Examples of homomorphism, endomorphism and automorphism
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If $(G, *)$ and $(H, +)$ are both groupoids and if $f: G to H$ is a surjective function such that $(forall x,y in G) f(x*y) = f(x) + f(y)$ than $f$ is called a homomorphism.
If $G=H$ the function is called endomorphism and if $f$ is bijective than it's called automorphism.
For automorphism an example would be the permutations of a set.
For homomorphism an example i found if a function $f(z) = |z|$ with property $f: C_0 to R_0$ where $C_0,R_0$ are the usual complex and real number sets but without the element $0$.
I can't find an example of endomorphism. I guess the permutations of a set are endomorphism since the function is bijective so in itself it is surjective but i need a pure example. Could you provide one or two :)?
abstract-algebra examples-counterexamples group-homomorphism
asked Jun 11 '17 at 20:44
diredragondiredragon
43
$endgroup$
add a comment |
$begingroup$
If $(G, *)$ and $(H, +)$ are both groupoids and if $f: G to H$ is a surjective function such that $(forall x,y in G) f(x*y) = f(x) + f(y)$ than $f$ is called a homomorphism.
If $G=H$ the function is called endomorphism and if $f$ is bijective than it's called automorphism.
For automorphism an example would be the permutations of a set.
For homomorphism an example i found if a function $f(z) = |z|$ with property $f: C_0 to R_0$ where $C_0,R_0$ are the usual complex and real number sets but without the element $0$.
I can't find an example of endomorphism. I guess the permutations of a set are endomorphism since the function is bijective so in itself it is surjective but i need a pure example. Could you provide one or two :)?
abstract-algebra examples-counterexamples group-homomorphism
asked Jun 11 '17 at 20:44
diredragondiredragon
43
$endgroup$
$begingroup$
The homomorphism is not surjective in general, there is a mistake in the definition.
$endgroup$
– Giuseppe Bargagnati
Jun 12 '17 at 11:51
add a comment |
$begingroup$
If $(G, *)$ and $(H, +)$ are both groupoids and if $f: G to H$ is a surjective function such that $(forall x,y in G) f(x*y) = f(x) + f(y)$ than $f$ is called a homomorphism.
If $G=H$ the function is called endomorphism and if $f$ is bijective than it's called automorphism.
For automorphism an example would be the permutations of a set.
For homomorphism an example i found if a function $f(z) = |z|$ with property $f: C_0 to R_0$ where $C_0,R_0$ are the usual complex and real number sets but without the element $0$.
I can't find an example of endomorphism. I guess the permutations of a set are endomorphism since the function is bijective so in itself it is surjective but i need a pure example. Could you provide one or two :)?
abstract-algebra examples-counterexamples group-homomorphism
asked Jun 11 '17 at 20:44
diredragondiredragon
43
$endgroup$
If $(G, *)$ and $(H, +)$ are both groupoids and if $f: G to H$ is a surjective function such that $(forall x,y in G) f(x*y) = f(x) + f(y)$ than $f$ is called a homomorphism.
If $G=H$ the function is called endomorphism and if $f$ is bijective than it's called automorphism.
For automorphism an example would be the permutations of a set.
For homomorphism an example i found if a function $f(z) = |z|$ with property $f: C_0 to R_0$ where $C_0,R_0$ are the usual complex and real number sets but without the element $0$.
I can't find an example of endomorphism. I guess the permutations of a set are endomorphism since the function is bijective so in itself it is surjective but i need a pure example. Could you provide one or two :)?
abstract-algebra examples-counterexamples group-homomorphism
abstract-algebra examples-counterexamples group-homomorphism
asked Jun 11 '17 at 20:44
diredragondiredragon
43
asked Jun 11 '17 at 20:44
diredragondiredragon
43
asked Jun 11 '17 at 20:44
diredragondiredragon
43
asked Jun 11 '17 at 20:44
diredragondiredragon
43
asked Jun 11 '17 at 20:44
diredragondiredragon
43
43
$begingroup$
The homomorphism is not surjective in general, there is a mistake in the definition.
$endgroup$
– Giuseppe Bargagnati
Jun 12 '17 at 11:51
add a comment |
$begingroup$
The homomorphism is not surjective in general, there is a mistake in the definition.
$endgroup$
– Giuseppe Bargagnati
Jun 12 '17 at 11:51
$begingroup$
The homomorphism is not surjective in general, there is a mistake in the definition.
$endgroup$
– Giuseppe Bargagnati
Jun 12 '17 at 11:51
$begingroup$
The homomorphism is not surjective in general, there is a mistake in the definition.
$endgroup$
– Giuseppe Bargagnati
Jun 12 '17 at 11:51
add a comment |
1 Answer
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$begingroup$
The function $f(x)=2x$ is an endomorphism of $mathbb Z$; it is obviously non surjective.
answered Jun 11 '17 at 20:49
Giuseppe BargagnatiGiuseppe Bargagnati
1,239514
$endgroup$
$begingroup$
Shouldn't the function BE surjective from the definition?
$endgroup$
– diredragon
Jun 11 '17 at 20:59
$begingroup$
No, shouldn't. $,$
$endgroup$
– Berci
Jun 11 '17 at 22:27
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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$begingroup$
The function $f(x)=2x$ is an endomorphism of $mathbb Z$; it is obviously non surjective.
answered Jun 11 '17 at 20:49
Giuseppe BargagnatiGiuseppe Bargagnati
1,239514
$endgroup$
$begingroup$
Shouldn't the function BE surjective from the definition?
$endgroup$
– diredragon
Jun 11 '17 at 20:59
$begingroup$
No, shouldn't. $,$
$endgroup$
– Berci
Jun 11 '17 at 22:27
add a comment |
$begingroup$
The function $f(x)=2x$ is an endomorphism of $mathbb Z$; it is obviously non surjective.
answered Jun 11 '17 at 20:49
Giuseppe BargagnatiGiuseppe Bargagnati
1,239514
$endgroup$
$begingroup$
Shouldn't the function BE surjective from the definition?
$endgroup$
– diredragon
Jun 11 '17 at 20:59
$begingroup$
No, shouldn't. $,$
$endgroup$
– Berci
Jun 11 '17 at 22:27
add a comment |
$begingroup$
The function $f(x)=2x$ is an endomorphism of $mathbb Z$; it is obviously non surjective.
answered Jun 11 '17 at 20:49
Giuseppe BargagnatiGiuseppe Bargagnati
1,239514
$endgroup$
The function $f(x)=2x$ is an endomorphism of $mathbb Z$; it is obviously non surjective.
answered Jun 11 '17 at 20:49
Giuseppe BargagnatiGiuseppe Bargagnati
1,239514
answered Jun 11 '17 at 20:49
Giuseppe BargagnatiGiuseppe Bargagnati
1,239514
answered Jun 11 '17 at 20:49
Giuseppe BargagnatiGiuseppe Bargagnati
1,239514
answered Jun 11 '17 at 20:49
Giuseppe BargagnatiGiuseppe Bargagnati
1,239514
1,239514
$begingroup$
Shouldn't the function BE surjective from the definition?
$endgroup$
– diredragon
Jun 11 '17 at 20:59
$begingroup$
No, shouldn't. $,$
$endgroup$
– Berci
Jun 11 '17 at 22:27
add a comment |
$begingroup$
Shouldn't the function BE surjective from the definition?
$endgroup$
– diredragon
Jun 11 '17 at 20:59
$begingroup$
No, shouldn't. $,$
$endgroup$
– Berci
Jun 11 '17 at 22:27
$begingroup$
Shouldn't the function BE surjective from the definition?
$endgroup$
– diredragon
Jun 11 '17 at 20:59
$begingroup$
Shouldn't the function BE surjective from the definition?
$endgroup$
– diredragon
Jun 11 '17 at 20:59
$begingroup$
No, shouldn't. $,$
$endgroup$
– Berci
Jun 11 '17 at 22:27
$begingroup$
No, shouldn't. $,$
$endgroup$
– Berci
Jun 11 '17 at 22:27
add a comment |
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$begingroup$
The homomorphism is not surjective in general, there is a mistake in the definition.
$endgroup$
– Giuseppe Bargagnati
Jun 12 '17 at 11:51