Algebra Solutions: 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 $i = k$ .

Algebra Solutions

Properties of inverses in a finite cyclic subgroup

No comments:

Post a Comment