Mixing
Is the inverse image of a group also a group for semigroup homomorphisms
$begingroup$
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.
abstract-algebra group-theory semigroups finite-semigroups
edited Jan 20 at 20:33
Shaun
9,366113684
asked Jan 20 at 20:17
StefanHStefanH
8,15652466
$endgroup$
add a comment |
$begingroup$
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.
abstract-algebra group-theory semigroups finite-semigroups
edited Jan 20 at 20:33
Shaun
9,366113684
asked Jan 20 at 20:17
StefanHStefanH
8,15652466
$endgroup$
$begingroup$
Hint: Try the semigroup $S=mathbb{N}$.
$endgroup$
– user10354138
Jan 20 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 at 20:21
add a comment |
$begingroup$
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.
abstract-algebra group-theory semigroups finite-semigroups
edited Jan 20 at 20:33
Shaun
9,366113684
asked Jan 20 at 20:17
StefanHStefanH
8,15652466
$endgroup$
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.
abstract-algebra group-theory semigroups finite-semigroups
abstract-algebra group-theory semigroups finite-semigroups
edited Jan 20 at 20:33
Shaun
9,366113684
asked Jan 20 at 20:17
StefanHStefanH
8,15652466
edited Jan 20 at 20:33
Shaun
9,366113684
asked Jan 20 at 20:17
StefanHStefanH
8,15652466
edited Jan 20 at 20:33
Shaun
9,366113684
edited Jan 20 at 20:33
Shaun
9,366113684
edited Jan 20 at 20:33
Shaun
9,366113684
9,366113684
asked Jan 20 at 20:17
StefanHStefanH
8,15652466
asked Jan 20 at 20:17
StefanHStefanH
8,15652466
asked Jan 20 at 20:17
StefanHStefanH
8,15652466
8,15652466
$begingroup$
Hint: Try the semigroup $S=mathbb{N}$.
$endgroup$
– user10354138
Jan 20 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 at 20:21
add a comment |
$begingroup$
Hint: Try the semigroup $S=mathbb{N}$.
$endgroup$
– user10354138
Jan 20 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 at 20:21
$begingroup$
Hint: Try the semigroup $S=mathbb{N}$.
$endgroup$
– user10354138
Jan 20 at 20:19
$begingroup$
Hint: Try the semigroup $S=mathbb{N}$.
$endgroup$
– user10354138
Jan 20 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 at 20:21
$begingroup$
@user10354138 I added the homomorphism to be surjective, to you still think that $S = mathbb N$ will work?
$endgroup$
– StefanH
Jan 20 at 20:21
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
edited Jan 20 at 20:34
Shaun
9,366113684
answered Jan 20 at 20:29
C MonsourC Monsour
6,2541325
$endgroup$
add a comment |
$begingroup$
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 $varphi(H) = G$.
answered Feb 3 at 13:34
J.-E. PinJ.-E. Pin
18.5k21754
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
edited Jan 20 at 20:34
Shaun
9,366113684
answered Jan 20 at 20:29
C MonsourC Monsour
6,2541325
$endgroup$
add a comment |
$begingroup$
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.
edited Jan 20 at 20:34
Shaun
9,366113684
answered Jan 20 at 20:29
C MonsourC Monsour
6,2541325
$endgroup$
add a comment |
$begingroup$
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.
edited Jan 20 at 20:34
Shaun
9,366113684
answered Jan 20 at 20:29
C MonsourC Monsour
6,2541325
$endgroup$
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.
edited Jan 20 at 20:34
Shaun
9,366113684
answered Jan 20 at 20:29
C MonsourC Monsour
6,2541325
edited Jan 20 at 20:34
Shaun
9,366113684
edited Jan 20 at 20:34
Shaun
9,366113684
edited Jan 20 at 20:34
Shaun
9,366113684
9,366113684
answered Jan 20 at 20:29
C MonsourC Monsour
6,2541325
answered Jan 20 at 20:29
C MonsourC Monsour
6,2541325
answered Jan 20 at 20:29
C MonsourC Monsour
6,2541325
6,2541325
add a comment |
add a comment |
$begingroup$
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 $varphi(H) = G$.
answered Feb 3 at 13:34
J.-E. PinJ.-E. Pin
18.5k21754
$endgroup$
add a comment |
$begingroup$
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 $varphi(H) = G$.
answered Feb 3 at 13:34
J.-E. PinJ.-E. Pin
18.5k21754
$endgroup$
add a comment |
$begingroup$
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 $varphi(H) = G$.
answered Feb 3 at 13:34
J.-E. PinJ.-E. Pin
18.5k21754
$endgroup$
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 $varphi(H) = G$.
answered Feb 3 at 13:34
J.-E. PinJ.-E. Pin
18.5k21754
answered Feb 3 at 13:34
J.-E. PinJ.-E. Pin
18.5k21754
answered Feb 3 at 13:34
J.-E. PinJ.-E. Pin
18.5k21754
answered Feb 3 at 13:34
J.-E. PinJ.-E. Pin
18.5k21754
18.5k21754
add a comment |
add a comment |
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$begingroup$
Hint: Try the semigroup $S=mathbb{N}$.
$endgroup$
– user10354138
Jan 20 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 at 20:21