Groups of order 306 has 4 types of group containing an element of order 9

There are at most 4 groups of order 306 = $2 \times 3^2 \times 17$, containing an element of order 9.

I want to prove the above statement. My trial is below;

Let $G$ be such group. If $G$ is abelian, by the fundamental theorem of abelian group, $$G \cong Z_{2} \times Z_{17} \times Z_{3} \times Z_{3} \textrm{ or } Z_{2} \times Z_{17} \times Z_{9}.$$ Thus, $G \cong Z_{2} \times Z_{17} \times Z_{9}$ is one of such group.

Suppose $G$ is nonabelian. Then, by the Sylow's third theorem, $n_3=1$ or $34$, $n_{17}=1$ or $18$, but not both are greater than 1. Thus there are three cases;

Case1 : $n_{3}=1, n_{17}=1$. In this case, we can assume $P_{3} \in Syl_{3}(G)$ is a cyclic group of order 9. Also, $\exists P_{17} \in Syl_{17}(G).$ And $P_3, P_{17}$ are normal subgroup of $G$. Hence $P_3 P_{17}$ is also normal subgroup of $G$. And for any Sylow 2 subgroup, say $P_{2}$, semidirect product of $P_{3}P_{17}$ and $P_{2}$ induces direct product of $P_{3}P_{17}$ and $P_{2}$, thus $G \cong P_{3}P_{17} \times P_{2}$, hence $P_{2}$ should be unique. Thus, this case is one of the nonabelian case containing element of order 9.

Case2 : $n_{3}=1, n_{17}=18$.

Case3 : $n_{3}=34, n_{17}=1$.

I don't know how to proceeds the other two cases. Could you give me some hints?

• In case 1, you should be getting a semidirect product $P_3P_{17} \rtimes C_2$. There are four possibilities here, with $n_2=1,3,17$ and $51$. The first of these is the abelian group, which you have already considered. Cases 2 and 3 cannot arise. – Derek Holt Jul 26 '17 at 10:19
• @DerekHolt Thank you for your hint. Now I've got why case2 or case 3 occurs; this is because homomorphism from $P_{17} \to Aut(P_{3})\cong C_{6}$ or $P_{3} \to Aut(P_{17}) \cong C_{16}$ is trivial, so it tells that Sylow 17 subgroup in Case2 and Sylow3 subgroup in Case3 is normal, contradiction. However, I still don't know other possibilities of $n_2$; I've got a map from $P_{17}P_{13}$ to $Aut(P_{2})$, it gives also the same trivial homomorphism. Could you give me more hints? – user124697 Jul 27 '17 at 2:57
• You have it the wrong way round. Note that $P_{17}P_3 \cong C_{17} \times C_9$, and $P_2 \cong C_2$. There are four possible homomorphisms $C_2 \to {\rm Aut}(C_{17} \times C_9)$ (the automorphism can invert or centralize the elements of $C_{17}$ or of $C_9$ independently), and these give rise to the four possible values of $n_2$. – Derek Holt Jul 27 '17 at 8:00
• @DerekHolt Thank you very much! Now I've understood your comments totally. – user124697 Jul 27 '17 at 9:21