This is Pinter 10.H.6.
Let $a$ have order $n$.
Let $k$ be an integer such that every prime factor of $k$ is a factor of $n$.
Prove: If $a$ has a kth root $b$, then $ord(b) = nk$.
Pinter will often set things up so that previous results in an exercise set will be helpful in the following exercises. So, here's a result from 10.H.5:
If $a$ has order $n$ and $a$ has a kth root $b$ then $b$ has order $nk/l$, where $n$ and $l$ are relatively prime.
Here's what I have so far.
$$ ord(a) = n \tag{1} $$
$$ a = b^k \tag{2} $$
By (1):
$$ a^n = e $$
Substitute (2):
$$ (b^k)^n = e $$
$$ b^{kn} = e $$
If $kn$ is the smallest number such that this is true, then $ord(b) = kn$.
Let's assume there is a number smaller than $kn$ such that $b$ raised to this power is equal to $e$.
We only need to consider factors of $kn$.
Let's let one of these numbers be $p$, a prime.
$$ p | k $$
$$ k = p k' \tag{4} $$
Let's assume:
$$ b^p = e. \tag{3} $$
Eq (2):
$$ a = b^k $$
Substitute (4):
$$ a = b^{pk'} $$
$$ a = (b^p)^{k'} $$
Substitute (3):
$$ a = (e)^{k'} $$
$$ a = e $$
This approach doesn't seem to help. :-)
Moreover, it doesn't use the fact that prime factors of $k$ are also factors of $n$.
What's a good approach to solving this problem?
Note: I'm primarily interested in solutions that only use what's available in Pinter's text up to this point and that solve the problem in the way that Pinter is likely expecting the reader to. Of course, feel free to post solutions that step outside of this constraint as they are often interesting.
It seems Pinter expects the reader to solve this based on how exercises 10.H.1, 10.H.2 and 10.H.3 were solved. He doesn't say this explicitly however we can't for sure assume this.
Approach 2
This approach continues right after (3) is established above.
By (3):
$$ b^p = e $$
Raise both sides to $k$:
$$ (b^p)^k = (e)^k $$
$$ (b^k)^p = e $$
Substitute (2):
$$ (a)^p = e $$
We know that $p$ is a factor of $k$ and also a factor of $n$ so:
$$ p < n $$
which violates (1).
I think I'd need to be a little more rigorous in showing that $p < n$.
Bill's approach
Here's a step-by-step explanation of Bill's answer below.
We're given:
$$ ord(a) = n \tag{1} $$
$$ a = b^k \tag{2} $$
By (1):
$$ a^n = e \tag{5} $$
Substitute (2):
$$ (b^k)^n = e $$
$$ b^{kn} = e \tag{6} $$
Eq (5):
$$ a^n = e $$
Substitute (6):
$$ a^n = b^{kn} \tag{7} $$
We'll be using the order test. I.e.:
$ord(a) = n$ iff $a^n = e$ but $a^{n/p} ≠ e$ for every prime $p|n$.
In other words, if $a^n = e$ and you can show that for every prime $p|n$ that $a^{n/p} ≠ e$ then $ord(a) = n$.
So for our case, if we can show that $b^{kn/p} ≠ e$ for each prime $p|kn$ then $ord(b) = kn$ by the order test.
The exercise mentions:
every prime factor of $k$ is a factor of $n$
Thus, if $p|kn$ then
$$p|n \text{ or } p|k \text{ (and thus p|n by the given statement)} \tag{8}$$
Eq (7):
$$ a^n = b^{kn} $$
Divide both powers by $p$:
$$ a^{n/p} = b^{kn/p} $$
We know that n/p is an integer by (8).
We know that kn/p is an integer by (8).
$n/p < n$ so $a^{n/p} ≠ e$ (Otherwise, (1) would be violated.)
So:
$$b^{kn/p} ≠ e$$
By order test:
$$ ord(b) = kn $$
ganeshie8's approach
Here's a step-by-step explanation of the answer provided by ganeshie8.
We'll be using 10.G.5:
Let $a \in G$. Let $ord(a) = n$. Then $ord(a^m) = lcm(m,n)/m$.
We're given the following:
$$ord(a) = n \tag{1}$$
$$a = b^k \tag{2}$$
$$k \text{ is an integer such that every prime factor of } k \text{ is a factor of }n \tag{3}$$
Let's begin. Let:
$$ord(b) = u \tag{7}$$
By 10.G.5:
$$ord(b^k) = lcm(k,u)/k$$
From basic number theory, $gcd(k,u) lcm(k,u) = ku$ thus $lcm(k,u)/k = u/gcd(k,u)$ so:
$$ord(b^k) = u/gcd(k,u)$$
Substitute (2):
$$ord(a) = u/gcd(k,u)$$
Substitute (1):
$$n = u/gcd(k,u) \tag{8}$$
$$n | u$$
Let:
$$u = na \tag{9}$$
Eq (8):
$$n = u/gcd(k,u)$$
Substitute (9):
$$n = na/gcd(k,na)$$
Divide by $n$:
$$1 = a/gcd(k,na)$$
$$gcd(k,na) = a$$
Divide by $a$:
$$gcd(k,na)/a = a/a$$
$$gcd(k/a,n) = 1$$
Thus $k/a$ and $n$ are relatively prime. By (3):
$$k/a = 1$$
$$k = a \tag{10}$$
Eq (9):
$$u = na$$
Substitute (10):
$$u = nk$$
Substitute (7):
$$ord(b) = nk$$
