Consider a Semigroup $(M, \ast)$ with a neutral element $e$. Now I have to prove that all left invertible elements of $(M, \ast)$ form a sub-semigroup.
A left invertible element is an element whose left inverse exists; i.e., $a$ is left invertible if there exists an element $a'$ such that $$a' \ast a=e.$$ Let us consider a set $A\subset M$ containing all left invertible elements. Let $a,b \in A;$ then there exist $a',b' \in M$ such that $$a' \ast a=e \ \ \text{and } \ b' \ast b=e$$ Now I have to prove that $a \ast b$ is also left invertible; i.e., there exists some $c \in M$ such that $$c \ast (a \ast b) = e.$$ Now I don't know how to prove it. Maybe my statement is wrong -- i.e., set of invertible is sub-semigroup -- or maybe there is another way to prove it.
In a semigroup with neutral element $e$, the left invertible elements form a sub-semigroup,
because if elements $a$ and $b$ have left inverses $a'$ and $b'$, respectively, in the semigroup,
then $a'*a$ and $b'*b=e$,
so $(b'*a')*(a*b)=b'*(a'*a)*b=b'*e*b=b'*b=e;$
i.e., $b'*a'$ is a left inverse of $a*b.$
Some refer to the opposite order of the inverses of $a$ and $b$ in the inverse of $a*b$ as the
Socks-Shoes property,
because you take off your socks and shoes in the opposite order you put them on.