First Isomorphism Theorem for semigroups

Given a semigroup S and a congruence \sigma on S, show that the mapping \pi_\sigma : S \rightarrow S/\sigma given by s \mapsto [s]_\sigma is a surjective semigroup homomorphism.

Let \varphi : S \rightarrow T be a homomorphism of semigroups. Define a relation \mathsf{ker}\ \varphi by (x,y) \in \mathsf{ker}\ \varphi if and only if \varphi(x) = \varphi(y). Show that \mathsf{ker}\ \varphi is a congruence on S. Show further that S/\mathsf{ker}\ \varphi \cong \mathsf{im}\ \varphi.

Recall that S/\sigma is a semigroup under the operator [x]_\sigma \star [y]_\sigma = [xy]_\sigma, which is well-defined (i.e. does not depend on the representatives chosen) since \sigma is a congruence. Now \pi_\sigma(st) = [st]_\sigma = [s]_\sigma[t]_\sigma = \pi_\sigma(s)\pi_\sigma(t), so that \pi_\sigma is a semigroup homomorphism. Moreover, we have [s]_\sigma = \pi_\sigma(s) for all s \in S, so that \pi_\sigma is surjective.

Now let \varphi and \mathsf{ker}\ \varphi be as described above. Certainly \mathsf{ker}\ \varphi is an equivalence. Now if (x,y) \in \mathsf{ker}\ \varphi and (z,w) \in \mathsf{ker}\ \varphi, then \varphi(x) = \varphi(y) and \varphi(z) = \varphi(w), so that \varphi(xz) = \varphi(x) \varphi(z) = \varphi(y) \varphi(w) = \varphi(yw). Hence (xz,yw) \in \mathsf{ker}\ \varphi, and so \mathsf{ker}\ \varphi is a congruence on S.

Now define \psi \subseteq S/\mathsf{ker}\ \varphi \times T by \psi = \{ ([x]_\mathsf{ker}\ \varphi, \varphi(x)) \ |\ x \in S \}. Certianly \psi is total, and we claim it is well defined. Indeed, if we have x,y \in S such that (x,y) \in \mathsf{ker}\ \varphi, then \varphi(x) = \varphi(y). So in fact \psi : S/\mathsf{ker}\ \varphi \rightarrow T is a function. Since \psi([x][y]) = \psi([xy]) = \varphi(xy) = \varphi(x)\varphi(y) = \psi([x])\psi([y]), \psi is a semigroup homomorphism. If \psi([x]) = \psi([y]), then we have \varphi(x) = \varphi(y), so that [x] = [y]. Hence psi is injective. Finally, if t \in \mathsf{im}\ \varphi, with t = \varphi(s), then t = \psi([s]). So \psi is onto the subset \mathsf{im}\ \varphi \subseteq T, and we have S/\mathsf{ker}\ \varphi \cong \mathsf{im}\ \varphi.

Post a comment or leave a trackback: Trackback URL.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: