Okay so we were told that the inverse Lagrange holds if G is finite and also a finitely generated abelian group (FGA).
We were given a proof but I am not sure I do get the steps so I will try to write here what I understood and hopefully you will help check this.
Let's say $|G| = n$ then we can analyse $n$ into discrete powers of prime numbers $p_i$: $$n = p_1^{k_1} p2^{k_2}...p_r^{k_r}$$
We can do the same for $d|n$ : $$d = p_1^{m_1}p_2^{m_2}...p_s^{m_s}$$
where $0\leq m_i \leq k_i$
Now, we split $k_i$ into a sum $k_i = k_{i1} + k_{i2} + .. + k_{iu}$
and we also split $m_i$ into a sum $m_i = m_{i1} + .. + m_{iu}$ such that $m_{ij} \leq k_{ij}$.
We rewrite n as $n = (p_1^{k_{11}} * .. *p1^{k_{1r}})*(p_2^{k_{21}}* .. *p_2^{k_{2r_2}})*....*(p_s^{k_{s1}} *..*p_s^{k_{sr_s}})$.
From the fundemental theorem of finitely generated abelian groups we have that :
$$G = (Z_{p_1^{k_{11}}} \times...\times Z_{p1^{k_{1r}}}) \times (Z_{p_2^{k_{21}}} \times.. \times p_2^{k_{2r_2}}) \times (Z{p_s^{k_{s1}}} \times ...\times Z_{p_s^{k_{sr_s}}}) $$
Since $m_{ij} \leq k_{ij}$ we get : $p^{m_{ij}}|p^{k_{ij}}$ and $Z_{p_i^{k_{ij}}}$ is cyclic so reverse Langrange holds : $\implies Z_{p_i^{m_{ij}}} \leq Z_{p_i^{k_{ij}}}$.
So we can (?) take the cross product of $Z_{p_i^{m_{ij}}}$'s and since every term is a subgroup of the corresponding $Z_{p_i^{m_{ij}}}$ in which we analysed G then this cross product must (?) be a subgroup of G and it has order d.
*E.G
Some explanations in the proof were not implicitly made by my professor but I tried to guess why they stand (I added some "?" to draw attention to these).
