Solvable groups of order 25920

I would like to prove the following statement:

Let $G$ be a finite solvable group of order $2^6.3^4.5$. If $O_{5^\prime}(G)\neq1$, then $G$ has an element of order $18$.

Also, I would like to know that whether I can find the structure ofsuch a group $G$? (For example, can I introduce these groups in GAP?)

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I am not sure whether you have asked the question as you intended, but I do not think that the statement is true as it stands. Let $G = A \times B,$ where $A$ is cyclic of order $5,$ and $B$ is the semi-direct product of an elementary Abelian group of order $64$ with a Sylow $3$-subgroup of ${\rm GL}(6,2)$ (with the action of the second group being given by its embedding in ${\rm GL}(6,2)).$ The latter Sylow $3$-subgroup is isomorphic to $C_{3} \wr C_{3},$ and each of its elements of order $9$ act fixed point freely on the elementry Abelian $2$-group ( strictly, without non-identity fixed points)- you can see this because the action of any element of order $9$ is faithful and irreducible. Hence $B$ contains no element of order $18,$ and $G$ doesn't either.
Thank you for your reply. In your example we have $O_5(G)\neq1$. What about the case when $O_5(G)=1$? Do you think it has a counterexample as well? – Tina Jun 8 '14 at 8:32
There are lots of counterexamples with $O_5(G)=1$, like $(3^4 \rtimes 5) \times S$, where $S$ is any group of order $64$, or $(2^4 \rtimes 5) \times 3^4 \times T$, where $T$ is any group of order $4$. – Derek Holt Jun 8 '14 at 9:54