Group of order $135$ abelian and not cyclic I am trying to solve the following:
Let $G$ be a group of order $135$. Show that if $G$ has more than one normal subgroup of order $3$, then $G$ is abelian and non-cyclic.
What I could do was:
$|G|=5.3^3$ Let $r_p=$number of $p-$Sylow subgroups. Then $$r_5 \equiv 1 (5), r_5|3^3 \space \implies r_5=1,$$ $$r_3 \equiv 1 (3), r_3|5 \space \implies r_3=1$$
Let $H$ and $K$ denote the $3-$ and $5-$Sylow subgroups respectively. Since $H,K \lhd G$ and $H \cap K=\{1\}$, then $HK=G$. But then $$G \cong H \times K$$
If $G$ is cyclic, then $H$ is cyclic, so $$G \cong Z_{27} \times Z_{5} \cong Z_{135}.$$ Since there is a unique subgroup of order $3$ in  $Z_{135}$, then $G$ has a unique subgroup of order $3$, which is a contradiction.
So $G$ has to be non-cyclic. I got stuck trying to show $G$ is abelian. I would appreciate suggestions with that part. All I could say up to now is that $G/K \cong Z_{5}$, which is abelian, so $[G,G] \subset H$.
 A: Let $A_1,A_2$ be distinct normal subgroups of order $3$ in $G$. Then $G \big/ A_i$ is of order $3^2 \times 5$ and so is a direct product of subgroups of orders $3^2$ and $5$. Groups of order $p^2$ are abelian so $G \big/ A_i$ is abelian. Therefore the commutator subgroup is inside both $A_i$, i.e. $[G,G] \leq A_1 \cap A_2 = \{e \}$. So $G$ is abelian. 
A: $H$ is abelian, so $G$ is abelian too. Indeed, suppose, that $H$ is a nonabelian, and $A_1,A_2$ - distinct normal subgroup of order 3 of $G$. Since $H$ is a unique Sylow 3-subgroup of $G$, then $A_1,A_2\leq H$. Since $H$ is a nonabelian, and of order $3^3$, then $|Z(H)|=3$, hence one of $A_1,A_2$ is not a subgroup of $Z(H)$. Let $A_1\not\leq Z(H)$. Since $|A_1|$ simple, then $A_1\cap Z(H)=1$ - contradiction, because in the finite nontrivial $p$-group each nontrivial normal subgroup has a nontrivial intersection with a center.
A: Say $H$ a subgroup of order $3$ (by Cauchy there is some). Since $|H\cap Z(G)|=1$ or $3$, if $H\unlhd G$ then necessarily $H$ is central (normal subgroups are union of conjugacy classes, whose sizes here can be only $1$, $3$ and higher values). Suppose there are distinct $H_1$ and $H_2$ normal of order $3$; then $H:=H_1H_2$ is normal of order $9$ and central, and $G/H$ is cyclic (as $3\nmid 5-1$). Therefore $G$ is Abelian (recall the "cyclic $G/Z(G)$ argument"). Moreover, $H\cong C_3\times C_3$ is not cyclic, so $G$ is not cyclic either.
