Searching for an alternative definition of group homomorphism
I am searching for an alternative definition of a group homomorphism using only their impact on subgroups. I thought of something like
Let $G$, $G'$ be groups. A map from $G$ to $G'$ is called homomorphism of groups if the image of every subgroup of $G$ is a subgroup in $G'$.
I guess this definition wouldn't work. It's maybe too weak to be a homomorphism.
Does anyone have an idea?
abstract-algebra group-theory
edited Dec 3 '18 at 9:29
José Carlos Santos
152k22123225
asked Dec 3 '18 at 9:26
Dominik
965
add a comment |
I am searching for an alternative definition of a group homomorphism using only their impact on subgroups. I thought of something like
Let $G$, $G'$ be groups. A map from $G$ to $G'$ is called homomorphism of groups if the image of every subgroup of $G$ is a subgroup in $G'$.
I guess this definition wouldn't work. It's maybe too weak to be a homomorphism.
Does anyone have an idea?
abstract-algebra group-theory
edited Dec 3 '18 at 9:29
José Carlos Santos
152k22123225
asked Dec 3 '18 at 9:26
Dominik
965
2
It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
– freakish
Dec 3 '18 at 9:33
1
Are you trying to do something that mimics the definition of a continuous map?
– Tobias Kildetoft
Dec 3 '18 at 9:44
1
@freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
– Dominik
Dec 3 '18 at 9:52
add a comment |
I am searching for an alternative definition of a group homomorphism using only their impact on subgroups. I thought of something like
Let $G$, $G'$ be groups. A map from $G$ to $G'$ is called homomorphism of groups if the image of every subgroup of $G$ is a subgroup in $G'$.
I guess this definition wouldn't work. It's maybe too weak to be a homomorphism.
Does anyone have an idea?
abstract-algebra group-theory
edited Dec 3 '18 at 9:29
José Carlos Santos
152k22123225
asked Dec 3 '18 at 9:26
Dominik
965
I am searching for an alternative definition of a group homomorphism using only their impact on subgroups. I thought of something like
Let $G$, $G'$ be groups. A map from $G$ to $G'$ is called homomorphism of groups if the image of every subgroup of $G$ is a subgroup in $G'$.
I guess this definition wouldn't work. It's maybe too weak to be a homomorphism.
Does anyone have an idea?
abstract-algebra group-theory
abstract-algebra group-theory
edited Dec 3 '18 at 9:29
José Carlos Santos
152k22123225
asked Dec 3 '18 at 9:26
Dominik
965
edited Dec 3 '18 at 9:29
José Carlos Santos
152k22123225
asked Dec 3 '18 at 9:26
Dominik
965
edited Dec 3 '18 at 9:29
José Carlos Santos
152k22123225
edited Dec 3 '18 at 9:29
José Carlos Santos
152k22123225
edited Dec 3 '18 at 9:29
José Carlos Santos
152k22123225
152k22123225
asked Dec 3 '18 at 9:26
Dominik
965
asked Dec 3 '18 at 9:26
Dominik
965
asked Dec 3 '18 at 9:26
Dominik
965
965
2
It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
– freakish
Dec 3 '18 at 9:33
1
Are you trying to do something that mimics the definition of a continuous map?
– Tobias Kildetoft
Dec 3 '18 at 9:44
1
@freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
– Dominik
Dec 3 '18 at 9:52
add a comment |
2
It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
– freakish
Dec 3 '18 at 9:33
1
Are you trying to do something that mimics the definition of a continuous map?
– Tobias Kildetoft
Dec 3 '18 at 9:44
1
@freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
– Dominik
Dec 3 '18 at 9:52
2
2
It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
– freakish
Dec 3 '18 at 9:33
It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
– freakish
Dec 3 '18 at 9:33
1
1
Are you trying to do something that mimics the definition of a continuous map?
– Tobias Kildetoft
Dec 3 '18 at 9:44
Are you trying to do something that mimics the definition of a continuous map?
– Tobias Kildetoft
Dec 3 '18 at 9:44
1
1
@freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
– Dominik
Dec 3 '18 at 9:52
@freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
– Dominik
Dec 3 '18 at 9:52
add a comment |
2 Answers
2
active
oldest
votes
Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.
Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.
answered Dec 3 '18 at 10:35
P Vanchinathan
14.9k12136
add a comment |
This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
x mapsto [x]$$ where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!
answered Dec 3 '18 at 10:18
Enkidu
95618
add a comment |
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2 Answers
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active
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2 Answers
2
active
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Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.
Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.
answered Dec 3 '18 at 10:35
P Vanchinathan
14.9k12136
add a comment |
Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.
Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.
answered Dec 3 '18 at 10:35
P Vanchinathan
14.9k12136
add a comment |
Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.
Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.
answered Dec 3 '18 at 10:35
P Vanchinathan
14.9k12136
Others have pointed out the trouble with this definition. This is the way one learns anyway. Try to see why this definition and not this gives a clear and critical understanding.
Let me point out another example where your suggested alternative definition fails: Take a group which is NOT abelian, say symmeric group on $n$ letters, . $n>3$. Consider the function $xmapsto x^{-1}$. As a function from a group to itself, it satisfies your condition, image of any subgroup is a subgroup (in fact it is the same subgroup). However it is not true in general that $(xy)^{-1}= x^{-1}y^{-1}$. So it is not a homomorphism.
answered Dec 3 '18 at 10:35
P Vanchinathan
14.9k12136
edited Dec 3 '18 at 10:42
answered Dec 3 '18 at 10:35
P Vanchinathan
14.9k12136
answered Dec 3 '18 at 10:35
P Vanchinathan
14.9k12136
answered Dec 3 '18 at 10:35
P Vanchinathan
14.9k12136
14.9k12136
add a comment |
add a comment |
This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
x mapsto [x]$$ where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!
answered Dec 3 '18 at 10:18
Enkidu
95618
add a comment |
This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
x mapsto [x]$$ where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!
answered Dec 3 '18 at 10:18
Enkidu
95618
add a comment |
This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
x mapsto [x]$$ where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!
answered Dec 3 '18 at 10:18
Enkidu
95618
This definition is far too weak! you need compatibility with the group structure, which gets completely forgotten by your definition, for example, consider the following map: $$mathbb{Z}/pmathbb{Z} to mathbb{Z}/qmathbb{Z}\
x mapsto [x]$$ where we interpret $mathbb{Z}/pmathbb{Z} $ as ${0,1,...,p-1}$.
Now if $q le p$ this is would be by your "definition" a "groupmorphism", but all the properties of the elements, like order gets messed up, and you lose all control!
answered Dec 3 '18 at 10:18
Enkidu
95618
answered Dec 3 '18 at 10:18
Enkidu
95618
answered Dec 3 '18 at 10:18
Enkidu
95618
answered Dec 3 '18 at 10:18
Enkidu
95618
95618
add a comment |
add a comment |
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2
It definitely is too weak. According to your definition any bijection $mathbb{Z}_ptomathbb{Z}_p$ that fixes $0$ would be a homomorphism. Why are you looking for an alternative if there is a simple classical definition?
– freakish
Dec 3 '18 at 9:33
1
Are you trying to do something that mimics the definition of a continuous map?
– Tobias Kildetoft
Dec 3 '18 at 9:44
1
@freakish When if first learned about normal subgroups I had absolutely no idea what this definition using conjugation really wants to tell me. It just seemed to be not very intuitive, but when I began to think of normal groups as the kernels of group homomorphism it really helped me a lot to understand what normal subgroups are all about.I hoped to find a definition which makes even clearer which structures of a group "survive" a homomorphism.
– Dominik
Dec 3 '18 at 9:52