Pinter 10.E.5: Let $a$ and $b$ commute. Prove: There is an element $c$ in $G$ whose order is $lcm(m,n)$.

Question

This is 10.E.5 from Pinter.

Let $a$ and $b$ be elements of a group $G$.

Let $\text{ord}(a) = m$ and $\text{ord}(b) = n$.

Let $a$ and $b$ commute.

Prove:

There is an element $c$ in $G$ whose order is $lcm(m,n)$.

(HINT: Use 10.E.4 together with 10.D.3. Let $c = a^i b$ where $a^i$ is a certain power of $a$.

Here is 10.E.4:

Let $a$ and $b$ commute.

If $m$ and $n$ are relatively prime, then $ord(ab) = mn$.

(HINT: Use 10.E.2)

Here is 10.D.3:

If $ord(a) = km$, then $ord(a^k) = m$.

Here's the start of one approach.

As Pinter suggests, let

$$c = a^i b$$

Let's start from what is to be proven:

$$ord(c) = lcm(m,n)$$

$$ord(a^i b) = lcm(m,n)$$

$$(a^i b)^{lcm(m,n)} = e$$

$$(a^i)^{lcm(m,n)} b^{lcm(m,n)} = e$$

$$a^{i lcm(m,n)} b^{lcm(m,n)} = e$$

$$a^{i lcm(m,n)} = b^{-lcm(m,n)}$$

$$a^i a^{lcm(m,n)} = b^{-lcm(m,n)}$$

$$a^i = b^{-lcm(m,n)} a^{-lcm(m,n)}$$

$$a^i = (ba)^{-lcm(m,n)}$$

In other words, we're aiming to solve for $a^i$.

However, this doesn't seem too helpful as it simplifies to $e$.

Moreover, we haven't used 10.E.5 or 10.D.3.

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It seems that this part of the hint:

$c = a^i b$ where $a^i$ is a certain power of $a$

is suggesting that we find an $a^i$ such that $gcd(a^i,b) = 1$.

And once that's done, apply 10.E.4:

$$ ord(a^i b) = ord(a^i) n $$

And hopefully, together with 10.D.3, change that to:

$$ ord(a^i b) = lcm(m,n)$$

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This question has been discussed in a few places on math.se.

Here's one approach.

The first part of the answer appears to prove 10.E.4 and solves the case for when $m$ and $n$ are relatively prime.

The second part of the answer does not appear to apply 10.D.3 as recommended by Pinter.

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There are many approaches illustrated on this post.

However, none appear to use the path suggested by Pinter.

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The answer here essentially restates part of Pinter's hint.

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Is there a clear approach which uses Pinter's suggestions to use 10.D.3, 10.E.4, and only what has been covered up to that point in the book?

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Bill's answer converted to Pinter's notation

• This is a conversion of the proof in Bill's answer to the notation in Pinter.
• It also makes many of the implied steps explicit.
• This will hopefully make the proof more familiar and digestible to folks coming to Pinter's book and abstract algebra perhaps for the first time.

We induct on

$$ ord(a) ord(b) $$

Part 1

If

$$ ord(a) ord(b) = 1 $$

the only possible values are:

$$ ord(a) = 1 $$

$$ ord(b) = 1 $$

so

$$ lcm(ord(a),ord(b)) = 1 $$

$$ ord(c) = 1 $$

which means

$$ c^1 = e $$

thus

$$ c = e $$

Part 2

Now let's consider the case where

$$ ord(a)ord(b) != 1$$

Split off $p^k$ in $ord(a)$, $ord(b)$. I.e.

$$ ord(a) = xP $$ $$ ord(b) = yP' $$

where

$$ P'|P $$

$$ P = p^k > 1 \tag{7} $$

$p$ is prime and $p∤x,y$.

Then, by 10.D.3:

$$ ord(a^P) = x \tag{1} $$ $$ ord(b^{P'}) = y \tag{2} $$

$$ ord(a^x) = P \tag{3} $$ $$ ord(b^y) = P' \tag{4} $$

By induction, there is a $c$ with

$$ ord(c) = lcm(x,y) $$

Substitute (1) and (2):

$$ ord(c) = lcm(ord(a^P),ord(b^{P'})) \tag{5} $$

$$ c = a^P b^{P'} \tag{6} $$

Substitute (6):

$$ ord(a^P b^{P'}) = lcm(ord(a^P),ord(b^{P'})) $$

So:

$$ ord(a^x c) $$

$$ lcm(ord(a^x),ord(c)) $$

Substitute (3) and (5):

$$ lcm(P, lcm(ord(a^P), ord(a^{P'}))) $$

$$ lcm(P, lcm(x,y)) $$

Since $P = p^k$ where $p$ is prime and $p$ is coprime to $x$ and $y$:

$$ P lcm(x,y) $$

$$ lcm(xP, yP') $$

$$ lcm(ord(a), ord(b)) $$

Thus:

$$ ord(a^x c) = lcm(ord(a), ord(b)) $$

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Pinter's approach

This is a start of an approach that closely follows Pinter's hint.

(I have this written up in a text file and didn't have time to format as LaTeX.)

Let ord(a) = m.
Let ord(b) = n.

Let i = gcd(m,n).           (1)

Let ord(a) = k i            (2)
========================================
Consider: ord(a^i b)

If ord(a^i) and ord(b) are relatively prime then by 10.E.4:

ord(a^i b) = ord(a^i) ord(b)        (3)
========================================
By eq (2) and 10.D.3:

ord(a^i) = k                (4)

Eq (2):

ord(a) = k i

Substitute (4):

ord(a) = ord(a^i) i

Solve for ord(a^i):

ord(a^i) = ord(a)/i         (5)
========================================
Eq (3):

ord(a^i b) = ord(a^i) ord(b)

Substitute (5):

ord(a^i b) = ord(a)/i ord(b)

ord(a^i b) = ord(a) ord(b) / i

Substitute (1):

ord(a^i b) = ord(a) ord(b) / gcd(m,n)

ord(a^i b) = ord(a) ord(b) / gcd(ord(a),ord(b))

ord(a^i b) = lcm(a,b)
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The only remaining part left to do is to handle the case where ord(a^i) and ord(b) are not relatively prime.

Answer 1

This is easily done by exploiting prime factorisations. With your notations, write

$$m= p_1^{k_1}\dots p_s^{k_s}; n = p_1^{l_1}\dots p_s^{l_s}$$

We can express $lcm(m,n)$ as

$$p_1^{\min\{k_1, l_1\}}\dots p_s^{\min\{k_s,l_s\}}$$

If we can find elements $c_i$ with $ord(c_i) = p_i^{\min\{k_i,l_i\}}$, then the element you are looking for is $c=\prod_i c_i$ because the order is multiplicative on commutating elements with coprime orders.

Put $$c_i =\begin{cases} a^{m/p^{k_i}} \quad k_i\leq l_i \\ b^{n/p^{l_i}} \quad l_i < k_i \end{cases}$$

and you see we are done.