I am trying to classify all groups of order 75. I am only allowed to use Sylow's theorems, and other basic facts (such as what groups of order $p^2$, $2p$ are isomorphic to, basic facts about direct products, etc).
My attempt was the following:
We have that $75 = 3 \cdot 5^2$.
Sylow's third theorem tells us that the number $n_5$ of Sylow $5$-subgroups is congruent to $1 \bmod 5$, and must divide $3$. Therefore $n_5 = 1$, so we must have a normal (by Sylow's second theorem) subgroup of order $5^2 = 25$.
Similarly we get that $n_3$, the number of Sylow $3$-subgroups, is either $1$ or $25$. In the case that $n_3 = 1$ it isn't hard to show that that the group is isomorphic to either $\mathbb Z_{25} \times \mathbb Z_3 \simeq \mathbb Z_{75}$, or to $(\mathbb Z_5 \times \mathbb Z_5) \times \mathbb Z_3 \simeq \mathbb Z_5 \times \mathbb Z_{15}$.
In the other case when $n_3 = 25$, we have $25$ subgroups where the pairwise intersection of any two is $\{e\}$. This accounts for $1 + 25 \times (3-1) = 51$ distinct elements (including the identity) of the group, leaving $24$ elements.
This is all I know for this case, and I don't know how to proceed. I appreciate any assistance.
There are three different groups of order $75$. Of course we have the $2$ abelian groups $C_5\times C_5\times C_3$ and $C_{25}\times C_3$. For the non-abelian, see
Construct a non-abelian group of order 75
A detailed proof about the full classification, using Sylow, is given here, page $3$, part 5.5.8. It shows in particular, that the non-abelian group constructed in the duplicate is the only one.