One homomorphism to another in group under additionIs the derivative function a group homomorphism on $G$?Homomorphism from C* into itself having ker = R>0Prove that there is no homomorphism from $Z_16 oplus Z_2$ onto $ Z_4 oplus Z_4$Group Homomorphism Questions (my attempts shown)Proving that a set is a group under additionWhy is the group $Z_6$ under addition isomorphic to the group $Z_7^*$ under multiplication?Group homomorphism induced by another homomorphismShow that the set of all group isomorphisms from $f:(Q,+)rightarrow (Q,+)$ is isomorphic with the $(Q^*,cdot)$ groupshowing $mathbbQ$ is isomorphic to $mathbbR$ or not as a group under additionGroup homomorphism and equivalence relation
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One homomorphism to another in group under addition
Is the derivative function a group homomorphism on $G$?Homomorphism from C* into itself having ker = R>0Prove that there is no homomorphism from $Z_16 oplus Z_2$ onto $ Z_4 oplus Z_4$Group Homomorphism Questions (my attempts shown)Proving that a set is a group under additionWhy is the group $Z_6$ under addition isomorphic to the group $Z_7^*$ under multiplication?Group homomorphism induced by another homomorphismShow that the set of all group isomorphisms from $f:(Q,+)rightarrow (Q,+)$ is isomorphic with the $(Q^*,cdot)$ groupshowing $mathbbQ$ is isomorphic to $mathbbR$ or not as a group under additionGroup homomorphism and equivalence relation
$begingroup$
Let $n geq 2$, and let $G = mathbbZ/nmathbbZ$, which is a group under addition. There is a function $phi_bara(barx):G rightarrow G$ for every $bara in G$, with $phi_bara(barx)=bara cdot barx$.
Now I'm asked to prove that every homomorphism $psi: Grightarrow G$ equals some $phi_bara$. I've already established $phi_bara$ is a homomorphism.
Since $psi$ is a homomorphism I know that $psi(barx+bary)=psi(barx)+psi(bary)$ but I have no idea how to tie this to $phi_bara$. Any suggestions?
abstract-algebra group-theory modular-arithmetic
asked Mar 16 at 19:56
MathbeginnerMathbeginner
33418
$endgroup$
add a comment |
$begingroup$
Let $n geq 2$, and let $G = mathbbZ/nmathbbZ$, which is a group under addition. There is a function $phi_bara(barx):G rightarrow G$ for every $bara in G$, with $phi_bara(barx)=bara cdot barx$.
Now I'm asked to prove that every homomorphism $psi: Grightarrow G$ equals some $phi_bara$. I've already established $phi_bara$ is a homomorphism.
Since $psi$ is a homomorphism I know that $psi(barx+bary)=psi(barx)+psi(bary)$ but I have no idea how to tie this to $phi_bara$. Any suggestions?
abstract-algebra group-theory modular-arithmetic
asked Mar 16 at 19:56
MathbeginnerMathbeginner
33418
$endgroup$
$begingroup$
Hint: $G$ is cyclic.
$endgroup$
– Shaun
Mar 16 at 20:00
add a comment |
$begingroup$
Let $n geq 2$, and let $G = mathbbZ/nmathbbZ$, which is a group under addition. There is a function $phi_bara(barx):G rightarrow G$ for every $bara in G$, with $phi_bara(barx)=bara cdot barx$.
Now I'm asked to prove that every homomorphism $psi: Grightarrow G$ equals some $phi_bara$. I've already established $phi_bara$ is a homomorphism.
Since $psi$ is a homomorphism I know that $psi(barx+bary)=psi(barx)+psi(bary)$ but I have no idea how to tie this to $phi_bara$. Any suggestions?
abstract-algebra group-theory modular-arithmetic
asked Mar 16 at 19:56
MathbeginnerMathbeginner
33418
$endgroup$
Let $n geq 2$, and let $G = mathbbZ/nmathbbZ$, which is a group under addition. There is a function $phi_bara(barx):G rightarrow G$ for every $bara in G$, with $phi_bara(barx)=bara cdot barx$.
Now I'm asked to prove that every homomorphism $psi: Grightarrow G$ equals some $phi_bara$. I've already established $phi_bara$ is a homomorphism.
Since $psi$ is a homomorphism I know that $psi(barx+bary)=psi(barx)+psi(bary)$ but I have no idea how to tie this to $phi_bara$. Any suggestions?
abstract-algebra group-theory modular-arithmetic
abstract-algebra group-theory modular-arithmetic
asked Mar 16 at 19:56
MathbeginnerMathbeginner
33418
asked Mar 16 at 19:56
MathbeginnerMathbeginner
33418
asked Mar 16 at 19:56
MathbeginnerMathbeginner
33418
asked Mar 16 at 19:56
MathbeginnerMathbeginner
33418
asked Mar 16 at 19:56
MathbeginnerMathbeginner
33418
33418
$begingroup$
Hint: $G$ is cyclic.
$endgroup$
– Shaun
Mar 16 at 20:00
add a comment |
$begingroup$
Hint: $G$ is cyclic.
$endgroup$
– Shaun
Mar 16 at 20:00
$begingroup$
Hint: $G$ is cyclic.
$endgroup$
– Shaun
Mar 16 at 20:00
$begingroup$
Hint: $G$ is cyclic.
$endgroup$
– Shaun
Mar 16 at 20:00
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hint: Given a homomorphism $f:Gto G$ write $bara=f(1)$. Now use the properties of a homomorphism.
answered Mar 16 at 20:01
MarkMark
10.4k1622
$endgroup$
$begingroup$
the only property I'm aware of is the one stated. How to apply this to your hint? I would say letting $bara=1$ would mean that you add $bar1$ to the $barx$, which would let you cycle through all $bara in G$.
$endgroup$
– Mathbeginner
Mar 16 at 20:06
$begingroup$
Well, for some $binmathbbZ_n$, what is $f(b)$? You can write $b=1+...+1$ (add $b$ times) and hence $f(b)=f(1)+...+f(1)=f(1)times b$. Of course this is all mod $n$.
$endgroup$
– Mark
Mar 16 at 20:08
$begingroup$
I think I understand what you mean. This would then lead to the conclusion that only $bara=bar1$ is isomorphic, since all larger $bara$ would 'skip' some elements in $G$. Correct?
$endgroup$
– Mathbeginner
Mar 16 at 20:10
$begingroup$
Not sure I understand your comment. But what we get is that letting $bara=f(1)$ we can conclude that $f(b)=f(1)times b=atimes b$ for all $binmathbbZ_n$. So $f=phi_bara$.
$endgroup$
– Mark
Mar 16 at 20:14
$begingroup$
Does this mean you use $f(b)=f(1 cdot b)=f(1) cdot f(b) = a cdot f(b)$? But how can you conclude $f(b)=b$?
$endgroup$
– Mathbeginner
Mar 16 at 20:18
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
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active
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active
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votes
$begingroup$
Hint: Given a homomorphism $f:Gto G$ write $bara=f(1)$. Now use the properties of a homomorphism.
answered Mar 16 at 20:01
MarkMark
10.4k1622
$endgroup$
$begingroup$
the only property I'm aware of is the one stated. How to apply this to your hint? I would say letting $bara=1$ would mean that you add $bar1$ to the $barx$, which would let you cycle through all $bara in G$.
$endgroup$
– Mathbeginner
Mar 16 at 20:06
$begingroup$
Well, for some $binmathbbZ_n$, what is $f(b)$? You can write $b=1+...+1$ (add $b$ times) and hence $f(b)=f(1)+...+f(1)=f(1)times b$. Of course this is all mod $n$.
$endgroup$
– Mark
Mar 16 at 20:08
$begingroup$
I think I understand what you mean. This would then lead to the conclusion that only $bara=bar1$ is isomorphic, since all larger $bara$ would 'skip' some elements in $G$. Correct?
$endgroup$
– Mathbeginner
Mar 16 at 20:10
$begingroup$
Not sure I understand your comment. But what we get is that letting $bara=f(1)$ we can conclude that $f(b)=f(1)times b=atimes b$ for all $binmathbbZ_n$. So $f=phi_bara$.
$endgroup$
– Mark
Mar 16 at 20:14
$begingroup$
Does this mean you use $f(b)=f(1 cdot b)=f(1) cdot f(b) = a cdot f(b)$? But how can you conclude $f(b)=b$?
$endgroup$
– Mathbeginner
Mar 16 at 20:18
|
show 2 more comments
$begingroup$
Hint: Given a homomorphism $f:Gto G$ write $bara=f(1)$. Now use the properties of a homomorphism.
answered Mar 16 at 20:01
MarkMark
10.4k1622
$endgroup$
$begingroup$
the only property I'm aware of is the one stated. How to apply this to your hint? I would say letting $bara=1$ would mean that you add $bar1$ to the $barx$, which would let you cycle through all $bara in G$.
$endgroup$
– Mathbeginner
Mar 16 at 20:06
$begingroup$
Well, for some $binmathbbZ_n$, what is $f(b)$? You can write $b=1+...+1$ (add $b$ times) and hence $f(b)=f(1)+...+f(1)=f(1)times b$. Of course this is all mod $n$.
$endgroup$
– Mark
Mar 16 at 20:08
$begingroup$
I think I understand what you mean. This would then lead to the conclusion that only $bara=bar1$ is isomorphic, since all larger $bara$ would 'skip' some elements in $G$. Correct?
$endgroup$
– Mathbeginner
Mar 16 at 20:10
$begingroup$
Not sure I understand your comment. But what we get is that letting $bara=f(1)$ we can conclude that $f(b)=f(1)times b=atimes b$ for all $binmathbbZ_n$. So $f=phi_bara$.
$endgroup$
– Mark
Mar 16 at 20:14
$begingroup$
Does this mean you use $f(b)=f(1 cdot b)=f(1) cdot f(b) = a cdot f(b)$? But how can you conclude $f(b)=b$?
$endgroup$
– Mathbeginner
Mar 16 at 20:18
|
show 2 more comments
$begingroup$
Hint: Given a homomorphism $f:Gto G$ write $bara=f(1)$. Now use the properties of a homomorphism.
answered Mar 16 at 20:01
MarkMark
10.4k1622
$endgroup$
Hint: Given a homomorphism $f:Gto G$ write $bara=f(1)$. Now use the properties of a homomorphism.
answered Mar 16 at 20:01
MarkMark
10.4k1622
answered Mar 16 at 20:01
MarkMark
10.4k1622
answered Mar 16 at 20:01
MarkMark
10.4k1622
answered Mar 16 at 20:01
MarkMark
10.4k1622
10.4k1622
$begingroup$
the only property I'm aware of is the one stated. How to apply this to your hint? I would say letting $bara=1$ would mean that you add $bar1$ to the $barx$, which would let you cycle through all $bara in G$.
$endgroup$
– Mathbeginner
Mar 16 at 20:06
$begingroup$
Well, for some $binmathbbZ_n$, what is $f(b)$? You can write $b=1+...+1$ (add $b$ times) and hence $f(b)=f(1)+...+f(1)=f(1)times b$. Of course this is all mod $n$.
$endgroup$
– Mark
Mar 16 at 20:08
$begingroup$
I think I understand what you mean. This would then lead to the conclusion that only $bara=bar1$ is isomorphic, since all larger $bara$ would 'skip' some elements in $G$. Correct?
$endgroup$
– Mathbeginner
Mar 16 at 20:10
$begingroup$
Not sure I understand your comment. But what we get is that letting $bara=f(1)$ we can conclude that $f(b)=f(1)times b=atimes b$ for all $binmathbbZ_n$. So $f=phi_bara$.
$endgroup$
– Mark
Mar 16 at 20:14
$begingroup$
Does this mean you use $f(b)=f(1 cdot b)=f(1) cdot f(b) = a cdot f(b)$? But how can you conclude $f(b)=b$?
$endgroup$
– Mathbeginner
Mar 16 at 20:18
|
show 2 more comments
$begingroup$
the only property I'm aware of is the one stated. How to apply this to your hint? I would say letting $bara=1$ would mean that you add $bar1$ to the $barx$, which would let you cycle through all $bara in G$.
$endgroup$
– Mathbeginner
Mar 16 at 20:06
$begingroup$
Well, for some $binmathbbZ_n$, what is $f(b)$? You can write $b=1+...+1$ (add $b$ times) and hence $f(b)=f(1)+...+f(1)=f(1)times b$. Of course this is all mod $n$.
$endgroup$
– Mark
Mar 16 at 20:08
$begingroup$
I think I understand what you mean. This would then lead to the conclusion that only $bara=bar1$ is isomorphic, since all larger $bara$ would 'skip' some elements in $G$. Correct?
$endgroup$
– Mathbeginner
Mar 16 at 20:10
$begingroup$
Not sure I understand your comment. But what we get is that letting $bara=f(1)$ we can conclude that $f(b)=f(1)times b=atimes b$ for all $binmathbbZ_n$. So $f=phi_bara$.
$endgroup$
– Mark
Mar 16 at 20:14
$begingroup$
Does this mean you use $f(b)=f(1 cdot b)=f(1) cdot f(b) = a cdot f(b)$? But how can you conclude $f(b)=b$?
$endgroup$
– Mathbeginner
Mar 16 at 20:18
$begingroup$
the only property I'm aware of is the one stated. How to apply this to your hint? I would say letting $bara=1$ would mean that you add $bar1$ to the $barx$, which would let you cycle through all $bara in G$.
$endgroup$
– Mathbeginner
Mar 16 at 20:06
$begingroup$
the only property I'm aware of is the one stated. How to apply this to your hint? I would say letting $bara=1$ would mean that you add $bar1$ to the $barx$, which would let you cycle through all $bara in G$.
$endgroup$
– Mathbeginner
Mar 16 at 20:06
$begingroup$
Well, for some $binmathbbZ_n$, what is $f(b)$? You can write $b=1+...+1$ (add $b$ times) and hence $f(b)=f(1)+...+f(1)=f(1)times b$. Of course this is all mod $n$.
$endgroup$
– Mark
Mar 16 at 20:08
$begingroup$
Well, for some $binmathbbZ_n$, what is $f(b)$? You can write $b=1+...+1$ (add $b$ times) and hence $f(b)=f(1)+...+f(1)=f(1)times b$. Of course this is all mod $n$.
$endgroup$
– Mark
Mar 16 at 20:08
$begingroup$
I think I understand what you mean. This would then lead to the conclusion that only $bara=bar1$ is isomorphic, since all larger $bara$ would 'skip' some elements in $G$. Correct?
$endgroup$
– Mathbeginner
Mar 16 at 20:10
$begingroup$
I think I understand what you mean. This would then lead to the conclusion that only $bara=bar1$ is isomorphic, since all larger $bara$ would 'skip' some elements in $G$. Correct?
$endgroup$
– Mathbeginner
Mar 16 at 20:10
$begingroup$
Not sure I understand your comment. But what we get is that letting $bara=f(1)$ we can conclude that $f(b)=f(1)times b=atimes b$ for all $binmathbbZ_n$. So $f=phi_bara$.
$endgroup$
– Mark
Mar 16 at 20:14
$begingroup$
Not sure I understand your comment. But what we get is that letting $bara=f(1)$ we can conclude that $f(b)=f(1)times b=atimes b$ for all $binmathbbZ_n$. So $f=phi_bara$.
$endgroup$
– Mark
Mar 16 at 20:14
$begingroup$
Does this mean you use $f(b)=f(1 cdot b)=f(1) cdot f(b) = a cdot f(b)$? But how can you conclude $f(b)=b$?
$endgroup$
– Mathbeginner
Mar 16 at 20:18
$begingroup$
Does this mean you use $f(b)=f(1 cdot b)=f(1) cdot f(b) = a cdot f(b)$? But how can you conclude $f(b)=b$?
$endgroup$
– Mathbeginner
Mar 16 at 20:18
|
show 2 more comments
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$begingroup$
Hint: $G$ is cyclic.
$endgroup$
– Shaun
Mar 16 at 20:00