structure of as an abelian group
is cyclic if and only if , or for an odd prime and an integer.
(2): Note first that the result is clear for , since then is the multiplicative group of the finite field and thus is cyclic (any finite subgroup of the multiplicative group of a field is cyclic). Also, . Since is abelian, it is the direct product of its -primary components for each prime ; we will show that each of those -primary components is cyclic. For , it suffices to find an element of of order . is such an element; see Lemma 3 below. For , consider the map
i.e. the reduction-by- map. This is a ring homomorphism; restricting it to gives a surjective group homomorphism . Since , it follows that the kernel of has order . Thus for , the -primary component of must map isomorphically into by order considerations. But is cyclic, so the -primary component is as well.
Thus each -primary component of is cyclic and thus is also cyclic.
(3): The result is true for , when , the Klein -group. So assume . has exact order in (see Lemma 4 below). Also by that lemma, is , so that and are two distinct elements of order . Thus is not cyclic, but has a cyclic subgroup of order ; the result follows. ∎
is clear, since
: Assume is cyclic. If is a power of , then by the theorem, it must be either or . Otherwise, if has two distinct odd prime factors , then contains the direct product . But the orders of these two factor groups are both even (they are and respectively), so their direct product is not cyclic. Thus can have at most one odd prime as a factor, so that for some integers , and
But the order of is even, so that (since the order of is also even for ) we must have or , so that or .
The above proof used the following lemmas, which we now prove:
Let be an odd prime and a positive integer. Then has exact order in the multiplicative group .
For , has exact order in the multiplicative group (which has order ). Additionally, .
The proof of this lemma is essentially identical to the proof of the preceding lemma. Again by the binomial theorem,
Now, is if , and is at least for . We thus get
Setting gives ; setting gives . (Note that since ). ∎
- 1 Dummit, D., Foote, R.M., Abstract Algebra, Third Edition, Wiley, 2004.
|Title||structure of as an abelian group|
|Date of creation||2013-03-22 18:42:40|
|Last modified on||2013-03-22 18:42:40|
|Last modified by||rm50 (10146)|