Order of a^k is a divisor of the order of a

Question

Exercise 10.D.2 from Pinter says:

Let a be any element of finite order of a group G. Prove the following:

The order of a^k is a divisor (factor) of the order of a.

Approach 1

Let ord(a) = n.

Let ord(a^k) = m.

We must show that m is a divisor of n. I.e. m|n.

Since ord(a^k) = m we know that:

(a^k)^m = e
a^(km) = e

By theorem 5:

Suppose an element a in a group has order n.

Then a^t = e iff t is a multiple of n.

("t is a multiple of n" means that t = nq for some integer q).

km = nq

Solving for m (the order of a^k):

m = nq/k

It isn't clear that this shows that m is a divisor of q.

Any suggestions for this approach?

Approach 2

(This is the approach suggested by the hint in the back of the book.)

Let ord(a) = n.

(a^k)^n
a^(nk)
(a^n)^k
e^k
e

At this point, the book says to use theorem 5.

By theorem 5:

nk = nq

Canceling n on both sides:

k = q

And that's all the book's hint has to offer. What's a good way forward?

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Note: there is a question on this site that specifically asks about Approach 2 shown above. What I'm asking here is, if Approach 1 is workable. It would also be nice to see Approach 2 completed as the full answer in the linked question was not shown.

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Approach based on 2 above and Bill's answer

Let ord(a) = n.

Let ord(a^k) = m.

(a^k)^n
a^(nk)
(a^n)^k
e^k
e

Thus

(a^k)^n = e

Let's state theorem 5 in these terms

Suppose an element a^k in a group has order m.

Then (a^k)^n = e iff n is a multiple of m.

Answer 1

$ (a^{\large k})^{\large n} = 1\,\Rightarrow\, {\rm ord}(a^{\large k})\mid n,\, $ QED, $ $ if you already know this Corollary on order (= Theorem 5?).

Else, repeating its proof: the set $\,J\,$ of $\,j\,$ with $\,a^{\large kj}=1$ is closed under subtraction thus its least positive element $\,( = {\rm ord}\ a^k)\,$ divides all $\,j\in J, \,$ including $\,n\in J$.

Answer 2

Not to trivialize what you've done, but, why not use the fact that $\vert a^k\vert=\dfrac n{\operatorname {gcd}(n,k)}$, where $n=\vert a\vert$.