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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.

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

2 Answers 2

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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.

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Your statement

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.

The correct statement is

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 $\varphi(H) = G$.

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