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.

Why is the generalized quaternion group $Q_n$ not a semidirect product?

share|improve this question
add comment

2 Answers

up vote 7 down vote accepted

How many elements of order 2 does a generalized quaternion 2-group have? How many elements of order 2 must each factor in the semi-direct product have?

Note that dicyclic groups (generalized quaternion groups that are not 2-groups) can be semi-direct products. The dicyclic group of order 24 is a semi-direct product of a group of a quaternion group of order 8 acting on a cyclic group of order 3.

share|improve this answer
    
Thanks Jack for the answer. Certainly, there are at most one element in Q_2^n which is from order two. So, you mean that having more than one element of order two is impossible? Is this true that with the last statement two factors have elements in share, so we have a contradiction? –  B. S. Mar 15 '11 at 7:00
    
@Babak: Exactly! The factors cannot share any elements of order 2, but they each must have one. There is no room for a semi-direct product. –  Jack Schmidt Mar 15 '11 at 7:11
add comment

One characterization of the generalized quaternion group is:

  • If $G$ is a non-abelian $p$-group which contains only-one subgroup of order $p$, then $G$ is generalized quaternion group [Hall-Theory of groups; Theorem 12.5.2].

So if we try to write the generalized quaternion group $Q_n$ as semi-direct product, then we should have a normal subgroup $N$, a subgroup $H$, with one necessary condition that $N\cap H=1$; which is not possible because of uniqueness of subgroup of order $p$; it will contained in all subgroups of $Q_n$. Hence the generalized quaternion group is not semi-direct product of smaller $p$-groups.

share|improve this answer
    
Dear Kahul, is this right: That normal subgroup of Q_n, forms Q_n's center? I saw this claimed by J.J. Rotman. If it is true how can I show that? Just by playing with the elements in the Q_n? Or by another way? –  B. S. Mar 15 '11 at 13:55
1  
@Babak: As $Q_n$ is a $p$-group, its any subgroup will also be a $p$-group, hence will contain a subgroup of order $p$. As there is unique subgroup of order $p$, this unique subgroup of order $p$ will be contained in every subgroup of $Q_n$; hence it is impossible to get a normal subgroup $N$, a subgroup $H$ of $Q_n$, such that $N\cap H=1$. So $Q_n$ can not be (internal) semi-direct product of its some subgroups. –  user8186 Mar 16 '11 at 8:12
    
Thanks Rahul for the answer. –  B. S. Mar 16 '11 at 11:38
add comment

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.