Let $G$ be a group, $x∈G$, $ a,b∈\Bbb Z$ and $a⊥b$. If $x^a=x^b$, then $x=1$.

Question

There is a missing step in this proof: https://math.stackexchange.com/a/106292/135812

Lemma

Let $G$ be a group, $x\in G$, $a,b\in \mathbb Z$ and $a\perp b$. If $x^a = x^b$, then $x=1$.

Proof: by Bezout's lemma, some $k,\ell\in\mathbb Z$ exist, such that $ak+b\ell=1$. Then $$ x = x^{ak+b\ell} = (x^a)^k \cdot (x^b)^\ell = 1^k \cdot 1^\ell = 1 $$

To my view it the fact that $x^a = 1$ and $x^b = 1$ is pulled out the air. I would like to try fix this. Thanks!

It's not true if $x^a=x^b$ only. For example, in a cyclic group $C_{10}$, then $x^3=x^{13}$. — Quang Hoang, Oct 12 '14 at 11:28

score 2 · Answer 1 · answered Oct 12 '14 at 11:29

2

This lemma is not true.

Take $G=\Bbb Z_{16}^*$, $x=\bar 5$, $a=3$, $b=11$.

Then, $x^3=\bar 5^3=\overline{125}=\overline{13}$, and $x^{11}=\overline{48828125}=\overline{13}$

answered Oct 12 '14 at 11:29

ajotatxe

65,084

score 2 · Accepted Answer · answered Oct 12 '14 at 11:33

2

One needs $x^a=x^b=1$. This is what was actually meant in the Lemma, I think.

answered Oct 12 '14 at 11:33

Martin Brandenburg

163,620

Let $G$ be a group, $x∈G$, $ a,b∈\Bbb Z$ and $a⊥b$. If $x^a=x^b$, then $x=1$.

Lemma

2 Answers2