Show that $|a^{k}|=|a^{n-k}|$

Question

Let G be a group and let $a$ be an element of G of order $n$. For each Integer $k$ between $1$ and $n$,

show that $\left | a^{k} \right |=\left | a^{n-k} \right |$

My attempt is as follows:

$\left | a \right |=\left | \left \langle a \right \rangle \right |=n$

Thus, $\left \langle a \right \rangle=\left \{ e,a^{1},a^{2},\cdot \cdot \cdot ,a^{n-1} \right \}$

The elements in $\left \langle a \right \rangle$ are of the form $a^{k} \forall k \in \mathbb{Z}^{+}$

Theorem: criterion for $a^{i}=a^{j}$

$a^{k}=a^{n-k}$ If and Only If $n\mid \left ( k-\left ( n-k \right ) \right )$

so, $a^{k}=a^{n-k}$ If and Only If $n\mid \left ( 2k-n \right )$

At this point, I am clearly stuck. I would like to request a hint. Thanks in advance.

no, why $a^k$ would generate $\langle a \rangle $ ? (it depends on $gcd(k,n)$) — reuns, Apr 18 '16 at 04:12
@user1952009 Thanks for pointing that out. It is true IFF gcd(n,k)=1 — Mathematicing, Apr 18 '16 at 04:13
Possible duplicate of An element of a group has the same order as its inverse — , Apr 18 '16 at 04:14
if you understand that, you understand why and when $a^k$ generates the same subgroup as $a^m$ — reuns, Apr 18 '16 at 04:14

score 2 · Accepted Answer · answered Apr 18 '16 at 04:44

Since $a^{n-k}a^k=a^n=e$, we have that $(a^k)^{-1}=a^{n-k}$. We know that $|b|=|b^{-1}|$, so $|a^k|=|a^{n-k}|$.

Proof that $|b|=|b^{-1}|$:

First note that $(b^{-1})^{|b|}=b^{-|b|}=(b^{|b|})^{-1}=e^{-1}=e$

Second note that if $(b^{-1})^n=e$ for $n<|b|$, then $(b^{n})^{-1}=e$, or $b^n=e^{-1}$, or $b^n=e$, a contradiction.

1 Answers1