How do I prove this seemingly obvious property of subgroups The statement is the following: Given an abelian group $G=\langle a_1,...,a_t\rangle$, and a subgroup $H$ of $G$, we need at most $t$ elements to generate $H$; i.e. $H=\langle b_1,...,b_t\rangle$ for some $b_1,...,b_t\in H$.
While this statement seems obvious I couln't find a way to prove it, even if I limited myself to finite groups. I tried a couple of approaches (see below). Any help will be appreciated.

Approach 1: I tried proving that if $\{ [a_1]_H,...,[a_k]_H\}$ is a minimal generating set for $G/H$, and $\{ b_1,...,b_t\}$ is a minimal generating set for $H$, then $\{ a_1,...,a_k,b_1,...,b_t\}$ is a minimal generating set for $G$.

Approach 2: Limiting myself to finite groups, I tried proving the statement using induction. I tried induction on the size of $G$, size of $H$, and size of the minimal sets generating $G$ and $H$. 

I already had a proof of this written down, so I have copied and pasted it. 
Let $K \le G$ with $G$ an (additive) abelian group generated by $x_1,\ldots,x_n$. We shall prove by induction on $n$ that $K$ can be generated
by at most $n$ elements. If $n=1$ then $G$ is cyclic and hence so
is $K$.  Suppose $n>1$, and let $H$ be the subgroup of $G$ generated by
$x_1,\ldots,x_{n-1}$. By induction, $K \cap H$ is generated by
$y_1,\ldots,y_{m-1}$, say, with $m \le n$. If $K \le H$, then  $K = K \cap H$
and we are done, so suppose not.
Then there exist elements of the form $h +  t x_n \in K$ with
$h \in H$ and $t \ne 0$. Since $-(h+t x_n) \in K$, we can assume that
$t > 0$. Choose such an element  $y_m = h +  t x_n \in K$
with $t$ minimal subject to $t > 0$.
We claim that $K$ is generated by $y_1,\ldots,y_m$, which will complete the
proof. Let $k \in K$. Then $k = h' + u x_n$ with $h' \in H$ and $u \in {\mathbb Z}$.
If $t$ does not divide $u$ then we can write $u = tq + r$ with $q,r \in {\mathbb Z}$
and $0 < r < t$, and then $k - qy_m = (h'-qh) + rx_n \in K$, contrary to
the choice of $t$. So $t|u$ and hence $u=tq$ and $k - qy_m \in K \cap H$.
But $K \cap H$ is generated by $y_1,\ldots,y_{m-1}$, so we are done.
