# Low Level Abstract Algebra Question

1. Oct 27, 2014

### PsychonautQQ

1. The problem statement, all variables and given/known data
Let ab=a and ba=b, show that a^2 = a and that b^2 = b

2. Relevant equations
none

3. The attempt at a solution
Not sure if I did this correct.. but here is what I did.

Given:
ab = a. Multiply both by left hand multiplication by a^-1
a^-1*a*b = 1. where a^-1*a is obviously the identity.
so b = 1.

Given:
ba = b. Multiply both by left hand multiplication by b^-1.
b^-1*b*a = 1
1a=1
a=1
so a =1.

If b = 1 and a=1, then b^2 = 1 and a^2 = 1, so a^2 = a and b^2 = 2. Did I do this correctly?

2. Oct 27, 2014

### RUber

What if you combined them. a=ab and b=ba.
ab=abba, ba=baab.
Can you do anything with those?
(Edit) Or maybe other combinations might work to show the point.

3. Oct 27, 2014

### Dick

What kind of algebraic structure are you dealing with? If it's a group under multiplication then it's as easy as you say. If it's not a group (and I suspect it's not) then you'd better tell us.

4. Oct 27, 2014

### gopher_p

If $a$ and $b$ are elements of a ring, then there is no guarantee they have multiplicative inverses. So you can't necessarily do what you did. For example, take $R$ to be the ring of $2\times2$ matrices with $$a=\begin{pmatrix}1&1\\0&0\end{pmatrix}\ \text{ and }\ b=\begin{pmatrix}0&0\\1&1\end{pmatrix}$$ You get $ab=a$ and $ba=b$, but neither matrix is invertible, and neither is the identity.

I recommend you start with $a^2=abab$ and $b^2=baba$ and see if you can't use $ab=a$ and $ba=b$ to whittle down the right-hand sides of those equalities until you're left with what you need.