# Quotient group of free groups

Let $G=\langle g_1,\ldots,g_k\rangle$ be a free abelian group generated with $g_1,\ldots,g_k$ and let $H=\langle g_{r+1},\ldots,g_k\rangle$ be a free abelian subgroup of $G$.

Is it then the case that $G/H\simeq\langle g_1,\ldots,g_r\rangle$?

The map $f:G \to G$ given by \begin{align*} f(g_i) &= \begin{cases} g_i & \text{if}\;\; i \leq r; \\ 0 & \text{if}\;\; i > r \end{cases} \end{align*} has kernel $\langle{g_{r+1}, \dots, g_k\rangle} = H$ (since $G$ is free). Hence $G/H = f(G) = \langle g_1, \dots, g_r\rangle\subset G$.