When is a group homomorphism “extendable”?
$begingroup$
$newcommand{iotaLift}{iota!uparrow!(S, H)}$
Say we have finite groups $Sleq G$ and $H$. If we have a homomorphism $phicolon Sto H$, it does not always have to be the case that we find a homomorphism from $Gto H$ that agrees with $phi$ on $S$. For instance, the identity homomorphism $G geq Sto S$ cannot be extended if (surjective) homomorphisms $Gto S$ don't even exist at all. This happens for instance if there is no $Nlhd G$ such that $S simeq G/N$.
Definition Let $S, G, H$ be finite groups, $iotacolon Sto G$ a monomorphism (that is, an injective group homomorphism). Define $$
iotaLift := {phicolon Sto H mid exists overlinephicolon Gto H: overlinephicirc iota = phi}
$$
to be the set of all such extendable homomorphisms.
- For which $H$ is $iotaLift$ trivial (contains only the zero homomorphism)?
- For which $H$ is $iotaLift = operatorname{Hom}_{underline{operatorname{Grp}}}(S, H)$?
- Does this set have any obvious algebraic structure, e.g. a partial order, or an algebraic operation that constructs new extendable functions from old ones?
- Does the complement of this set?
(I know that the last two points are rather open-ended, and hope this is still acceptable for M.SE.)
Examples
- Assume the injection $iota$ “splits”, i.e., admits an inverse $sigma: sigmacirciota = operatorname{id}_S$. Then every $phi$ can be extended to $phicircsigma$, which indeed satisfies $phicircsigmacirciota=phi$. So $iotaLift = operatorname{Hom}_{underline{operatorname{Grp}}}(S, H)$.
- Similar to the case mentioned above, Let $G$ be simple, $S$ any subgroup, and $H$ a group such that $G$ does not embed into $H$. Therefore, the only homomorphisms $Gto H$ must be trivial, and so is every restriction onto $S$. Ergo, $iotaLift={(smapsto 1)}$.
Consider the (only nontrivial) embedding $C_2to C_4$ with arbitrary $H$. Since morphisms $C_nto H$ correspond with elements $hin H$ of order dividing $n$, extendable morphisms $C_2to H$ correspond to elements who have a square root (including the identity).
If we have an extendable function, we can compose it with any element of $operatorname{Aut}(H)$ that is not trivial on its image to obtain a new function, whose extandability is easily checked.
group-theory soft-question finite-groups definition
asked Jan 15 at 22:32
LukeLuke
1,2391026
$endgroup$
|
show 1 more comment
$begingroup$
$newcommand{iotaLift}{iota!uparrow!(S, H)}$
Say we have finite groups $Sleq G$ and $H$. If we have a homomorphism $phicolon Sto H$, it does not always have to be the case that we find a homomorphism from $Gto H$ that agrees with $phi$ on $S$. For instance, the identity homomorphism $G geq Sto S$ cannot be extended if (surjective) homomorphisms $Gto S$ don't even exist at all. This happens for instance if there is no $Nlhd G$ such that $S simeq G/N$.
Definition Let $S, G, H$ be finite groups, $iotacolon Sto G$ a monomorphism (that is, an injective group homomorphism). Define $$
iotaLift := {phicolon Sto H mid exists overlinephicolon Gto H: overlinephicirc iota = phi}
$$
to be the set of all such extendable homomorphisms.
- For which $H$ is $iotaLift$ trivial (contains only the zero homomorphism)?
- For which $H$ is $iotaLift = operatorname{Hom}_{underline{operatorname{Grp}}}(S, H)$?
- Does this set have any obvious algebraic structure, e.g. a partial order, or an algebraic operation that constructs new extendable functions from old ones?
- Does the complement of this set?
(I know that the last two points are rather open-ended, and hope this is still acceptable for M.SE.)
Examples
- Assume the injection $iota$ “splits”, i.e., admits an inverse $sigma: sigmacirciota = operatorname{id}_S$. Then every $phi$ can be extended to $phicircsigma$, which indeed satisfies $phicircsigmacirciota=phi$. So $iotaLift = operatorname{Hom}_{underline{operatorname{Grp}}}(S, H)$.
- Similar to the case mentioned above, Let $G$ be simple, $S$ any subgroup, and $H$ a group such that $G$ does not embed into $H$. Therefore, the only homomorphisms $Gto H$ must be trivial, and so is every restriction onto $S$. Ergo, $iotaLift={(smapsto 1)}$.
Consider the (only nontrivial) embedding $C_2to C_4$ with arbitrary $H$. Since morphisms $C_nto H$ correspond with elements $hin H$ of order dividing $n$, extendable morphisms $C_2to H$ correspond to elements who have a square root (including the identity).
If we have an extendable function, we can compose it with any element of $operatorname{Aut}(H)$ that is not trivial on its image to obtain a new function, whose extandability is easily checked.
group-theory soft-question finite-groups definition
asked Jan 15 at 22:32
LukeLuke
1,2391026
$endgroup$
1
$begingroup$
The last sentence in your first paragraph is wrong. Take $S=Z_2$ as subgroup of the symmetric group $S_3$ in $G = S_3times Z_2$.
$endgroup$
– j.p.
Jan 16 at 6:58
$begingroup$
@j.p. that concerns only the “any only if” part, correct?
$endgroup$
– Luke
Jan 17 at 21:00
$begingroup$
Yes, as the kernel $N$ of the extended identity (assuming such an extension exists) has the property $Ssimeq G/N$.
$endgroup$
– j.p.
Jan 18 at 8:30
$begingroup$
@j.p. thanks, I corrected it.
$endgroup$
– Luke
Jan 18 at 14:44
$begingroup$
Wait, so $iota$ can be any monomorphism? Not just the inclusion map? And is $iota$ fixed for all four of those questions?
$endgroup$
– Cardioid_Ass_22
Jan 31 at 17:11
|
show 1 more comment
$begingroup$
$newcommand{iotaLift}{iota!uparrow!(S, H)}$
Say we have finite groups $Sleq G$ and $H$. If we have a homomorphism $phicolon Sto H$, it does not always have to be the case that we find a homomorphism from $Gto H$ that agrees with $phi$ on $S$. For instance, the identity homomorphism $G geq Sto S$ cannot be extended if (surjective) homomorphisms $Gto S$ don't even exist at all. This happens for instance if there is no $Nlhd G$ such that $S simeq G/N$.
Definition Let $S, G, H$ be finite groups, $iotacolon Sto G$ a monomorphism (that is, an injective group homomorphism). Define $$
iotaLift := {phicolon Sto H mid exists overlinephicolon Gto H: overlinephicirc iota = phi}
$$
to be the set of all such extendable homomorphisms.
- For which $H$ is $iotaLift$ trivial (contains only the zero homomorphism)?
- For which $H$ is $iotaLift = operatorname{Hom}_{underline{operatorname{Grp}}}(S, H)$?
- Does this set have any obvious algebraic structure, e.g. a partial order, or an algebraic operation that constructs new extendable functions from old ones?
- Does the complement of this set?
(I know that the last two points are rather open-ended, and hope this is still acceptable for M.SE.)
Examples
- Assume the injection $iota$ “splits”, i.e., admits an inverse $sigma: sigmacirciota = operatorname{id}_S$. Then every $phi$ can be extended to $phicircsigma$, which indeed satisfies $phicircsigmacirciota=phi$. So $iotaLift = operatorname{Hom}_{underline{operatorname{Grp}}}(S, H)$.
- Similar to the case mentioned above, Let $G$ be simple, $S$ any subgroup, and $H$ a group such that $G$ does not embed into $H$. Therefore, the only homomorphisms $Gto H$ must be trivial, and so is every restriction onto $S$. Ergo, $iotaLift={(smapsto 1)}$.
Consider the (only nontrivial) embedding $C_2to C_4$ with arbitrary $H$. Since morphisms $C_nto H$ correspond with elements $hin H$ of order dividing $n$, extendable morphisms $C_2to H$ correspond to elements who have a square root (including the identity).
If we have an extendable function, we can compose it with any element of $operatorname{Aut}(H)$ that is not trivial on its image to obtain a new function, whose extandability is easily checked.
group-theory soft-question finite-groups definition
asked Jan 15 at 22:32
LukeLuke
1,2391026
$endgroup$
$newcommand{iotaLift}{iota!uparrow!(S, H)}$
Say we have finite groups $Sleq G$ and $H$. If we have a homomorphism $phicolon Sto H$, it does not always have to be the case that we find a homomorphism from $Gto H$ that agrees with $phi$ on $S$. For instance, the identity homomorphism $G geq Sto S$ cannot be extended if (surjective) homomorphisms $Gto S$ don't even exist at all. This happens for instance if there is no $Nlhd G$ such that $S simeq G/N$.
Definition Let $S, G, H$ be finite groups, $iotacolon Sto G$ a monomorphism (that is, an injective group homomorphism). Define $$
iotaLift := {phicolon Sto H mid exists overlinephicolon Gto H: overlinephicirc iota = phi}
$$
to be the set of all such extendable homomorphisms.
- For which $H$ is $iotaLift$ trivial (contains only the zero homomorphism)?
- For which $H$ is $iotaLift = operatorname{Hom}_{underline{operatorname{Grp}}}(S, H)$?
- Does this set have any obvious algebraic structure, e.g. a partial order, or an algebraic operation that constructs new extendable functions from old ones?
- Does the complement of this set?
(I know that the last two points are rather open-ended, and hope this is still acceptable for M.SE.)
Examples
- Assume the injection $iota$ “splits”, i.e., admits an inverse $sigma: sigmacirciota = operatorname{id}_S$. Then every $phi$ can be extended to $phicircsigma$, which indeed satisfies $phicircsigmacirciota=phi$. So $iotaLift = operatorname{Hom}_{underline{operatorname{Grp}}}(S, H)$.
- Similar to the case mentioned above, Let $G$ be simple, $S$ any subgroup, and $H$ a group such that $G$ does not embed into $H$. Therefore, the only homomorphisms $Gto H$ must be trivial, and so is every restriction onto $S$. Ergo, $iotaLift={(smapsto 1)}$.
Consider the (only nontrivial) embedding $C_2to C_4$ with arbitrary $H$. Since morphisms $C_nto H$ correspond with elements $hin H$ of order dividing $n$, extendable morphisms $C_2to H$ correspond to elements who have a square root (including the identity).
If we have an extendable function, we can compose it with any element of $operatorname{Aut}(H)$ that is not trivial on its image to obtain a new function, whose extandability is easily checked.
group-theory soft-question finite-groups definition
group-theory soft-question finite-groups definition
asked Jan 15 at 22:32
LukeLuke
1,2391026
asked Jan 15 at 22:32
LukeLuke
1,2391026
edited Jan 18 at 14:43
Luke
asked Jan 15 at 22:32
LukeLuke
1,2391026
asked Jan 15 at 22:32
LukeLuke
1,2391026
asked Jan 15 at 22:32
LukeLuke
1,2391026
1,2391026
1
$begingroup$
The last sentence in your first paragraph is wrong. Take $S=Z_2$ as subgroup of the symmetric group $S_3$ in $G = S_3times Z_2$.
$endgroup$
– j.p.
Jan 16 at 6:58
$begingroup$
@j.p. that concerns only the “any only if” part, correct?
$endgroup$
– Luke
Jan 17 at 21:00
$begingroup$
Yes, as the kernel $N$ of the extended identity (assuming such an extension exists) has the property $Ssimeq G/N$.
$endgroup$
– j.p.
Jan 18 at 8:30
$begingroup$
@j.p. thanks, I corrected it.
$endgroup$
– Luke
Jan 18 at 14:44
$begingroup$
Wait, so $iota$ can be any monomorphism? Not just the inclusion map? And is $iota$ fixed for all four of those questions?
$endgroup$
– Cardioid_Ass_22
Jan 31 at 17:11
|
show 1 more comment
1
$begingroup$
The last sentence in your first paragraph is wrong. Take $S=Z_2$ as subgroup of the symmetric group $S_3$ in $G = S_3times Z_2$.
$endgroup$
– j.p.
Jan 16 at 6:58
$begingroup$
@j.p. that concerns only the “any only if” part, correct?
$endgroup$
– Luke
Jan 17 at 21:00
$begingroup$
Yes, as the kernel $N$ of the extended identity (assuming such an extension exists) has the property $Ssimeq G/N$.
$endgroup$
– j.p.
Jan 18 at 8:30
$begingroup$
@j.p. thanks, I corrected it.
$endgroup$
– Luke
Jan 18 at 14:44
$begingroup$
Wait, so $iota$ can be any monomorphism? Not just the inclusion map? And is $iota$ fixed for all four of those questions?
$endgroup$
– Cardioid_Ass_22
Jan 31 at 17:11
1
1
$begingroup$
The last sentence in your first paragraph is wrong. Take $S=Z_2$ as subgroup of the symmetric group $S_3$ in $G = S_3times Z_2$.
$endgroup$
– j.p.
Jan 16 at 6:58
$begingroup$
The last sentence in your first paragraph is wrong. Take $S=Z_2$ as subgroup of the symmetric group $S_3$ in $G = S_3times Z_2$.
$endgroup$
– j.p.
Jan 16 at 6:58
$begingroup$
@j.p. that concerns only the “any only if” part, correct?
$endgroup$
– Luke
Jan 17 at 21:00
$begingroup$
@j.p. that concerns only the “any only if” part, correct?
$endgroup$
– Luke
Jan 17 at 21:00
$begingroup$
Yes, as the kernel $N$ of the extended identity (assuming such an extension exists) has the property $Ssimeq G/N$.
$endgroup$
– j.p.
Jan 18 at 8:30
$begingroup$
Yes, as the kernel $N$ of the extended identity (assuming such an extension exists) has the property $Ssimeq G/N$.
$endgroup$
– j.p.
Jan 18 at 8:30
$begingroup$
@j.p. thanks, I corrected it.
$endgroup$
– Luke
Jan 18 at 14:44
$begingroup$
@j.p. thanks, I corrected it.
$endgroup$
– Luke
Jan 18 at 14:44
$begingroup$
Wait, so $iota$ can be any monomorphism? Not just the inclusion map? And is $iota$ fixed for all four of those questions?
$endgroup$
– Cardioid_Ass_22
Jan 31 at 17:11
$begingroup$
Wait, so $iota$ can be any monomorphism? Not just the inclusion map? And is $iota$ fixed for all four of those questions?
$endgroup$
– Cardioid_Ass_22
Jan 31 at 17:11
|
show 1 more comment
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$begingroup$
The last sentence in your first paragraph is wrong. Take $S=Z_2$ as subgroup of the symmetric group $S_3$ in $G = S_3times Z_2$.
$endgroup$
– j.p.
Jan 16 at 6:58
$begingroup$
@j.p. that concerns only the “any only if” part, correct?
$endgroup$
– Luke
Jan 17 at 21:00
$begingroup$
Yes, as the kernel $N$ of the extended identity (assuming such an extension exists) has the property $Ssimeq G/N$.
$endgroup$
– j.p.
Jan 18 at 8:30
$begingroup$
@j.p. thanks, I corrected it.
$endgroup$
– Luke
Jan 18 at 14:44
$begingroup$
Wait, so $iota$ can be any monomorphism? Not just the inclusion map? And is $iota$ fixed for all four of those questions?
$endgroup$
– Cardioid_Ass_22
Jan 31 at 17:11