Why is the generalized quaternion group $Q_n$ not a semidirect product?
Why is the generalized quaternion group $Q_n$ not a semi-direct product?
2 Answers
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.
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0Thanks 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? – 2011-03-15
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0@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. – 2011-03-15
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.
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0Dear 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? – 2011-03-15
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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. – 2011-03-16
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0Thanks Rahul for the answer. – 2011-03-16