TITLE: Can I always get the generators of a subgroup?
QUESTION [5 upvotes]: Let's say I have a group with a presentation in generators and relations:
$$G = \langle g_1, g_2, ...\mid r_1, r_2, ...\rangle$$
And I have $H$ a subgroup of $G$. If I take $h_1, h_2, ...$ the generators that are elements of $H$ and $q_1, q_2, ...$ the relations which relate only to the $h_i$. Can I say that:
$$H = \langle h_1, h_2, ... \mid q_1, q_2, ... \rangle$$

If that's not true, can I go in the other direction? In other words, can I say that if:
$$H = \langle h_1, h_2, ... \mid q_1, q_2, ... \rangle$$
then I can "extend" this to a presentation for $G$:
$$G = \langle g_1, g_2, ...\mid r_1, r_2, ...\rangle$$
such that each $h_i$ is one of the $g_i$ and each $q_i$ is one of the $r_i$?

REPLY [8 votes]: Here is a small counterexample for your first question.
$G = \langle g,h \mid g^4, g^2h^{-1} \rangle.$
Then $G$ is cyclic of order $4$ and the subgroup $H = \langle h \rangle$ has order $2$, but none of the relators of $G$ involve only generators of $H$.