Group divisibility question

Question

I have the following question which I can't make sense of, here is the entire question:

If $G$ is a group, $b\in G, o(b)=k$ and $b^n = e$, show that $k|n$

What is $o(b)$? Please help.

Answer 1

$o(b)$ is generally taken to be the least positive integer $m$ such that $b^m = e$, and it appears that is how $o(b)$ is defined in the present question. Granting this, and setting $k = o(b)$, we may use the Euclidean division algorithm in $\Bbb Z$ and hence write $n = qk + r$ with $0 \le r < k$. Then

$e = b^n = b^{(qk + r)} = b^{qk}b^r = (b^k)^q b^r = e^q b^r = eb^r = b^r. \tag{1}$

If $r \ne 0$, (1) contradicts $o(b) = k$ since $r < k$. Thus $r = 0$, $n = qk$ and so $k \mid n$. QED.

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!