I came across the following problem in category theory I'm working on:
Problem: Let $\mathbb{S}_3$ by the symmetric group of order $3$. Then:
a) Describe Hom$_{Groups}(\mathbb{S}_3, \mathbb{S}_3)$.
b) Assume that the category $Groups$ is pre-additive. Conclude that Hom$_{Groups}(\mathbb{S}_3, \mathbb{S}_3)$ must be a domain (i.e. $ab = 0$ implies $a = 0$ or $b = 0$), although not necessarily commutative.
c) Wedderburn's little theorem states that every finite domain is a field. Apply this to b) and conclude that $Groups$ is not pre-additive.
Attempt: For a), I found out that Hom$_{Groups}(\mathbb{S}_3, \mathbb{S}_3)$ contains $10$ arrows. There are $6$ automorphisms, the unique morphism sending every element to the identity, and then the $3$ projections onto the $3$ subgroups of order $2$.
However, what I do not understand is, why not the projection onto the alternating subgroup of order $3$? And how can we see that the above projections are morphisms?
b) For this part, I'm assuming the operation is $\circ$ for which we have to prove the statement, since if $Groups$ is assumed pre-additive, then every $Hom(A,B)$ is an abelian group with respect to $+$.
But I'm not sure how to show that it is a domain. Say $f, g \in Hom_{Groups}(\mathbb{S}_3, \mathbb{S}_3) $ and let $$ f \circ g = 0. $$ I'm confused here. Is this $0$ the morphism which sends every element in $\mathbb{S}_3$ to the identity? If $f$ is an automorphism, then we can compose with $f^{-1}$ and show that $g = 0$. But how to prove the statement for the other cases?
Help/clarification is appreciated.
I do not know exactly what you mean by "projection" here. But bear in mind that the kernel of a homomorphism is a normal subgroup and $S_3$ has no normal subgroups of order two.
You only need to prove this in the case that $f$ and $g$ are nontrivial homomorphisms with nontrivial kernel. Then they are of the form $f(\sigma) = \tau_f^{\mathrm{sgn}(\sigma)}$ for some fixed transposition $\tau_f$, so you should be able to show that $f \circ g = f$.