order of non abelian group can't be what?

Solution 1:

$|G|=25$ is not possible since groups of order $p^2$ are always abelian.
$|G|=125$ is possible since for any prime $p$ there exist a non-abelian group of order $p^3$.
In a group of order $35$ the $7$ group and the $5$ group are normal by the Sylow theorems and hence this group is cyclic.
In a group of order $55$, the $11$-group is normal, but the $5$-group does not have to be normal and therefore there is a non-commutative group of order $55$.

Other words for $35,55$. In both cases the group must be a semi-direct product $$C_q\rtimes C_p,$$ where $p<q$. In the case $q=7$ this action must be trivial since $5$ do not divide $7-1$. However, for $q=11$ there is a non-trivial action of $C_5$ on $C_{11}$ and therefore the group do not have to be commutative.

Solution 2:

Hint: 1)Note that $o(G)=5^2$,what do u know about groups of order $p^2$.

2)Note that $o(G)=5*11 $ what do u know about groups of order $pq$.

3)Note that $o(G)=5*7$ again what do u know about groups of order $pq$.

4)Note that $o(G)=5^3$,what do u know about groups of order $p^3$.

Added:1)Every group of order $p^2$ is abelian.

2,3)Suppose $o(G)=pq$, $p<q$ and p does not divide $q-1$ then every group of order $pq$ is cyclic.Suppose p divides $q-1$ then there is a unique nonabelian group of order $pq$,obviously upto isomorphism.

4)For any prime there exist a nonabelian group of order $p^3$,namely Heisenberg Group with entries from a field of order $p$.