Let $G,H$ groups, $H$ abelian, prove $defHom{operatorname{Hom}}Hom(G,H)$, with the entrywise sum...
$begingroup$
I'm working on a proof to show that for a group $G$ and an abelian group $H$, the set of all homomorphisms $defHom{operatorname{Hom}}Hom(G,H)$ from $G$ to $H$ is an abelian group. I just want to verify that my proof is valid. I proceed by trying to show this group satisfies 4 conditions.
First, I defined the operation on $Hom(G,H)$ as $(f+g)(u)$, $u in G$, $f(u) + g(u)$. Is this a valid first move?
Then, associativity is easy to show. $((f+g)+h)(u) = (f+g)(u) + h(u) = f(u) + g(u) + h(u) = (f+(g+h))(u)$.
It is similarly easy to verify that this group is abelian: $(f+g)(u) = f(u) + g(u) = g(u) + f(u)$ since $H$ is abelian, so $(f+g)(u) = (g+f)(u)$.
Is it necessary to show closure, i.e. the product of homomorphisms into an abelian group is a homomorphism? This also follows since $H$ is abelian.
The trivial (zero) homomorphism is the homomorphism that maps every element of G to the identity element of $H$. Since $Hom(G,H)$ includes all homomorphisms, the identity homomorphism is also in $Hom(G,H)$.
I'm having trouble finding the inverse homomorphism. Let's say $f$ is in $Hom(G,H)$. If we take $g$ to be $-f$, $(f+g)(u) = f(u) - f(u) = 1$? What does it mean for a homomorphism to be an inverse of another?
Thanks for your help.
abelian-groups abstract-algebra
edited Mar 12 '17 at 2:03
Santropedro
3101316
asked Oct 28 '14 at 19:58
jstnchngjstnchng
524214
$endgroup$
|
show 1 more comment
$begingroup$
I'm working on a proof to show that for a group $G$ and an abelian group $H$, the set of all homomorphisms $defHom{operatorname{Hom}}Hom(G,H)$ from $G$ to $H$ is an abelian group. I just want to verify that my proof is valid. I proceed by trying to show this group satisfies 4 conditions.
First, I defined the operation on $Hom(G,H)$ as $(f+g)(u)$, $u in G$, $f(u) + g(u)$. Is this a valid first move?
Then, associativity is easy to show. $((f+g)+h)(u) = (f+g)(u) + h(u) = f(u) + g(u) + h(u) = (f+(g+h))(u)$.
It is similarly easy to verify that this group is abelian: $(f+g)(u) = f(u) + g(u) = g(u) + f(u)$ since $H$ is abelian, so $(f+g)(u) = (g+f)(u)$.
Is it necessary to show closure, i.e. the product of homomorphisms into an abelian group is a homomorphism? This also follows since $H$ is abelian.
The trivial (zero) homomorphism is the homomorphism that maps every element of G to the identity element of $H$. Since $Hom(G,H)$ includes all homomorphisms, the identity homomorphism is also in $Hom(G,H)$.
I'm having trouble finding the inverse homomorphism. Let's say $f$ is in $Hom(G,H)$. If we take $g$ to be $-f$, $(f+g)(u) = f(u) - f(u) = 1$? What does it mean for a homomorphism to be an inverse of another?
Thanks for your help.
abelian-groups abstract-algebra
edited Mar 12 '17 at 2:03
Santropedro
3101316
asked Oct 28 '14 at 19:58
jstnchngjstnchng
524214
$endgroup$
2
$begingroup$
Yes, you need to show closure. If you are writing your group $H$ additively, why do you get $f(u)-f(u)=1$ rather than $0$?
$endgroup$
– James
Oct 28 '14 at 20:00
$begingroup$
Oops sorry, I meant 0.
$endgroup$
– jstnchng
Oct 28 '14 at 20:03
1
$begingroup$
Is $G$ abelian too?
$endgroup$
– brick
Oct 28 '14 at 20:06
1
$begingroup$
@brick it doesn't matter.
$endgroup$
– Adam Hughes
Oct 28 '14 at 20:10
$begingroup$
Yeah we need H is abelian to show that Hom(G,H) is abelian, and also when demonstrating closure.
$endgroup$
– jstnchng
Oct 28 '14 at 20:11
|
show 1 more comment
$begingroup$
I'm working on a proof to show that for a group $G$ and an abelian group $H$, the set of all homomorphisms $defHom{operatorname{Hom}}Hom(G,H)$ from $G$ to $H$ is an abelian group. I just want to verify that my proof is valid. I proceed by trying to show this group satisfies 4 conditions.
First, I defined the operation on $Hom(G,H)$ as $(f+g)(u)$, $u in G$, $f(u) + g(u)$. Is this a valid first move?
Then, associativity is easy to show. $((f+g)+h)(u) = (f+g)(u) + h(u) = f(u) + g(u) + h(u) = (f+(g+h))(u)$.
It is similarly easy to verify that this group is abelian: $(f+g)(u) = f(u) + g(u) = g(u) + f(u)$ since $H$ is abelian, so $(f+g)(u) = (g+f)(u)$.
Is it necessary to show closure, i.e. the product of homomorphisms into an abelian group is a homomorphism? This also follows since $H$ is abelian.
The trivial (zero) homomorphism is the homomorphism that maps every element of G to the identity element of $H$. Since $Hom(G,H)$ includes all homomorphisms, the identity homomorphism is also in $Hom(G,H)$.
I'm having trouble finding the inverse homomorphism. Let's say $f$ is in $Hom(G,H)$. If we take $g$ to be $-f$, $(f+g)(u) = f(u) - f(u) = 1$? What does it mean for a homomorphism to be an inverse of another?
Thanks for your help.
abelian-groups abstract-algebra
edited Mar 12 '17 at 2:03
Santropedro
3101316
asked Oct 28 '14 at 19:58
jstnchngjstnchng
524214
$endgroup$
I'm working on a proof to show that for a group $G$ and an abelian group $H$, the set of all homomorphisms $defHom{operatorname{Hom}}Hom(G,H)$ from $G$ to $H$ is an abelian group. I just want to verify that my proof is valid. I proceed by trying to show this group satisfies 4 conditions.
First, I defined the operation on $Hom(G,H)$ as $(f+g)(u)$, $u in G$, $f(u) + g(u)$. Is this a valid first move?
Then, associativity is easy to show. $((f+g)+h)(u) = (f+g)(u) + h(u) = f(u) + g(u) + h(u) = (f+(g+h))(u)$.
It is similarly easy to verify that this group is abelian: $(f+g)(u) = f(u) + g(u) = g(u) + f(u)$ since $H$ is abelian, so $(f+g)(u) = (g+f)(u)$.
Is it necessary to show closure, i.e. the product of homomorphisms into an abelian group is a homomorphism? This also follows since $H$ is abelian.
The trivial (zero) homomorphism is the homomorphism that maps every element of G to the identity element of $H$. Since $Hom(G,H)$ includes all homomorphisms, the identity homomorphism is also in $Hom(G,H)$.
I'm having trouble finding the inverse homomorphism. Let's say $f$ is in $Hom(G,H)$. If we take $g$ to be $-f$, $(f+g)(u) = f(u) - f(u) = 1$? What does it mean for a homomorphism to be an inverse of another?
Thanks for your help.
abelian-groups abstract-algebra
abelian-groups abstract-algebra
edited Mar 12 '17 at 2:03
Santropedro
3101316
asked Oct 28 '14 at 19:58
jstnchngjstnchng
524214
edited Mar 12 '17 at 2:03
Santropedro
3101316
asked Oct 28 '14 at 19:58
jstnchngjstnchng
524214
edited Mar 12 '17 at 2:03
Santropedro
3101316
edited Mar 12 '17 at 2:03
Santropedro
3101316
edited Mar 12 '17 at 2:03
Santropedro
3101316
3101316
asked Oct 28 '14 at 19:58
jstnchngjstnchng
524214
asked Oct 28 '14 at 19:58
jstnchngjstnchng
524214
asked Oct 28 '14 at 19:58
jstnchngjstnchng
524214
524214
2
$begingroup$
Yes, you need to show closure. If you are writing your group $H$ additively, why do you get $f(u)-f(u)=1$ rather than $0$?
$endgroup$
– James
Oct 28 '14 at 20:00
$begingroup$
Oops sorry, I meant 0.
$endgroup$
– jstnchng
Oct 28 '14 at 20:03
1
$begingroup$
Is $G$ abelian too?
$endgroup$
– brick
Oct 28 '14 at 20:06
1
$begingroup$
@brick it doesn't matter.
$endgroup$
– Adam Hughes
Oct 28 '14 at 20:10
$begingroup$
Yeah we need H is abelian to show that Hom(G,H) is abelian, and also when demonstrating closure.
$endgroup$
– jstnchng
Oct 28 '14 at 20:11
|
show 1 more comment
2
$begingroup$
Yes, you need to show closure. If you are writing your group $H$ additively, why do you get $f(u)-f(u)=1$ rather than $0$?
$endgroup$
– James
Oct 28 '14 at 20:00
$begingroup$
Oops sorry, I meant 0.
$endgroup$
– jstnchng
Oct 28 '14 at 20:03
1
$begingroup$
Is $G$ abelian too?
$endgroup$
– brick
Oct 28 '14 at 20:06
1
$begingroup$
@brick it doesn't matter.
$endgroup$
– Adam Hughes
Oct 28 '14 at 20:10
$begingroup$
Yeah we need H is abelian to show that Hom(G,H) is abelian, and also when demonstrating closure.
$endgroup$
– jstnchng
Oct 28 '14 at 20:11
2
2
$begingroup$
Yes, you need to show closure. If you are writing your group $H$ additively, why do you get $f(u)-f(u)=1$ rather than $0$?
$endgroup$
– James
Oct 28 '14 at 20:00
$begingroup$
Yes, you need to show closure. If you are writing your group $H$ additively, why do you get $f(u)-f(u)=1$ rather than $0$?
$endgroup$
– James
Oct 28 '14 at 20:00
$begingroup$
Oops sorry, I meant 0.
$endgroup$
– jstnchng
Oct 28 '14 at 20:03
$begingroup$
Oops sorry, I meant 0.
$endgroup$
– jstnchng
Oct 28 '14 at 20:03
1
1
$begingroup$
Is $G$ abelian too?
$endgroup$
– brick
Oct 28 '14 at 20:06
$begingroup$
Is $G$ abelian too?
$endgroup$
– brick
Oct 28 '14 at 20:06
1
1
$begingroup$
@brick it doesn't matter.
$endgroup$
– Adam Hughes
Oct 28 '14 at 20:10
$begingroup$
@brick it doesn't matter.
$endgroup$
– Adam Hughes
Oct 28 '14 at 20:10
$begingroup$
Yeah we need H is abelian to show that Hom(G,H) is abelian, and also when demonstrating closure.
$endgroup$
– jstnchng
Oct 28 '14 at 20:11
$begingroup$
Yeah we need H is abelian to show that Hom(G,H) is abelian, and also when demonstrating closure.
$endgroup$
– jstnchng
Oct 28 '14 at 20:11
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Note: I'm using additive notation for both groups G and H, however, I assume ONLY $H$ abelian, I will not use abelian for $G$, tell me if I mistake that. but it totally can be done without using $G$ abelian.
Proof: Let $A=Hom(G,H)$, with the sum $(f+g)(x)=f(x)+g(x)$. Let's prove $A$ it's a abelian group:
Closed: $(f+g)(x+y)=f(x+y)+g(x+y)=f(x)+f(y)+g(x)+g(y)$
Now IMPORTANT, we use "$H$ is abelian" hypothesis, if not, we can't do this step, it's simply false.
$=f(x)+g(x)+f(y)+g(y)=(f+g)(x)+(f+g)(y)$ that proves the sum of homomorphism is homomorphism, so it's closed.
Associative: I copy and improve your proof $((f+g)+h)(u)=(f+g)(u)+h(u)=f(u)+g(u)+h(u)=(f+(g+h))(u)$
Abelian: $(f+g)(u)=f(u)+g(u)=g(u)+f(u)$ IMPORTANT: This uses $H$ abelian.
Inverses: Let $f in Hom(G,H)$ then the only candidate to be inverse that could work, is the entrywise inverse $-f$ defined by $-f(x)=f(-x)$ . So let's check it does the job first:
$(f+(-f))(x)=f(x)+(-f)(x)=f(x)+f(-x)=f(x-x)=f(0)=0$ And the sum in the other direction has the same proof (or you remember this group $A$ is abelian).
And $-f$ is a homomorphism: $-f(x+y)=f(-(x+y))=f(-y-x)=f(-y)+f(-x)=f(-x)+f(-y)
=-f(x)+-f(y)$ Note this also uses $H$ to be abelian.
answered Mar 12 '17 at 2:38
SantropedroSantropedro
3101316
$endgroup$
add a comment |
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$begingroup$
Note: I'm using additive notation for both groups G and H, however, I assume ONLY $H$ abelian, I will not use abelian for $G$, tell me if I mistake that. but it totally can be done without using $G$ abelian.
Proof: Let $A=Hom(G,H)$, with the sum $(f+g)(x)=f(x)+g(x)$. Let's prove $A$ it's a abelian group:
Closed: $(f+g)(x+y)=f(x+y)+g(x+y)=f(x)+f(y)+g(x)+g(y)$
Now IMPORTANT, we use "$H$ is abelian" hypothesis, if not, we can't do this step, it's simply false.
$=f(x)+g(x)+f(y)+g(y)=(f+g)(x)+(f+g)(y)$ that proves the sum of homomorphism is homomorphism, so it's closed.
Associative: I copy and improve your proof $((f+g)+h)(u)=(f+g)(u)+h(u)=f(u)+g(u)+h(u)=(f+(g+h))(u)$
Abelian: $(f+g)(u)=f(u)+g(u)=g(u)+f(u)$ IMPORTANT: This uses $H$ abelian.
Inverses: Let $f in Hom(G,H)$ then the only candidate to be inverse that could work, is the entrywise inverse $-f$ defined by $-f(x)=f(-x)$ . So let's check it does the job first:
$(f+(-f))(x)=f(x)+(-f)(x)=f(x)+f(-x)=f(x-x)=f(0)=0$ And the sum in the other direction has the same proof (or you remember this group $A$ is abelian).
And $-f$ is a homomorphism: $-f(x+y)=f(-(x+y))=f(-y-x)=f(-y)+f(-x)=f(-x)+f(-y)
=-f(x)+-f(y)$ Note this also uses $H$ to be abelian.
answered Mar 12 '17 at 2:38
SantropedroSantropedro
3101316
$endgroup$
add a comment |
$begingroup$
Note: I'm using additive notation for both groups G and H, however, I assume ONLY $H$ abelian, I will not use abelian for $G$, tell me if I mistake that. but it totally can be done without using $G$ abelian.
Proof: Let $A=Hom(G,H)$, with the sum $(f+g)(x)=f(x)+g(x)$. Let's prove $A$ it's a abelian group:
Closed: $(f+g)(x+y)=f(x+y)+g(x+y)=f(x)+f(y)+g(x)+g(y)$
Now IMPORTANT, we use "$H$ is abelian" hypothesis, if not, we can't do this step, it's simply false.
$=f(x)+g(x)+f(y)+g(y)=(f+g)(x)+(f+g)(y)$ that proves the sum of homomorphism is homomorphism, so it's closed.
Associative: I copy and improve your proof $((f+g)+h)(u)=(f+g)(u)+h(u)=f(u)+g(u)+h(u)=(f+(g+h))(u)$
Abelian: $(f+g)(u)=f(u)+g(u)=g(u)+f(u)$ IMPORTANT: This uses $H$ abelian.
Inverses: Let $f in Hom(G,H)$ then the only candidate to be inverse that could work, is the entrywise inverse $-f$ defined by $-f(x)=f(-x)$ . So let's check it does the job first:
$(f+(-f))(x)=f(x)+(-f)(x)=f(x)+f(-x)=f(x-x)=f(0)=0$ And the sum in the other direction has the same proof (or you remember this group $A$ is abelian).
And $-f$ is a homomorphism: $-f(x+y)=f(-(x+y))=f(-y-x)=f(-y)+f(-x)=f(-x)+f(-y)
=-f(x)+-f(y)$ Note this also uses $H$ to be abelian.
answered Mar 12 '17 at 2:38
SantropedroSantropedro
3101316
$endgroup$
add a comment |
$begingroup$
Note: I'm using additive notation for both groups G and H, however, I assume ONLY $H$ abelian, I will not use abelian for $G$, tell me if I mistake that. but it totally can be done without using $G$ abelian.
Proof: Let $A=Hom(G,H)$, with the sum $(f+g)(x)=f(x)+g(x)$. Let's prove $A$ it's a abelian group:
Closed: $(f+g)(x+y)=f(x+y)+g(x+y)=f(x)+f(y)+g(x)+g(y)$
Now IMPORTANT, we use "$H$ is abelian" hypothesis, if not, we can't do this step, it's simply false.
$=f(x)+g(x)+f(y)+g(y)=(f+g)(x)+(f+g)(y)$ that proves the sum of homomorphism is homomorphism, so it's closed.
Associative: I copy and improve your proof $((f+g)+h)(u)=(f+g)(u)+h(u)=f(u)+g(u)+h(u)=(f+(g+h))(u)$
Abelian: $(f+g)(u)=f(u)+g(u)=g(u)+f(u)$ IMPORTANT: This uses $H$ abelian.
Inverses: Let $f in Hom(G,H)$ then the only candidate to be inverse that could work, is the entrywise inverse $-f$ defined by $-f(x)=f(-x)$ . So let's check it does the job first:
$(f+(-f))(x)=f(x)+(-f)(x)=f(x)+f(-x)=f(x-x)=f(0)=0$ And the sum in the other direction has the same proof (or you remember this group $A$ is abelian).
And $-f$ is a homomorphism: $-f(x+y)=f(-(x+y))=f(-y-x)=f(-y)+f(-x)=f(-x)+f(-y)
=-f(x)+-f(y)$ Note this also uses $H$ to be abelian.
answered Mar 12 '17 at 2:38
SantropedroSantropedro
3101316
$endgroup$
Note: I'm using additive notation for both groups G and H, however, I assume ONLY $H$ abelian, I will not use abelian for $G$, tell me if I mistake that. but it totally can be done without using $G$ abelian.
Proof: Let $A=Hom(G,H)$, with the sum $(f+g)(x)=f(x)+g(x)$. Let's prove $A$ it's a abelian group:
Closed: $(f+g)(x+y)=f(x+y)+g(x+y)=f(x)+f(y)+g(x)+g(y)$
Now IMPORTANT, we use "$H$ is abelian" hypothesis, if not, we can't do this step, it's simply false.
$=f(x)+g(x)+f(y)+g(y)=(f+g)(x)+(f+g)(y)$ that proves the sum of homomorphism is homomorphism, so it's closed.
Associative: I copy and improve your proof $((f+g)+h)(u)=(f+g)(u)+h(u)=f(u)+g(u)+h(u)=(f+(g+h))(u)$
Abelian: $(f+g)(u)=f(u)+g(u)=g(u)+f(u)$ IMPORTANT: This uses $H$ abelian.
Inverses: Let $f in Hom(G,H)$ then the only candidate to be inverse that could work, is the entrywise inverse $-f$ defined by $-f(x)=f(-x)$ . So let's check it does the job first:
$(f+(-f))(x)=f(x)+(-f)(x)=f(x)+f(-x)=f(x-x)=f(0)=0$ And the sum in the other direction has the same proof (or you remember this group $A$ is abelian).
And $-f$ is a homomorphism: $-f(x+y)=f(-(x+y))=f(-y-x)=f(-y)+f(-x)=f(-x)+f(-y)
=-f(x)+-f(y)$ Note this also uses $H$ to be abelian.
answered Mar 12 '17 at 2:38
SantropedroSantropedro
3101316
edited Mar 12 '17 at 16:43
answered Mar 12 '17 at 2:38
SantropedroSantropedro
3101316
answered Mar 12 '17 at 2:38
SantropedroSantropedro
3101316
answered Mar 12 '17 at 2:38
SantropedroSantropedro
3101316
3101316
add a comment |
add a comment |
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$begingroup$
Yes, you need to show closure. If you are writing your group $H$ additively, why do you get $f(u)-f(u)=1$ rather than $0$?
$endgroup$
– James
Oct 28 '14 at 20:00
$begingroup$
Oops sorry, I meant 0.
$endgroup$
– jstnchng
Oct 28 '14 at 20:03
1
$begingroup$
Is $G$ abelian too?
$endgroup$
– brick
Oct 28 '14 at 20:06
1
$begingroup$
@brick it doesn't matter.
$endgroup$
– Adam Hughes
Oct 28 '14 at 20:10
$begingroup$
Yeah we need H is abelian to show that Hom(G,H) is abelian, and also when demonstrating closure.
$endgroup$
– jstnchng
Oct 28 '14 at 20:11