Prove or disprove that if a and b have inverses with respect to ∗, then so does a∗b (where * is an associative binary operation with an identity e)?

Question

This is part 2 of a question. Part 1 I was proving that the inverse of an element $a$ is unique with respect to $\cdot$ and I think I solved that one.

Part 3 of the question is the opposite of part 2: Prove or disprove that if $a\cdot b$ has an inverse with respect to $\cdot$, then so do $a$ and $b$.

I am a bit unsure of where to begin with this, as I'm pretty new to binary operations. I've started by just making a list of what I know:

$a\cdot a^{-1} = e$

$b\cdot b^{-1} = e$

$a\cdot e = a$

$b\cdot e = b$

where $a^{-1}$ is the inverse of $a$, etc. I don't necessarily need the whole problem solved, but could someone point me in the right direction? Much appreciated

Answer 1

It's a standard trick. I'm not sure how to hint without immediately given it away but.....

You want $(ab)*K = 1$

You could rewrite that as $a(bK) = 1$ which means $bK = a^{-1}$ (because $a$ has an inverse).

You can also do the operation on both sides and get something $(ab)*K =1$ so $m*(ab)*K = (ma)*(bK) = m*1 = m$.

Can you brainstorm anything?

Hint: Two. What if the was a system you knew.

If $4\cdot \frac 14 = 1$ and

$7\cdot \frac 17 = 1$ and $(4\cdot 7)\cdot x = 1$ how might you go about solving for $x$. (Try to pretend you don't know that $4\cdot 7 = 28$)

Anyway

$ab*x = 1\implies a^{-1}*abx = a^{-1}*1\implies e*bx = bx = a^{-1}\implies b^{-1}bx = b^{-1}a^{-1}\implies e*x = x = b^{-1}a^{-1}$ and ... that's it.

Verify

Do $(ab)*(b^{-1}a^{-1})=e$? Well, $(ab)*(b^{-1}a^{-1}) = a*(bb^{-1})*a^{-1} = a*e*a^{-1}=aa^{-1}= e$. That's it.

It's just unpeeling.