Algebra Solutions

Characterize the elements of cyclic subgroups

Let $G$ be a group, and let $x \in G$ be an element of finite order; say $|x| = n$ . Use the Division Algorithm to show that any integral power of $x$ is equal to one of the elements in the set $A = \{ 1, x, x^2, \ldots, x^{n-1} \}$ . Conclude that $A$ is precisely the set of distinct elements of the cyclic subgroup of $G$ generated by $x$ .

Let $k \in \mathbb{Z}$ . By the Division Algorithm, there exist unique integers $(q,r)$ such that $k = qn + r$ and $0 \leq r < |n|$ . Thus $x^k = x^{qn + r} = (x^n)^q x^r = x^r$ , where $x^r \in A$ .
Hence the cyclic subgroup of $G$ generated by $x$ is $A$ ; moreover, by a previous result, the elements of $A$ are all distinct.

The order of a group element is smaller than the cardinality of the group

Let $G$ be a group and $x \in G$ an element of order $n < \infty$ . Prove that the elements $1, x, x^2, \ldots, x^{n-1}$ are all distinct. Deduce that $|x| \leq |G|$ .

Suppose to the contrary that $x^a = x^b$ for some $0 \leq a < b \leq n-1$ . Then we have $x^{b-a} = 1$ , with $1 \leq b-a < n$ . However, recall that $n$ is by definition the least integer $k$ such that $x^k = 1$ , so we have a contradiction. Thus all the $x^i$ , $0 \leq i \leq n-1$ , are distinct.
In particular, we have $\{ x^i \ |\ 0 \leq i \leq n-1 \} \subseteq G$ , so that $|x| = n \leq |G|$ .

The powers of a group element of infinite order are pairwise distinct

Let $G$ be a group, and let $x \in G$ be an element of infinite order. Prove that the elements $x^k$ with $k \in \mathbb{Z}$ are all distinct.

Suppose to the contrary that $x^a = x^b$ for some (distinct) integers $a$ and $b$ ; without loss of generality say $a < b$ . Then we have $x^{b-a} = 1$ and $0 < b-a$ . This is a contradiction since $x$ has infinite order; thus no such $a$ and $b$ exist.

Properties of inverses in a finite cyclic subgroup

Let $G$ be a group and $x \in G$ an element of finite order, say $|x| = n$ .

Prove that if $n = 2k+1$ is odd and $1 \leq i < n$ , then $x^i \neq x^{-i}$ .
Prove that if $n = 2k$ is even and $1 \leq i < n$ , then $x^i = x^{-i}$ if and only if $i = k$ .

We begin with a lemma.
Lemma 1. Let $n,k \in \mathbb{Z}^+$ . If $2 \leq k < n$ and $n|k$ , then $n = k$ . Proof: We have $k = nq$ for some $q \in \mathbb{N}$ . If $q \geq 2$ , then since $2k \leq 2n$ , we have $k < 2k \leq 2n \leq k$ , a contradiction. Thus $q = 1$ and we have $k = n$ . $\square$
Now for the main results.

If $x^i = x^{-i}$ for some $1 \leq i < n$ , we have $x^{2i} = 1$ and thus $n|2i$ by a lemma to a previous example. Since $n$ is odd, we have $n|i$ . By the lemma, $i = n$ , a contradiction.
$(\Leftarrow)$ Clearly, if $i = k$ then $x^{2i} = 1$ , so that $x^i = x^{-i}$ . $(\Rightarrow)$ Suppose $x^i = x^{-i}$ for some $1 \leq i < n$ . Then $x^{2i} = 1$ . By the Division Algorithm we have $2i = qn + r$ for some integers $q$ and $r$ with $0 \leq r 0$ , we have a contradiction, so that $r = 0$ . Hence $2i = qn$ . Now we must have $q > 0$ , and since $0 < 2i < 2n$ we have $0 < qn < 2n$ and hence $q=1$ . So $2i = 2k$ and thus $i = k$ .