Quotient group of finitely generated group is finitely generated

Question

I am studying elementary level of algebra, and I'm trying to prove that a quotient group of finitely generated group is finitely generated. It is intuitively true, but I can't get an idea to prove it.

Answer 1

Suppose that a group $G$ is finitely generated. Let $g_1,\cdots,g_n \in G$ be generators, and let $H \trianglelefteq G$ be a normal subgroup. Can you show that $g_1H,\cdots,g_nH$ generate $G/H$? Hint: write an arbitrary element of $G/H$ as $gH$ for some $g \in G$, and then use the fact that the $g_1,\cdots,g_n$ generate $G$.

Another more involved way to prove it which is categorically more intuitive (but less intuitive at the beginning) is that saying that a subset $S \subseteq G$ generates $G$ means that one can form the free product $K \overset{\mathrm{def}}{=} \mathbb Z * \cdots * \mathbb Z$ ($|S|$ copies of $\mathbb Z$) and the map $K \to G$ sends the word $\omega_i$ to $g_i$ where $S = \{g_1,\cdots,g_n\}$ is the generating set. In this setting, saying that $S$ generates $G$ is the same as saying that the induced map $K \to G$ defined by $S$ is surjective. But then clearly if $K \to G$ is surjective, so is the composition $K \to G \to G/H$.

You sort of already know this argument if you know real vector spaces and quotient spaces of them: if $S = \{v_1,\cdots,v_n\}$ is a subset of a vector space $V$, you can form the map $\mathbb R^n \to V$ where the unit basis vector $e_i$ of $\mathbb R^n$ is mapped to $v_i$. A finite subset $S$ of $V$ generates $V$ if and only if the induced map $\mathbb R^n \to V$ is surjective. It is then clear that if $S$ generates $V$, then $S$ generates $V/W$ for any subspace $W$, because the composition $\mathbb R^n \to V \to V/W$ remains surjective. If you replace $\mathbb R$ by $\mathbb Z$, you obtain the equivalent argument for $\mathbb Z$-modules, also known as abelian groups (and $\mathbb R^n$ becomes $\mathbb Z^n$ in the argument). But for general groups, the free object is not as pretty as $\mathbb R^n$ or $\mathbb Z^n$, that's why this free product appears to get every possible product of elements of the group.

Hope that helps,