Inverse composition (GRE problem)
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Inverse composition (GRE problem)

October 24, 2019

Hello to all of you math-inclined people! One of my subscribers asked me to solve one GRE problem. But before I show you the solution, I want to discuss one important mathematical concept. It’s about the following formula: The inverse of a sequence of actions is equal to the sequence of inverse actions but in inverse order. I want to visualise this using the Cambridge hill, which is behind me. First I’m going to climb the Cambridge hill
– action x Then I want to have a seat, but before this
I take out a newspaper to sit on – action y Finally I sit down – action z now I want to come back, so I will invert
xyz. Step by step: First I stand up – action z^(-1)
Then I put back my newspaper – action y^(-1) And finally I go down from the top – action
x^(-1) I performe the composition z^(-1)y^(-1)x^(-1)
– the inverse actions in inverse order. And the number of actions doesn’t really matter In this problem they ask to express the inverse of z. Well, if that’s something we have to do, then let’s do it. For this we should invert the product of all these 4 multipliers
According to the theorem which we have already discussed we should invert these multipliers
and put them in reverse order So, y-cubed gives us y to the negative third
power x-cubed gives us x to the negative third power
then the inverse of y and finally, x to the negative two power gives
us x squared Of course, here we are using the formula that’s
discussed in another video Two operations: raising to the power and inversion
can be done in any order: you can first raise x to the power of n and then invert it or
vice versa: first invert x and then raise to the power
This is precisely how the negative power of an element is defined in group theory
It’s an important formula which derives from the associative law and which we are going
to use actively here We’ve completed the first task. They asked
us to express, so we expressed. What’s next? The second task is to put this expression
in the simplest form. You can see we have FOUR multipliers while each answer contains
only 2, at least they are presented as products of 2 factors
To understand what the further steps are we should set a goal.
So, our goal is to obtain the answer in the form of a product of some power of x and some
power of y. It could be x to some power, y to some power
or y to some power. x to some power – these are not necessarily the same because x, y
and their powers do not necessarily commute So, we should reduce the number of multipliers
from 4 to 2 From this point such problems cause diffuclties.
We have a lot of constraints (namely 3) and we don’t know exactly where to start
The expression x squared in terms of y, some information about the order of y, and some
information about the relation between x and y (called conjugation)… loads of stuff
For example, we can start with x squared and replace it with y cubed
Or we can take y to the negative third which resembles y cubed and express this power in
terms of x Or we can express the inverse of y using this
equality After each such substitution the number of
possible options increases imminently In many tricky mathematical solutions of nontrivial
problems you can observe certain mathematical chaos which somehow suddenly results in the
answer Some mathematicians claim that this is a form
of art. But if we are honest .. They play cunning. Because usually every step is part
of a hidden algorithm supported by one or several notions
So, one aim of this video is not to demonstrate the shortest solution (maybe you can find
a shorter one) but to substantiate every step as part of an algorithm
The first step has already been made – it was application of the formula to invert our
product Our goal helps us to make the second step
We really want to reduce the number of multipliers, so lets’ start, for example with the first
one You might ask “why not the last one ?”, we’ll
get back to this later Using the “orange formula” and this equality
we can invert it and express y to the negative third as x to the negative 2
“This part (x^(-3)y^(-1)x^2 ) remains but x to the negative 2 and x to the negative
third gives x to the negative 5 according to associativity
Note that we have replaced 4 factors with 3 What next? We can continue the idea of substitution. Lets’ consider x squared. You will see why
it doesn’t work and why we didn’t choose it in the beginning.
So we will take the “purple path” and after simplifications we finally obtain
Indeed, we’ve obtained the right form of the expression but you can check that it does
not coincide with any of the answers, at least at first sight
For now we will consider this path a dead end But using the fact that this problem is a test one we can just use backsolving and simply
check every answer Let’s check if option (A) works, then if option
(B) works, (c) as well etc. Write a comment below if you manage to find the answer this
way Lets’ chose another path, a blue one.
Of course we could work with x to the negative 5 but there is a problem:
-5 is an odd number but in the statement we have 2 – an even power of x.
we can raise the equality to the power of n which can be either positive or negative
as a result we can get an even power of x, which couldn’t be equal to (-5)
So, it doesn’t work We are left with the inverse of y but since
direct substitution doesn’t reduce the number of multipliers let’s try a small transformation
Let’s consider this expression. In Abstract Algebra It’s called the conjugation of y
We could try to separate this conjugation structure in our expression if not precisely
then at least something very similar So, what do we need: we want to see the inverse
of y, with the inverse of x to the left and х to the right
To make the equality true we add x to the negative 4 on the left (which together with
the inverse of x makes x to the negative 5) and another x on the right (which together
with the first x makes x squared) And now we finally see some expression which
is similar to the conjugate of y It turns out that conjugation of an element
has one useful property If we raise this to the power of n we can
just raise the central y to this power It means we can obtain the following expression.
By the way, I visualize this property in another video
Since it works for all integers n we can write it for n=-1
So, this could be replaced with y In fact the last equality could be obtained
in a different way We can invert our conjugation structure using
the inversion formula As you see the result is the same
Finally we obtain this expression and we are one step away from the answer
What have we achieved? It’s very similar to this expression we had
before But the one importaint difference is that
now we have even power where we need it From this we derive x to the negative 4
And y to the negative 6 comes from this equality, so it’s also equal to identity
So , the left multiplayer is equal to identity and finally you can see that option (D) is
the right answer If the solution is clear I have a question for you It turns out that the second condition is
not necessary. One of the discussed formulae can help to understand why. Please
write in comments how to do it! Don’t forget to press like or dislike button If you don’t want to miss my next video, please subscribe to my channel And remember that your every comment is extremely appreciated!

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  1. Thank you for the great video and so clear explanation of the problem!

    As concerns the redundancy of the second condition.
    We can derive it from the other two conditions as follows.
    Let's raise the expression y^-1=x^-1 y x to the third power:

    y^-3=(x^-1 y x) (x^-1 y x) (x^-1 y x)=x^-1 y^3 x.

    Now let's remember that x^2=y^3, so:

    x^-2=x^-1 x^2 x = x^2, so x^4=y^6=1.

    Hope to see more videos like this soon ­čÖé

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