order of non abelian group can't be what? Let $G$ be a non abelian group; then its order can be:


*

*$25$

*$55$

*$35$

*$125$
I think the order cannot be $25$ and $35$. But from option $55$ and $125$ which one is not possible? 
Why not $25$ because every group of order $p^2$ is abelian. Here $p$ is $5$. Why not $35$ because $35=5*7$ and $5$ does not divide $7-1$. There there is only one group which is cyclic up to isomorphism of order $35$.
 A: *

*$|G|=25$ 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.
A: 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 $p$.
A: Take G=HF(5), Heisenberg group over feild of order 5 gives non Abelian group of order 125
