Mathematics Prelims

February 3, 2009

Cyclic Groups Part I – The Order of an Element

Filed under: Abstract Algebra,Algebra — cjohnson @ 3:14 pm

We mentioned last time that if we have a group G and a subset (not necessarily a subgroup) of elements S \subseteq G, then there was a unique minimal subgroup of G which contained S.  This subgroup is the intersection of all subgroups which contain S and is denoted \langle S \rangle.  In the case when S is a singleton (i.e., contains only one element), say S = \{ x \}, then we just write \langle x \rangle for the smallest group containing x.  Groups generated by a single element like this are referred to as cyclic.

Recalling the order of a group is simply the cardinality of the underlying set, we define the order of an element of the group to be the order of the subgroup generated by that element; |x| = |\langle x \rangle|.  Also recall that an alternative way of describing a group generated by a set was to talk about all of the ways to multiply (apply the group operation) to elements of the set.  In the case of a cyclic group, where the group is generated by a single element, this means that every element in the group is actually a power of the generator.  That is, if y \in \langle x \rangle, then there exists a k \in \mathbb{Z} such that y = x^k.

If |x| = n < \infty, then n is actually the smallest natural number such that x^n = 1.  To see this, suppose there was another number, m < n with x^m = 1.  Then we would have for every k that k = mq + r, where 0 \leq r < m (by Euclid’s algorithm).  This gives x^k = x^{mq + r} = x^{mq} x^r = (x^m)^q x^4 = 1^q x^r = x^r.  If this were the case, then the group would in fact only have order r.

Likewise, if |x| = n < \infty and x^m = 1, then n | m.  Again, use Euclid’s algorithm to write m = nq + r where 0 \leq r < n.  We have again 1 = x^m = x^{nq + r} = x^{nq} x^r = (x^n)^q x^r = x^r, but this contradicts the minimality of n if r \neq 0, so we conclude r = 0 and n | m.


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