$(ab)^2=a^2b^2$ doesn't hold for semigroup.

Question

I am trying to prove that $(ab)^2=a^2b^2$ doesn't hold for a semigroup $S$. It is easy to prove(using inverse property) that the above condition holds for an abelian group. However, for semigroup, it doesn't hold. Is it because that the inverse element doesn't exist in $S$? How to prove formally?

Edit: I found the above question in an undergraduate book. I think that it should have been stated as

Give a counter example to show that $(ab)^2=a^2b^2$ doesn't necessarily hold in a semigroup.

The answer is as stated by Arthur and Thomas in the comments.

Answer 1

In fact, it doesn't even have to be true in a group. If we take $S_3$ as our group and take $a$ as (12) and $b$ as (123), then this doesn't hold.

Answer 2

I am trying to prove that $(ab)^2=a^2b^2$ doesn't hold for a semigroup $S$.

You mean "it doesn't have to hold".

It is easy to prove(using inverse property) that the above condition holds for an abelian group.

Indeed. Moreover you don't have to look outside of groups for a counterexample. In fact:

Lemma. A group $G$ is abelian if and only if for any $a,b\in G$ we have $(ab)^2=a^2b^2$.

Proof. "$\Rightarrow$" obvious.

"$\Leftarrow$" $(ab)^2=a^2b^2$ can be written equivalently as

$$abab=aabb$$

Now just multiply by $a^{-1}$ from the left and $b^{-1}$ from the right (both exist because we are dealing with a group) to obtain $ba=ab$. $\Box$

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So all you have to do is to pick your favourite non-abelian group, since every group is a semigroup. For example the symmetric group $S_n$ for $n\geq 3$ does the job.

Answer 3

As other people have mentioned, all you need is one example.

You could take $S$ to be the free semigroup of words with characters $a$ and $b$ with the operation of concatenation of words.

https://en.wikipedia.org/wiki/Free_monoid