Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?)

share|improve this question

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
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 at 8:32
1  
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 at 9:54
    
Thanks a million for your help. –  Tina Jun 8 at 10:03

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.