Prove $o(u^{-1})=o(u)$ for all $u \in G$ and $o(uh)=o(hu)$ for all $u,h \in G$.

Question

Let $G$ be a group and $g \in G-\{1_G\}$.

Let $n=o(u)$. Then $u^n=1$ and so 1$=u^nu^{-n}=u^{-n}=(u^{-1})^n$ so $o(u^{-1})|n$. How do I show $n|o(u^{-1})$?

For the second bit I am not sure.

You showed that $o(u^{-1}) | o(u)$ for all $u\in G$. Well $u^{-1} \in G$, so $o((u^{-1})^{-1}) | o(u^{-1})$, but $(u^{-1})^{-1} = u$! So by symmetry $o(u) | o(u^{-1})$ too. — Adam Francey, Jan 24 '16 at 22:26
Is it true that the order of $ab$ is always equal to the order of $ba$? and An element of a group has the same order as its inverse. — Martin Sleziak, Jan 24 '16 at 22:42

Adam Francey · Answer 1 · 2016-01-25T07:25:47.320

1

$\textbf{Claim}$: $\quad o(u) = o(u^{-1})$ for all $u \in G$.

$\textbf{Proof}$:

Let $u \in G$.

Let $n = o(u)$ and $m = o(u^{-1})$.

Then $1 = 11 = u^n (u^{-1})^m$.

If $n>m$, then $1 = u^{n-m}$, which cannot be since $o(u) = n$.

If $n<m$, then $1 = (u^{-1})^{m-n}$, which cannot be since $o(u^{-1}) = m$.

So it can be concluded that $n = m$.

$\textbf{Claim}$: $\quad o(uh) = o(hu)$ for all $u,h = G$.

$\textbf{Proof}$:

Let $u,h \in G$.

Let $n = o(uh)$ and $m = o(hu)$.

Then $hu = h(uh)^nu = (hu)^nhu \Rightarrow (hu)^n = 1$. So $m|n$.

Similarly, $uh = u(hu)^mh = (uh)^muh \Rightarrow (uh)^m= 1$. So $n|m$.

Therefore $n = m$.

edited Jan 25 '16 at 07:25

answered Jan 24 '16 at 21:53

Adam Francey

How did $h(uh)^nu$ turn into $(hu)^nhu$? – Jones Jan 24 '16 at 22:03
For the first one is there anything I can do to my solution? In the solutions it said by symmetry $n|o(u^{-1})$ but I don't see it. – Jones Jan 24 '16 at 22:04
To see why try it with $n = 3$, so $h(uh)^3u = h(uhuhuh)u = (huhuhu)hu = (hu)^3hu$, now convince yourself that it works for any $n$. – Adam Francey Jan 24 '16 at 22:06
Why can you move the elements? – Jones Jan 24 '16 at 22:07
No elements are moving, the brackets are just in different places. – Adam Francey Jan 24 '16 at 22:10
I see it thanks – Jones Jan 24 '16 at 22:12

1 Answers1