# Is the inverse image of a group also a group for semigroup homomorphisms

If $$\varphi : S \to T$$ is a surjective semigroup homomorphism between semigroups and $$G \subseteq T$$ is a group, then is $$\varphi^{-1}(G)$$ also a group?

I know that this result holds if $$S$$ and $$T$$ are finite, as then I can find an idempotent element in $$\varphi^{-1}(1_G)$$ and everything follows from that. But I cannot prove it for infinite semigroups, nor can I find a counter-example.

• Hint: Try the semigroup $S=\mathbb{N}$. Jan 20, 2019 at 20:19
• @user10354138 I added the homomorphism to be surjective, to you still think that $S = \mathbb N$ will work? Jan 20, 2019 at 20:21

Project $$\Bbb{N}$$ on $$\{0,1\}=\Bbb{Z}/(2)$$, sending evens to $$0$$ and odds to $$1$$. So, no, the inverse image of a group under a semigroup homomorphism need not be a group, even if it is a surjective monoid homomorphism.
I know that this result holds if $$S$$ and $$T$$ are finite, as then I can find an idempotent element in $$\varphi^{-1}(1_G)$$ and everything follows from that.
is unfortunately wrong. Let $$T$$ be the trivial group $$1$$, let $$S$$ be any finite non-group semigroup and let $$\varphi:S \to T$$ be the trivial map. Then $$\varphi^{-1}(1) = S$$ and thus $$\varphi^{-1}(1)$$ is not a group.
Let $$S$$ be finite semigroup and let $$\varphi : S \to T$$ be a surjective semigroup morphism. Then for every group $$G \subseteq T$$, there exists a group $$H \subseteq S$$ such that $$\varphi(H) = G$$.