{H}$, thus $\overline{x} = \pi(x) \in \overline{H}$, so $x \in \pi^{-1}(\overline{H}) = H$. This shows that
$$S \subseteq H \leq G.$$
Since $G$ is the smallest subgroup of $G$ which contains $S$, this implies that $H = G$, therefore $\overline{H} =  \pi (H) = \pi(G) = \overline{G}$. This shows that $\overline{G}$ is the smallest subgroup of $\overline{G}$ which contains $\overline{S}$, so
$$\overline{G} = \langle \overline{S} angle.$$

\end{enumerate}

\end{proof}

Exercise 3.1.16 ($G = \langle S \rangle \Rightarrow \overline{G} = \langle \overline{S} \rangle$)

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Exercise 3.1.16 ($G = \langle S \rangle \Rightarrow \overline{G} = \langle \overline{S} \rangle$)

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