Disclaimer: these are my personal solutions; they have not been reviewed, so be careful relying on them as there might be errors. Some competitions may have official solutions from the contest publishers which you may wish to refer to instead if you’re looking for verified solutions.
Problem 5
Two elements $A$, $B$ in a group $G$ have the property $ABA^{-1}B = 1$, where $1$ denotes the identity element in $G$.
- Show that $A B^2 = B^{-2} A$.
- Show that $A B^n = B^{-n} A$ for any integer $n$.
- Find $u$ and $v$ so that $(B^a A^b)(B^c A^d) = B^u A^v$.
Hints
Hint #1 for (a)
This may require a multi-step process; first, try solving for $A$ and see where you get to.
Hint #2 for (a)
First, try to prove the initial case $AB = B^{-1}A$. Then, combine it with the result from hint #1 above and see where that leads.
Hint for (b)
Can you use the result from part (a) to help you extend it to arbitrary $n$?
Hint for (c)
Use the result from part (b) as well as the hint #1 from part (a) and see if you can apply both of them. You may need to apply a property of groups to rearrange the equation to help you make these transformations.
Solution for (a)
Open to see the solution; try solving it first or see the hints above!
The goal is to show that $AB^2 = B^{-2}A$. Let’s start with the given equation, find a way to simplify it to compute $A$, and then see what we can do with that.
$$ \begin{align*} A B A^{-1} B & = 1 & \quad \text{given} \\ A B A^{-1} B B^{-1} & = B^{-1} & \quad \text{right-multiply by $B^{-1}$} \\ A B A^{-1} & = B^{-1} & \quad \text{simplify} \\ A B A^{-1} A & = B^{-1} A & \quad \text{right-multiply by $A$} \\ A B & = B^{-1} A & \quad \text{simplify (1)} \\ A B B^{-1} & = B^{-1} A B^{-1} & \quad \text{right-multiply by $B^{-1}$} \\ A & = B^{-1} A B^{-1} & \quad \text{simplify (2)} \\ \end{align*} $$Now, let’s substitute the value for $A$ from eq. $(2)$ into the right side of eq. $(1)$:
$$ \begin{align*} AB & = B^{-1} A & \quad \text{eq. $(1)$ above} \\ AB & = B^{-1} B^{-1} A B^{-1} & \quad\text{substitute $A$ in rhs from eq. $(2)$} \\ AB & = B^{-2} A B^{-1} & \quad\text{simplify} \\ ABB & = B^{-2} A B^{-1} B & \quad\text{right-multiply by $B$} \\ AB^2 & = B^{-2} A & \quad\text{simplify} \quad \blacksquare \\ \end{align*} $$Solution for (b)
Open to see the solution; try solving it first or see the hints above!
In part (a), we showed that $AB^2 = B^{-2}A$, and we showed it by first finding that $AB = B^{-1}A$ and then extending it by substituting $A$ on the right-hand side. We can easily extend this to $AB^3$, etc. manually, but what we need to do here is to prove the general case for any $n$.
Thus, we will prove it by induction:
- first, we will prove this for a base case, e.g., $n = 1$;
- then we will assume that this holds for some arbitrary $n$, and given that, we will prove it for $n+1$.
Let’s start with $n=1$, which takes the form $AB = B^{-1}A$. This was already proven in part (a), so we’re done.
$$ \begin{align*} A B^n & = B^{-n} A & \quad \text{assumption to prove inductive case} \\ A B^n & = B^{-n} B^{-1} A B^{-1} & \quad \text{substitute for $A$ from eq. $(2)$ in part (a)} \\ A B^n B & = B^{-n} B^{-1} A B^{-1} B & \quad \text{right-multiply by $B$} \\ A B^{n+1} & = B^{-(n+1)} A & \quad \text{simplify} \quad \blacksquare \\ \end{align*} $$Solution for (c)
Open to see the solution; try solving it first or see the hints above!
$$ \begin{align*} (B^a A^b)(B^c A^d) & = B^u A^v & \quad \text{given} \\ B^a (A^b B^c) A^d & = B^u A^v & \quad \text{reassociate} \\ B^a (A^{b-1} A B^c) A^d & = B^u A^v & \quad \text{factor out $A$} \\ B^a (A^{b-1} B^{-c} A) A^d & = B^u A^v & \quad \text{using result from part (b)} \\ B^a (A^{b-2} A B^{-c} A) A^d & = B^u A^v & \quad \text{factor out another $A$} \\ B^a (A^{b-2} B^{c} A^2) A^d & = B^u A^v & \quad \text{using result from part (b)} \\ \end{align*} $$At this point, we see a curious pattern: we can keep factoring out an $A$ term from $A^b$ and transfer it to the right side of the $B^{\pm c}$ term, but each time we do, the sign of the exponent of $B$ changes! Thus, if $b$ is even, it remains as $B^c$, but if $b$ is odd, it becomes $B^{-c}$.
Hence, let’s define
$$ \begin{align*} c' = \begin{cases} c & \text{if $b \mod 2 \equiv 0$} \\ -c & \text{if $b \mod 2 \equiv 1$} \end{cases} \end{align*} $$The exponent of $A$ will move over as-is, so the end result will be:
$$ B^a B^{c’} A^b A^d = B^u A^v $$
which can be simplified to:
$$ B^{a + c’} A^{b + d} = B^u A^v $$
so we have
$$ \boxed{u = a + c’, \quad v = b + d} \quad \text{with $c’ = \pm c$ as defined above.} \quad \blacksquare $$