0
$\begingroup$

I came across this theorem when studying thess lecture notes


Theorem: Prove that

  1. The endomorphisms of a semigroup S form a monoid.

  2. The automorphisms of a semigroup S form a group.


I do not have even a small idea of how to construct the proof. Can somebody help?

Thanks.

$\endgroup$
2
  • 2
    $\begingroup$ The endomorphisms/automorphisms of any object in any category whatsoever always form a monoid/group. That you're working in the category of semigroups is irrelevant. $\endgroup$
    – Zhen Lin
    Commented Mar 17, 2012 at 17:58
  • $\begingroup$ @ZhenLin:Please can you provide a reference to justify your statement? $\endgroup$ Commented Mar 18, 2012 at 3:31

2 Answers 2

3
$\begingroup$

Lemma 1. The composition of endomorphisms of a semigroup is an endomorphism; hence the set of endomorphisms is a semigroup.

Proof. If $f,g\colon S\to S$ are endomorphisms, then plainly $g\circ f$ is a function from $S$ to $S$. To show it is an endomorphisms, we have $$g\circ f(st) = g(f(st)) = g(f(s)f(t)) = g(f(s))g(f(t)) = (g\circ f)(s)(g\circ f)(t),$$ since each of $f$ and $g$ are homomorphisms. Therefore, since composition of functions is associative, the set of endomorphisms is certainly a semigroup. $\Box$

Lemma 2. The identity map is an endomorphism (in fact, an automorphism) that acts as the identity of the semigroup of endomorphisms.

Proof. $\mathrm{id}_S\colon S\to S$ is an automorphism; for every function (not just any endomorphism) $f\colon S\to S$, we have $f\circ\mathrm{id}_S = \mathrm{id}_S\circ f = f$. $\Box$

Corollary 3. The set of endomorphisms of a semigroup is a monoid under function composition.

Proof. By Lemma 1, it is a semigroup. By Lemma 2, it contains an identity. Hence, it is a monoid. $\Box$

Lemma 4. The set of automorphisms of a semigroup is a submonoid of the monoid of endomorphisms.

Proof. You need to show that the composition of two automorphisms is an automorphism. Since an automorphism is a bijective endomorphism, you only need to observe that the composition of two bijective functions is bijective, and the composition of two endomorphisms is an endomorphism. This shows the automorphisms form a subsemigroup. Since the identity is an automorphism, it is in fact a submonoid. $\Box$

Corollary 5. The monoid of automorphisms of a semigroup is a group.

Proof. We only need to show that an automorphism has an inverse that is an automorphism. An automorphism is a bijective homomorphism; it certainly has a (set-theoretic) inverse. It is straightforward to show this set-theoretic inverse is also a homomorphism (if you haven't done so already). $\Box$

$\endgroup$
0
1
$\begingroup$

Ignoring the semigroup structure on $S$ for a moment, consider the set $F(S) := \{ f: S\to S\}$ of functions from the set $S$ to itself. Composing two such functions yields another. Can you prove that the operation of composition makes $F(S)$ into a monoid? Now think about the subsets of $F(S)$ defined by your (1) and (2). What changes?

$\endgroup$
1
  • 1
    $\begingroup$ You still have to consider the semigroup structure of $S$, since an endomorphism is a homomorphism. So you'd have to establish that the composition of two such functions is not only another function, but also a homomorphism -- use the definition of homomorphism for that. Also, a monoid has a special element and you'd need to establish that it is there too -- again check the definition of monoid. Likewise, for automorphisms you'd need to prove that the extra structure of a group also satisfies the criteria for homomorphisms. (Assuming this is homework, we are only giving hints.) $\endgroup$ Commented Mar 17, 2012 at 18:35

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .