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Is a function that preserves normal subgroups a group morphism?
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Proposition 2.10.4 in Michael Artin's Algebra says that the inverse image of a normal subgroup under a group morphism is a normal subgroup (of the morphism's domain, of course)
Motivated by analogy with topology, my question is then if the converse implication holds: given a function between groups such that inverse images of normal subgroups are always normal, is such a function in fact a group morphism?
abstract-algebra group-theory
asked 1 hour ago
alkchf
40836
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up vote
1
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Proposition 2.10.4 in Michael Artin's Algebra says that the inverse image of a normal subgroup under a group morphism is a normal subgroup (of the morphism's domain, of course)
Motivated by analogy with topology, my question is then if the converse implication holds: given a function between groups such that inverse images of normal subgroups are always normal, is such a function in fact a group morphism?
abstract-algebra group-theory
asked 1 hour ago
alkchf
40836
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Proposition 2.10.4 in Michael Artin's Algebra says that the inverse image of a normal subgroup under a group morphism is a normal subgroup (of the morphism's domain, of course)
Motivated by analogy with topology, my question is then if the converse implication holds: given a function between groups such that inverse images of normal subgroups are always normal, is such a function in fact a group morphism?
abstract-algebra group-theory
asked 1 hour ago
alkchf
40836
Proposition 2.10.4 in Michael Artin's Algebra says that the inverse image of a normal subgroup under a group morphism is a normal subgroup (of the morphism's domain, of course)
Motivated by analogy with topology, my question is then if the converse implication holds: given a function between groups such that inverse images of normal subgroups are always normal, is such a function in fact a group morphism?
abstract-algebra group-theory
abstract-algebra group-theory
asked 1 hour ago
alkchf
40836
asked 1 hour ago
alkchf
40836
asked 1 hour ago
alkchf
40836
asked 1 hour ago
alkchf
40836
asked 1 hour ago
alkchf
40836
40836
add a comment |Â
add a comment |Â
1 Answer
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No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.
answered 1 hour ago
Mees de Vries
14.9k12450
I think you also need $f$ to be bijective for your example to work.
– C Monsour
37 mins ago
@CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
– Mees de Vries
35 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.
answered 1 hour ago
Mees de Vries
14.9k12450
I think you also need $f$ to be bijective for your example to work.
– C Monsour
37 mins ago
@CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
– Mees de Vries
35 mins ago
add a comment |Â
up vote
5
down vote
accepted
No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.
answered 1 hour ago
Mees de Vries
14.9k12450
I think you also need $f$ to be bijective for your example to work.
– C Monsour
37 mins ago
@CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
– Mees de Vries
35 mins ago
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.
answered 1 hour ago
Mees de Vries
14.9k12450
No, and here is a silly reason: there are many groups whose only normal subgroups are the trivial group $1$ and the entire group! Such groups are called simple. Basic examples include the cyclic groups $C_p$ for primes $p$, and the alternating groups $A_n$ for $n neq 4$. If you take any two big simple groups $G, G'$, and any mapping $f: G to G'$ such that $f(a) = 1 iff a = 1$, then $f$ will satisfy your requirements. Obviously $f$ can still be a very weird function then; certainly it will not need to be a group homomorphism.
answered 1 hour ago
Mees de Vries
14.9k12450
answered 1 hour ago
Mees de Vries
14.9k12450
answered 1 hour ago
Mees de Vries
14.9k12450
answered 1 hour ago
Mees de Vries
14.9k12450
14.9k12450
I think you also need $f$ to be bijective for your example to work.
– C Monsour
37 mins ago
@CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
– Mees de Vries
35 mins ago
add a comment |Â
I think you also need $f$ to be bijective for your example to work.
– C Monsour
37 mins ago
@CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
– Mees de Vries
35 mins ago
I think you also need $f$ to be bijective for your example to work.
– C Monsour
37 mins ago
I think you also need $f$ to be bijective for your example to work.
– C Monsour
37 mins ago
@CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
– Mees de Vries
35 mins ago
@CMonsour, no, this is not necessary. Note that we are talking about inverse images under $f$, not images. So e.g. the mapping $f: A_5 to A_5$ defined by $f(mathrmid) = mathrmid$ and $f(alpha) = (1, 2)$ when $alpha neq mathrmid$ is an example.
– Mees de Vries
35 mins ago
add a comment |Â
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