# Solving equations in groups

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• September 24th 2012, 08:24 PM
jzellt
Solving equations in groups
I will denote x-inverse as x' and the identity as e.

I'm asked to solve the two equations below simultaneously:

(1) ax^2 = b and (2) x^3 = e

Here's what i did:

ax^3 = bx (by multiplying (1) on the right by x)

Now, ae = bx (since x^3 = e)

b'a = x (by multiplying b' on the left)

BUT, if we plug in the above x into (1) we get a(b'a)(b'a) = b which isn't true..????

I've tried manipulating both (1) and (2) in many different ways but always come up with x = b'a.

What have I done wrong? Can someone show how to do this correctly please. Thanks a lot!
• September 24th 2012, 09:03 PM
johnsomeone
Re: Solving equations in groups
What does it mean to "solve an equation (or set of equations) for x"? It means, that, assuming the equation(s) hold, you've found out what x must be.
What you've proved is that, **IF** 1 & 2 hold for some x, THEN it must be that x = b'a.
If x = b'a doesn't actually make 1 & 2 hold in your group, then it means that the equations have no solution. It doesn't mean that you made a mistake. You didn't.

Ex: Suppose this was a set of equations in a finite group where 3 does not divide the order of the group, and that a is not equal to b. Could this set of equations have a solution?
No. Since Equation 2 requires that ord(x) divides 3, thus ord(x) is either 3 or 1. But it can't be 3, since that doesn't divide the group order (Lagrange's Thm with <x>).
Thus ord(x) = 1, so x = e. But then ax = b says a(e) = b, so a = b. But that's a contradiction, since a not equal to b by premise.
Thus these equations have no solution in a finite group where 3 does not divide the order of the group, and where a is not equal to b.
Some equations simply don't have solutions - that's true even with the "ordinary" real numbers.
• September 24th 2012, 09:05 PM
FernandoRevilla
Re: Solving equations in groups
Quote:

Originally Posted by jzellt
What have I done wrong? Can someone show how to do this correctly please. Thanks a lot!

You have correctly proved that if $x$ is a solution of the system, then necessarily $x=b^{-1}a.$ But this is not sufficient, for example choose the group $(\mathbb{R}-\{0\},\cdot)$ , $a=1$ and $b=-1.$ Then, $\begin{Bmatrix} ax^2=b\\x^3=e\end{matrix} \Leftrightarrow \begin{Bmatrix} x^2=-1\\x^3=1\end{matrix}$ , and the system has no solution.

Edited: Sorry, I didn't see johnsomeone's post.
• September 25th 2012, 09:49 AM
Deveno
Re: Solving equations in groups
for completeness' sake, let's look at a group G where the two equations DO actually have a solution:

G = S3, a = (1 2), b = (2 3), x = (1 3 2).

then x2 = (1 3 2)(1 3 2) = (1 2 3), and ax2 = (1 2)(1 2 3) = (2 3) = b, and x3 = e.

now let's look at b-1a:

b-1a = (2 3)-1(1 2) = (2 3)(1 2) = (1 3 2) = x.

another example: G = Z6 under addition mod 6:

let a = 5, b = 3, x = 2.

then a + (x + x) = 5 + 2 + 2 = 9 = 3 (mod 6) = b, and x + x + x = 2 + 2 + 2 = 6 = 0 (mod 6).

and -b + a = (-3) + 5 = 2 = x (we get the same answer if we write -b = 3: 3 + 5 = 8 = 2 (mod 6)).

now, for each of these groups, let's COMPUTE:

ab-1ab-1a.

in S3, this is:

(1 2)(2 3)(1 2)(2 3)(1 2) =

(1 2)[(2 3)(1 2)][(2 3)(1 2)] =

(1 2)(1 3 2)(1 3 2) =

(1 2)(1 2 3) = (2 3) = b.

in Z6:

a - b + a - b + a =

5 + 3 + 5 + 3 + 5 =

5 + 8 + 8 = 5 + 2 + 2 (since 8 = 2 (mod 6))

= 5 + 4 = 9 = 3 (mod 6) = b.

note that whether or not G is abelian did not affect our solution (although "the math is easier" in an abelian group).

in other words there is no reason for you to say that ab-1ab-1a ≠ b. it might.
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