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The title has it all:

Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups?

If yes, then $S$ is a subdirect product of subdirectly irreducible groups, and hence we would get a positive answer to a related question I posted yesterday (here), where Keith Kearnes has shown (here) that the answer is indeed yes in the commutative setting (for commutative semigroups, the embeddability condition is equivalent to cancellativity).

To begin, we have an embedding $q$ of $S$ into a group $G$. By Birkhoff's subdirect representation theorem (and the fact that every homomorphic image of a group within the category of magmas is a group), there is a subdirect representation $p$ of $G$ into a product $\prod_{i \in I} G_i$ of subdirectly irreducible groups. If only we could guarantee that $\pi_j \circ p \circ q[S]$ is a group for each $j \in I$, where $\pi_j$ is the canonical projection $\prod_{i \in I} G_i \to G_j$, then we would be done. This is the case in the aforementioned answer by Keith Kearnes, the key point being that, in the commutative setting, every subdirectly irreducible group embeds into a Prüfer group (and each non-empty subsemigroup of a Prüfer group is still a group).

The title has it all:

Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups?

If yes, then $S$ is a subdirect product of subdirectly irreducible groups, and hence we would obtain a positive (albeit partial) answer to a related question I posted yesterday (here), where Keith Kearnes has shown (here) that the answer is indeed yes in the commutative setting (for commutative semigroups, the embeddability condition is equivalent to cancellativity).

To begin, we have an embedding $q$ of $S$ into a group $G$. By Birkhoff's subdirect representation theorem (and the fact that every homomorphic image of a group within the category of magmas is a group), there is a subdirect representation $p$ of $G$ into the direct product of a family $(G_i)_{i \in I}$ of subdirectly irreducible groups. If only we could guarantee that $\pi_j \circ p \circ q[S]$ is a group for each $j \in I$, where $\pi_j$ is the canonical projection $\prod_{i \in I} G_i \to G_j$, then we would be done. This is the case in the aforementioned answer by Keith Kearnes, the key point being that, in the commutative setting, every subdirectly irreducible group embeds into a Prüfer group (and each non-empty subsemigroup of a Prüfer group is still a group).

The title has it all:

Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups?

If yes, then $S$ is a subdirect product of subdirectly irreducible groups, and hence we would get a positive answer to a related question I posted yesterday (here), where Keith Kearnes has shown (here) that the answer is indeed yes in the commutative setting (for commutative semigroups, the embeddability condition is equivalent to cancellativity).

To begin, we have an embedding $q$ of $S$ into a group $G$. By Birkhoff's subdirect representation theorem (and the fact that every homomorphic image of a group within the category of magmas is a group), there is a subdirect representation $p$ of $G$ into a product $\prod_{i \in I} G_i$ of subdirectly irreducible groups. If only we could guarantee that $\pi_j \circ p \circ q[S]$ is a group for each $j \in I$, where $\pi_j$ is the canonical projection $\prod_{i \in I} G_i \to G_j$, then we would be done. This is the case in the aforementioned answer by Keith Kearnes, the key point being that, in the commutative setting, one can take the groups $G_i$ to be Prüfer groups (and every non-empty subsemigroup of a Prüfer group is still a group).

The title has it all:

Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups?

If yes, then $S$ is a subdirect product of subdirectly irreducible groups, and hence we would get a positive answer to a related question I posted yesterday (here), where Keith Kearnes has shown (here) that the answer is indeed yes in the commutative setting (for commutative semigroups, the embeddability condition is equivalent to cancellativity).

To begin, we have an embedding $q$ of $S$ into a group $G$. By Birkhoff's subdirect representation theorem (and the fact that every homomorphic image of a group within the category of magmas is a group), there is a subdirect representation $p$ of $G$ into a product $\prod_{i \in I} G_i$ of subdirectly irreducible groups. If only we could guarantee that $\pi_j \circ p \circ q[S]$ is a group for each $j \in I$, where $\pi_j$ is the canonical projection $\prod_{i \in I} G_i \to G_j$, then we would be done. This is the case in the aforementioned answer by Keith Kearnes, the key point being that, in the commutative setting, every subdirectly irreducible group embeds into a Prüfer group (and each non-empty subsemigroup of a Prüfer group is still a group).

The title has it all:

Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups?

If yes, then $S$ is a subdirect product of subdirectly irreducible groups, and hence we would get a positive answer to a related question I posted yesterday (here), where Keith Kearnes has shown (here) that the answer is indeed yes in the commutative setting (for commutative semigroups, the embeddability condition is equivalent to cancellativity).

To begin, we have an embedding $q$ of $S$ into a group $G$. By Birkhoff's subdirect representation theorem (and the fact that every homomorphic image of a group within the category of magmas is a group), there is a subdirect representation $p$ of $G$ into a product $\prod_{i \in I} G_i$ of subdirectly irreducible groups. If only we could guarantee that $\pi_i \circ p \circ q[S]$ is a group, we would be done. This is the case in the aforementioned answer by Keith Kearnes, the key point being that, in the commutative setting, one can take the groups $G_i$ to be Prüfer groups (and every non-empty subsemigroup of a Prüfer group is still a group).

The title has it all:

Q. If a semigroup $S$ embeds into a group, then is $S$ (isomorphic to) a subdirect product of groups?

If yes, then $S$ is a subdirect product of subdirectly irreducible groups, and hence we would get a positive answer to a related question I posted yesterday (here), where Keith Kearnes has shown (here) that the answer is indeed yes in the commutative setting (for commutative semigroups, the embeddability condition is equivalent to cancellativity).

To begin, we have an embedding $q$ of $S$ into a group $G$. By Birkhoff's subdirect representation theorem (and the fact that every homomorphic image of a group within the category of magmas is a group), there is a subdirect representation $p$ of $G$ into a product $\prod_{i \in I} G_i$ of subdirectly irreducible groups. If only we could guarantee that $\pi_j \circ p \circ q[S]$ is a group for each $j \in I$, where $\pi_j$ is the canonical projection $\prod_{i \in I} G_i \to G_j$, then we would be done. This is the case in the aforementioned answer by Keith Kearnes, the key point being that, in the commutative setting, one can take the groups $G_i$ to be Prüfer groups (and every non-empty subsemigroup of a Prüfer group is still a group).