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 $n | m$.