Is there a function such that $f(xyz)=f(x)f(y)f(z)$, for any $x,y,zin G$ but $f$ is not an endomorphism?Can you always find a surjective endomorphism of groups such that it is not injective?If $G$ is a finite group, then $operatornameordS(a)=operatornameord(G)/operatornameOrd(C(a))$Elementary Properties of cyclic groupsFinite order of an element $f(g)$ of a homomorphism $f$Is there any group $G$ with $ord(G)=20$ so that $varphi:Grightarrow mathbbZ_15$ is epimorphism?For which finite groups $G$ and $H$ does only one homomorphism $Grightarrow H$ exist?Is there any finite group G such that G has no subgroup of order $d_1$, $d_2$ but have a subgroup of order $d_1 d_2$Prove that $G$ is abelian if there exists $min mathbbN$, so that $f(x)=x^m$, $f in Aut(G)$The number of idempotent endomorphisms of a finite groupExistence of a group morphism between an arbitrary group and an abelian group.
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Is there a function such that $f(xyz)=f(x)f(y)f(z)$, for any $x,y,zin G$ but $f$ is not an endomorphism?
Can you always find a surjective endomorphism of groups such that it is not injective?If $G$ is a finite group, then $operatornameordS(a)=operatornameord(G)/operatornameOrd(C(a))$Elementary Properties of cyclic groupsFinite order of an element $f(g)$ of a homomorphism $f$Is there any group $G$ with $ord(G)=20$ so that $varphi:Grightarrow mathbbZ_15$ is epimorphism?For which finite groups $G$ and $H$ does only one homomorphism $Grightarrow H$ exist?Is there any finite group G such that G has no subgroup of order $d_1$, $d_2$ but have a subgroup of order $d_1 d_2$Prove that $G$ is abelian if there exists $min mathbbN$, so that $f(x)=x^m$, $f in Aut(G)$The number of idempotent endomorphisms of a finite groupExistence of a group morphism between an arbitrary group and an abelian group.
$begingroup$
I am required to show that, for a $ninmathbbN$ and $(G,cdot)$ a finite group of order $n$, if $n$ is even there is a function $f:G rightarrow G$ such that $f(xyz)=f(x)f(y)f(z)$, for any $x,y,zin G$ but $f$ is not an endomorphism. Is this true for $n$ odd?
I chose $DeclareMathOperatorordordf(x)=ord(x)$, but I am unsure that $ord(xy)=ord(x)ord(y)$ for any $x,yin G$.
At least I obtained that it is not a morphism if it respects that property, because indeed as for $e=a^kla^pha^-1$, where $(k,p)=1$ and $kl+ph=1$, $f(e)=f(a^kla^pha^-1)=-klph$, which is false if $f$ is a morphism.
abstract-algebra group-theory finite-groups
edited 2 days ago
egreg
184k1486205
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am required to show that, for a $ninmathbbN$ and $(G,cdot)$ a finite group of order $n$, if $n$ is even there is a function $f:G rightarrow G$ such that $f(xyz)=f(x)f(y)f(z)$, for any $x,y,zin G$ but $f$ is not an endomorphism. Is this true for $n$ odd?
I chose $DeclareMathOperatorordordf(x)=ord(x)$, but I am unsure that $ord(xy)=ord(x)ord(y)$ for any $x,yin G$.
At least I obtained that it is not a morphism if it respects that property, because indeed as for $e=a^kla^pha^-1$, where $(k,p)=1$ and $kl+ph=1$, $f(e)=f(a^kla^pha^-1)=-klph$, which is false if $f$ is a morphism.
abstract-algebra group-theory finite-groups
edited 2 days ago
egreg
184k1486205
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
What have you tried? It's false for $n=1$, and most likely you've tried some small values such as $n=2,3$: what did you get? And write "every" instead of "a" which is unclear as it may mean "some". Next, $f(x)=mathrmord(x)$ is not a self-map of $G$, but is a map from $G$ to $mathbfN$.
$endgroup$
– YCor
2 days ago
$begingroup$
Your proposed function doesn’t work because in general, the order of a product is not the product of their orders. For a simple counterexample, consider the case in which $x=y^-1$, where $xy=e$.
$endgroup$
– Arturo Magidin
2 days ago
$begingroup$
First of all, $ord$ is not a function from $mathbb G$ into $mathbb G$.
$endgroup$
– enedil
2 days ago
add a comment |
$begingroup$
I am required to show that, for a $ninmathbbN$ and $(G,cdot)$ a finite group of order $n$, if $n$ is even there is a function $f:G rightarrow G$ such that $f(xyz)=f(x)f(y)f(z)$, for any $x,y,zin G$ but $f$ is not an endomorphism. Is this true for $n$ odd?
I chose $DeclareMathOperatorordordf(x)=ord(x)$, but I am unsure that $ord(xy)=ord(x)ord(y)$ for any $x,yin G$.
At least I obtained that it is not a morphism if it respects that property, because indeed as for $e=a^kla^pha^-1$, where $(k,p)=1$ and $kl+ph=1$, $f(e)=f(a^kla^pha^-1)=-klph$, which is false if $f$ is a morphism.
abstract-algebra group-theory finite-groups
edited 2 days ago
egreg
184k1486205
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am required to show that, for a $ninmathbbN$ and $(G,cdot)$ a finite group of order $n$, if $n$ is even there is a function $f:G rightarrow G$ such that $f(xyz)=f(x)f(y)f(z)$, for any $x,y,zin G$ but $f$ is not an endomorphism. Is this true for $n$ odd?
I chose $DeclareMathOperatorordordf(x)=ord(x)$, but I am unsure that $ord(xy)=ord(x)ord(y)$ for any $x,yin G$.
At least I obtained that it is not a morphism if it respects that property, because indeed as for $e=a^kla^pha^-1$, where $(k,p)=1$ and $kl+ph=1$, $f(e)=f(a^kla^pha^-1)=-klph$, which is false if $f$ is a morphism.
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
edited 2 days ago
egreg
184k1486205
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
egreg
184k1486205
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
egreg
184k1486205
edited 2 days ago
egreg
184k1486205
edited 2 days ago
egreg
184k1486205
184k1486205
asked 2 days ago
user651754user651754
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
user651754user651754
454
asked 2 days ago
user651754user651754
454
454
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user651754 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$begingroup$
What have you tried? It's false for $n=1$, and most likely you've tried some small values such as $n=2,3$: what did you get? And write "every" instead of "a" which is unclear as it may mean "some". Next, $f(x)=mathrmord(x)$ is not a self-map of $G$, but is a map from $G$ to $mathbfN$.
$endgroup$
– YCor
2 days ago
$begingroup$
Your proposed function doesn’t work because in general, the order of a product is not the product of their orders. For a simple counterexample, consider the case in which $x=y^-1$, where $xy=e$.
$endgroup$
– Arturo Magidin
2 days ago
$begingroup$
First of all, $ord$ is not a function from $mathbb G$ into $mathbb G$.
$endgroup$
– enedil
2 days ago
add a comment |
$begingroup$
What have you tried? It's false for $n=1$, and most likely you've tried some small values such as $n=2,3$: what did you get? And write "every" instead of "a" which is unclear as it may mean "some". Next, $f(x)=mathrmord(x)$ is not a self-map of $G$, but is a map from $G$ to $mathbfN$.
$endgroup$
– YCor
2 days ago
$begingroup$
Your proposed function doesn’t work because in general, the order of a product is not the product of their orders. For a simple counterexample, consider the case in which $x=y^-1$, where $xy=e$.
$endgroup$
– Arturo Magidin
2 days ago
$begingroup$
First of all, $ord$ is not a function from $mathbb G$ into $mathbb G$.
$endgroup$
– enedil
2 days ago
$begingroup$
What have you tried? It's false for $n=1$, and most likely you've tried some small values such as $n=2,3$: what did you get? And write "every" instead of "a" which is unclear as it may mean "some". Next, $f(x)=mathrmord(x)$ is not a self-map of $G$, but is a map from $G$ to $mathbfN$.
$endgroup$
– YCor
2 days ago
$begingroup$
What have you tried? It's false for $n=1$, and most likely you've tried some small values such as $n=2,3$: what did you get? And write "every" instead of "a" which is unclear as it may mean "some". Next, $f(x)=mathrmord(x)$ is not a self-map of $G$, but is a map from $G$ to $mathbfN$.
$endgroup$
– YCor
2 days ago
$begingroup$
Your proposed function doesn’t work because in general, the order of a product is not the product of their orders. For a simple counterexample, consider the case in which $x=y^-1$, where $xy=e$.
$endgroup$
– Arturo Magidin
2 days ago
$begingroup$
Your proposed function doesn’t work because in general, the order of a product is not the product of their orders. For a simple counterexample, consider the case in which $x=y^-1$, where $xy=e$.
$endgroup$
– Arturo Magidin
2 days ago
$begingroup$
First of all, $ord$ is not a function from $mathbb G$ into $mathbb G$.
$endgroup$
– enedil
2 days ago
$begingroup$
First of all, $ord$ is not a function from $mathbb G$ into $mathbb G$.
$endgroup$
– enedil
2 days ago
add a comment |
1 Answer
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$begingroup$
First of all note that $f(e)=f(e)^3$ and so $f(e)^2=e$. This gives us a hint about $f(e)$.
So if $n$ is even then pick $gin G$ such that $g^2=e$, $gneq e$ (which exists by Cauchy's theorem) and define $f(x)=g$, i.e. the constant function at $g$. Obviously $f(xyz)=f(x)f(y)f(z)$ regardless of what $x,y,z$ are. But since $gneq e$ then $f$ is not a group homomorphism.
Now if $n$ is odd then the condition $f(e)^2=e$ implies that $f(e)=e$. And therefore for any $x,y$ we have
$$f(xy)=f(exy)=f(e)f(x)f(y)=f(x)f(y)$$
and so $f$ is a group homomorphism.
but I am unsure that $ord(xy)=ord(x)ord(y)$ for any $x,yin G$.
If $G$ is non-trivial then that never holds for all $x,y$. Consider $xneq e$ and $y=x^-1$. You can also easily check that $ord$ does not satisfy the original condition as well, i.e. consider $ord(exx^-1)$.
Also talking about $ord$ does not make sense to begin with since the codomain of $ord$ is $mathbbN$, it's not even a group. Not to mention that domain and codomain are different although that assumption is unnecessary.
edited yesterday
J. W. Tanner
3,0981320
answered 2 days ago
freakishfreakish
12.7k1631
$endgroup$
add a comment |
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$begingroup$
First of all note that $f(e)=f(e)^3$ and so $f(e)^2=e$. This gives us a hint about $f(e)$.
So if $n$ is even then pick $gin G$ such that $g^2=e$, $gneq e$ (which exists by Cauchy's theorem) and define $f(x)=g$, i.e. the constant function at $g$. Obviously $f(xyz)=f(x)f(y)f(z)$ regardless of what $x,y,z$ are. But since $gneq e$ then $f$ is not a group homomorphism.
Now if $n$ is odd then the condition $f(e)^2=e$ implies that $f(e)=e$. And therefore for any $x,y$ we have
$$f(xy)=f(exy)=f(e)f(x)f(y)=f(x)f(y)$$
and so $f$ is a group homomorphism.
but I am unsure that $ord(xy)=ord(x)ord(y)$ for any $x,yin G$.
If $G$ is non-trivial then that never holds for all $x,y$. Consider $xneq e$ and $y=x^-1$. You can also easily check that $ord$ does not satisfy the original condition as well, i.e. consider $ord(exx^-1)$.
Also talking about $ord$ does not make sense to begin with since the codomain of $ord$ is $mathbbN$, it's not even a group. Not to mention that domain and codomain are different although that assumption is unnecessary.
edited yesterday
J. W. Tanner
3,0981320
answered 2 days ago
freakishfreakish
12.7k1631
$endgroup$
add a comment |
$begingroup$
First of all note that $f(e)=f(e)^3$ and so $f(e)^2=e$. This gives us a hint about $f(e)$.
So if $n$ is even then pick $gin G$ such that $g^2=e$, $gneq e$ (which exists by Cauchy's theorem) and define $f(x)=g$, i.e. the constant function at $g$. Obviously $f(xyz)=f(x)f(y)f(z)$ regardless of what $x,y,z$ are. But since $gneq e$ then $f$ is not a group homomorphism.
Now if $n$ is odd then the condition $f(e)^2=e$ implies that $f(e)=e$. And therefore for any $x,y$ we have
$$f(xy)=f(exy)=f(e)f(x)f(y)=f(x)f(y)$$
and so $f$ is a group homomorphism.
but I am unsure that $ord(xy)=ord(x)ord(y)$ for any $x,yin G$.
If $G$ is non-trivial then that never holds for all $x,y$. Consider $xneq e$ and $y=x^-1$. You can also easily check that $ord$ does not satisfy the original condition as well, i.e. consider $ord(exx^-1)$.
Also talking about $ord$ does not make sense to begin with since the codomain of $ord$ is $mathbbN$, it's not even a group. Not to mention that domain and codomain are different although that assumption is unnecessary.
edited yesterday
J. W. Tanner
3,0981320
answered 2 days ago
freakishfreakish
12.7k1631
$endgroup$
add a comment |
$begingroup$
First of all note that $f(e)=f(e)^3$ and so $f(e)^2=e$. This gives us a hint about $f(e)$.
So if $n$ is even then pick $gin G$ such that $g^2=e$, $gneq e$ (which exists by Cauchy's theorem) and define $f(x)=g$, i.e. the constant function at $g$. Obviously $f(xyz)=f(x)f(y)f(z)$ regardless of what $x,y,z$ are. But since $gneq e$ then $f$ is not a group homomorphism.
Now if $n$ is odd then the condition $f(e)^2=e$ implies that $f(e)=e$. And therefore for any $x,y$ we have
$$f(xy)=f(exy)=f(e)f(x)f(y)=f(x)f(y)$$
and so $f$ is a group homomorphism.
but I am unsure that $ord(xy)=ord(x)ord(y)$ for any $x,yin G$.
If $G$ is non-trivial then that never holds for all $x,y$. Consider $xneq e$ and $y=x^-1$. You can also easily check that $ord$ does not satisfy the original condition as well, i.e. consider $ord(exx^-1)$.
Also talking about $ord$ does not make sense to begin with since the codomain of $ord$ is $mathbbN$, it's not even a group. Not to mention that domain and codomain are different although that assumption is unnecessary.
edited yesterday
J. W. Tanner
3,0981320
answered 2 days ago
freakishfreakish
12.7k1631
$endgroup$
First of all note that $f(e)=f(e)^3$ and so $f(e)^2=e$. This gives us a hint about $f(e)$.
So if $n$ is even then pick $gin G$ such that $g^2=e$, $gneq e$ (which exists by Cauchy's theorem) and define $f(x)=g$, i.e. the constant function at $g$. Obviously $f(xyz)=f(x)f(y)f(z)$ regardless of what $x,y,z$ are. But since $gneq e$ then $f$ is not a group homomorphism.
Now if $n$ is odd then the condition $f(e)^2=e$ implies that $f(e)=e$. And therefore for any $x,y$ we have
$$f(xy)=f(exy)=f(e)f(x)f(y)=f(x)f(y)$$
and so $f$ is a group homomorphism.
but I am unsure that $ord(xy)=ord(x)ord(y)$ for any $x,yin G$.
If $G$ is non-trivial then that never holds for all $x,y$. Consider $xneq e$ and $y=x^-1$. You can also easily check that $ord$ does not satisfy the original condition as well, i.e. consider $ord(exx^-1)$.
Also talking about $ord$ does not make sense to begin with since the codomain of $ord$ is $mathbbN$, it's not even a group. Not to mention that domain and codomain are different although that assumption is unnecessary.
edited yesterday
J. W. Tanner
3,0981320
answered 2 days ago
freakishfreakish
12.7k1631
edited yesterday
J. W. Tanner
3,0981320
edited yesterday
J. W. Tanner
3,0981320
edited yesterday
J. W. Tanner
3,0981320
3,0981320
answered 2 days ago
freakishfreakish
12.7k1631
answered 2 days ago
freakishfreakish
12.7k1631
answered 2 days ago
freakishfreakish
12.7k1631
12.7k1631
add a comment |
add a comment |
user651754 is a new contributor. Be nice, and check out our Code of Conduct.
user651754 is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
What have you tried? It's false for $n=1$, and most likely you've tried some small values such as $n=2,3$: what did you get? And write "every" instead of "a" which is unclear as it may mean "some". Next, $f(x)=mathrmord(x)$ is not a self-map of $G$, but is a map from $G$ to $mathbfN$.
$endgroup$
– YCor
2 days ago
$begingroup$
Your proposed function doesn’t work because in general, the order of a product is not the product of their orders. For a simple counterexample, consider the case in which $x=y^-1$, where $xy=e$.
$endgroup$
– Arturo Magidin
2 days ago
$begingroup$
First of all, $ord$ is not a function from $mathbb G$ into $mathbb G$.
$endgroup$
– enedil
2 days ago