If a group $G$ is generated by $n$ elements, can every subgroup of $G$ by generated by $\leq n$ elements? I wonder if the statement in the title is true, so if:
$$G=\langle g_1,g_2,\ldots,g_n\rangle$$
means that we can write every subgroup $H$ of $G$ as
$$H=\langle h_1,h_2,\ldots,h_m\rangle$$
for some $h_i\in G$ and $m\leq n$.
What I'm thinking so far: if $n=1$, then:
$$H=\langle h_1,h_2\rangle=\langle g^a,g^b\rangle=\langle g^{\gcd(a,b)}\rangle$$
So the statement is true for $n=1$. 
However for $n>1$ I'm having trouble proving/disproving it. I think it comes down to some number theory stuff that I'm not aware of. For example for the case $n=2$: 
Say $|g_1|=n, |g_2|=m$
$$H=\langle h_1,h_2,h_3\rangle =\langle g_1^{a_1} g_2^{a_2}, g_1^{b_1} g_2^{b_2}, g_1^{c_1} g_2^{c_2}\rangle$$
So I think that proving that this can be generated by only $2$ of the elements comes down to proving that there are $x,y$ s.t.
$$xa_1+yb_1\equiv c_1\bmod n$$
$$xa_2+yb_2\equiv c_2\bmod m$$
However, I don't know the conditions that have to be satisfied for this system to have solutions. 
By the way, I'm not sure/given that this is actually true.

edit: by the way, I think that in writing down that system I accidentally assumed the group to be Abelian. Now I'm interested in the result both for Abelian and non-Abelian groups

The statement is true for abelian groups: every abelian group generated by $n$ elements is a quotient of $\mathbb{Z}^n$ and every subgroup of $\mathbb{Z}^n$ is free generated by at most $n$ elements. This can be found in any text elementary group theory.
It is false for non abelian groups. For example the permutation group $S_n$ is generated by two elements: the cycle $(12\ldots n)$ and the trasposition $(12)$. Every finite group $G$ is a subgroup of some $S_n$ but not every finite group is generated by two elements
