I am visiting this thread:
$G$ is Abelian iff all the Sylow subgroups of $G$ are normal
I am wondering whether more can be said about the lower bounds of number of subgroups of $G$ when $G$ is of that order, which is $p_1 ^2 p_2 ^2 p_3 ^2 ... p_n ^2$, and it satisfies the said property, which is Abelian and/or having normal Sylow subgroups only. Maybe we can use the fact that Abelian groups are uniquely represented as direct product or something?
I know for a fact that exactly counting the number of subgroups of even an Abelian group is not easy, but this case seems like a special case to at least estimate how many subgroups $G$ at least has.
Thanks in advance.
For any abelian group $G$ with $|G| = p_1^{a_1}\cdots p_n^{a_n}$, where $p_1,\dots p_n$ are distinct primes, we have $G = P_1\times\cdots\times P_n$ with $|P_i| = p_i^{a_i}$. Thus, to count the number of subgroups of $G$, we only need to consider each $P_i$. You can have a try.
The "if and only if" statement holds only for $a_i\le 2$ for each $i$, because groups of order $p^2$ is abelian, but for $p^3$ we have two another non-abelian groups.