Classifying groups of order 90. Since $3\cdot 3\cdot 2\cdot 5=90$, we know that we have a $3$-Sylow subgroup $P_3$ of order $9$, a $2$-sylow subgroup $P_2$ of order $2$, a $ 5$-Sylow subgroup $P_5$ of order $5$. 
I know that $P_5 \cong Z_5$ and $P_2 \cong Z_2$, right? But I'm not sure what $P_3$ is isomorphic to, because we cannot necessarily conclude that it is cyclic...since it might have $4$ different elements of order $3$. 
So when I'm looking at the different cases for the semidirect products (for example if I look at the case when all of the sylow subgroups are normal), I will just say $G \cong P_3 \times Z_{10}$, right? 
I am just asking to make sure if I'm doing it correctly (for this specific case). 
Thanks in advance 
 A: Here are some hints to help along:


*

*There is a normal subgroup of index 2 (use the regular representation, and look for an odd permutation).

*All groups of order 45 are abelian (use Sylow's theorems, and the fact groups of order $p^2$ are abelian). 

*Your group is a semidirect product of a group of order 45 and one of order 2.

*Consider the possible order-2 actions on an abelian group of order 45 (call it H). There are 2⋅2 possible actions when H is cyclic, and 2⋅3 when H is noncyclic.  [To see this, decompose $H$ as a direct product, remembering it is abelian.]


Thus there are 4+6=10 possible groups. It is fairly easy to show all are distinct.
A: Some ideas:
Let $\,G\,$ be a group with $\;|G|=90=2\cdot3^2\cdot5\;$ , $\,n_p:=\,$ be the number of Sylow $\,p$-subgroups of $\,G\,$. We denote by $\,P_p, Q_p, R_p\,$ , etc., the different Sylow $\,p$-subgroups:
$$[G:N_G(P_3)]=n_3\in\{1\,,\,10\}$$
Denote $\,M:=P_3\cap Q_3\,$ , and since $\,|M|=1\,,\,3\;$ (why can't this be $\,9\,$ ?) , we get
$$|P_3Q_3|=\frac{|P_3|\,|Q_3|}{|M|}=\frac{81}{|M|}\ge\frac{81}{3}=27$$
so by Lagrange's theorem, $\,|\langle P_3\,,\,Q_3\rangle|=45\,,\,90\,$ (why cannot this be $\,30\,$?) : 
$$(i)\;\;\;\;\;|\langle P_3\,,\,Q_3\rangle|=45\implies [G:\langle P_3\,,\,Q_3\rangle]=2\implies \langle P_3\,,\,Q_3\rangle\lhd G$$
$$(ii)\;\;\;\;\;|\langle P_3\,,\,Q_3\rangle|=90\implies M\lhd P_3\,,\,Q_3 \;\;(\text{why?})\implies M\lhd\langle P_3,Q_3\rangle=G$$
So we already have that a group of order $\,90\,$ cannot be simple, but not only that: both $\,P_3\,,\,Q_3\lhd\langle P_3\,,\,Q_3\rangle\,$ and from here we get that in case (i) there's one single Sylow $\,3$-subgroup which is then normal.
Fill in details and try to take it from here at least for some cases.
