Mixing
kernel of homomorphism and commutator subgroup
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Let $G$ be a group generated by two elements $a$ and $b$. Suppose that there is a surjective group homomorphism $phi: Gto mathbbZoplus mathbbZ$. Why is $kerphi=[G,G]$?
group-theory
asked 1 hour ago
John N.
37519
add a comment |Â
up vote
2
down vote
favorite
Let $G$ be a group generated by two elements $a$ and $b$. Suppose that there is a surjective group homomorphism $phi: Gto mathbbZoplus mathbbZ$. Why is $kerphi=[G,G]$?
group-theory
asked 1 hour ago
John N.
37519
What is the definition of the commutator subgroup $[G,G]$?
– crispypizza
1 hour ago
@crispypizza it's the group generated by the elements [g,h]:=g^-1h^-1gh, with g,h in G
– John N.
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $G$ be a group generated by two elements $a$ and $b$. Suppose that there is a surjective group homomorphism $phi: Gto mathbbZoplus mathbbZ$. Why is $kerphi=[G,G]$?
group-theory
asked 1 hour ago
John N.
37519
Let $G$ be a group generated by two elements $a$ and $b$. Suppose that there is a surjective group homomorphism $phi: Gto mathbbZoplus mathbbZ$. Why is $kerphi=[G,G]$?
group-theory
group-theory
asked 1 hour ago
John N.
37519
asked 1 hour ago
John N.
37519
asked 1 hour ago
John N.
37519
asked 1 hour ago
John N.
37519
asked 1 hour ago
John N.
37519
37519
What is the definition of the commutator subgroup $[G,G]$?
– crispypizza
1 hour ago
@crispypizza it's the group generated by the elements [g,h]:=g^-1h^-1gh, with g,h in G
– John N.
1 hour ago
add a comment |Â
What is the definition of the commutator subgroup $[G,G]$?
– crispypizza
1 hour ago
@crispypizza it's the group generated by the elements [g,h]:=g^-1h^-1gh, with g,h in G
– John N.
1 hour ago
What is the definition of the commutator subgroup $[G,G]$?
– crispypizza
1 hour ago
What is the definition of the commutator subgroup $[G,G]$?
– crispypizza
1 hour ago
@crispypizza it's the group generated by the elements [g,h]:=g^-1h^-1gh, with g,h in G
– John N.
1 hour ago
@crispypizza it's the group generated by the elements [g,h]:=g^-1h^-1gh, with g,h in G
– John N.
1 hour ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
$G/[G,G]$ is an abelian group generated by two elements. It is thus a quotient of $mathbbZtimesmathbbZ$. So you then have a sequence
$$ mathbbZ^2 rightarrow G/[G,G]xrightarrowphi mathbbZ^2 $$
where both maps are surjective.
Can you finish from here?
answered 34 mins ago
Steve D
2,8452622
add a comment |Â
up vote
2
down vote
First of all by the first isomorphism theorem you know that $G/Kerphi$ is isomorphic to $Imphi$ which is an abelian group. Hence $[G,G]leq Kerphi$.
The other direction is a bit harder. Every element in $G$ can be written as the product of powers of $a$ and powers of $b$. $mathbbZtimesmathbbZ$ is abelian and hence every element in $Imphi$ can be written in the form $Tphi(a)+Sphi(b)$ when $T,SinmathbbZ$. From here you get that $phi(a),phi(b)$ generate $mathbbZtimesmathbbZ$ (as $phi$ is surjective), and because they are two elements we can conclude that they form a basis. So $Tphi(a)+Sphi(b)=0$ iff $T=S=0$. So $Kerphi$ is the set of all elements of the form $a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. It is left to show that every such element is in $[G,G]$.
So let $g=a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. We know that the quotient group $G/[G,G]$ is abelian, and hence:
$g[G,G]=a^n_1b^m_1...a^n_kb^m_k[G,G]=a^n_1+...+n_kb^m_1+...+m_k[G,G]=[G,G]$
So $Kerphileq[G,G]$ and there is the result.
answered 15 mins ago
Mark
1,983111
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
$G/[G,G]$ is an abelian group generated by two elements. It is thus a quotient of $mathbbZtimesmathbbZ$. So you then have a sequence
$$ mathbbZ^2 rightarrow G/[G,G]xrightarrowphi mathbbZ^2 $$
where both maps are surjective.
Can you finish from here?
answered 34 mins ago
Steve D
2,8452622
add a comment |Â
up vote
3
down vote
accepted
$G/[G,G]$ is an abelian group generated by two elements. It is thus a quotient of $mathbbZtimesmathbbZ$. So you then have a sequence
$$ mathbbZ^2 rightarrow G/[G,G]xrightarrowphi mathbbZ^2 $$
where both maps are surjective.
Can you finish from here?
answered 34 mins ago
Steve D
2,8452622
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
$G/[G,G]$ is an abelian group generated by two elements. It is thus a quotient of $mathbbZtimesmathbbZ$. So you then have a sequence
$$ mathbbZ^2 rightarrow G/[G,G]xrightarrowphi mathbbZ^2 $$
where both maps are surjective.
Can you finish from here?
answered 34 mins ago
Steve D
2,8452622
$G/[G,G]$ is an abelian group generated by two elements. It is thus a quotient of $mathbbZtimesmathbbZ$. So you then have a sequence
$$ mathbbZ^2 rightarrow G/[G,G]xrightarrowphi mathbbZ^2 $$
where both maps are surjective.
Can you finish from here?
answered 34 mins ago
Steve D
2,8452622
answered 34 mins ago
Steve D
2,8452622
answered 34 mins ago
Steve D
2,8452622
answered 34 mins ago
Steve D
2,8452622
2,8452622
add a comment |Â
add a comment |Â
up vote
2
down vote
First of all by the first isomorphism theorem you know that $G/Kerphi$ is isomorphic to $Imphi$ which is an abelian group. Hence $[G,G]leq Kerphi$.
The other direction is a bit harder. Every element in $G$ can be written as the product of powers of $a$ and powers of $b$. $mathbbZtimesmathbbZ$ is abelian and hence every element in $Imphi$ can be written in the form $Tphi(a)+Sphi(b)$ when $T,SinmathbbZ$. From here you get that $phi(a),phi(b)$ generate $mathbbZtimesmathbbZ$ (as $phi$ is surjective), and because they are two elements we can conclude that they form a basis. So $Tphi(a)+Sphi(b)=0$ iff $T=S=0$. So $Kerphi$ is the set of all elements of the form $a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. It is left to show that every such element is in $[G,G]$.
So let $g=a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. We know that the quotient group $G/[G,G]$ is abelian, and hence:
$g[G,G]=a^n_1b^m_1...a^n_kb^m_k[G,G]=a^n_1+...+n_kb^m_1+...+m_k[G,G]=[G,G]$
So $Kerphileq[G,G]$ and there is the result.
answered 15 mins ago
Mark
1,983111
add a comment |Â
up vote
2
down vote
First of all by the first isomorphism theorem you know that $G/Kerphi$ is isomorphic to $Imphi$ which is an abelian group. Hence $[G,G]leq Kerphi$.
The other direction is a bit harder. Every element in $G$ can be written as the product of powers of $a$ and powers of $b$. $mathbbZtimesmathbbZ$ is abelian and hence every element in $Imphi$ can be written in the form $Tphi(a)+Sphi(b)$ when $T,SinmathbbZ$. From here you get that $phi(a),phi(b)$ generate $mathbbZtimesmathbbZ$ (as $phi$ is surjective), and because they are two elements we can conclude that they form a basis. So $Tphi(a)+Sphi(b)=0$ iff $T=S=0$. So $Kerphi$ is the set of all elements of the form $a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. It is left to show that every such element is in $[G,G]$.
So let $g=a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. We know that the quotient group $G/[G,G]$ is abelian, and hence:
$g[G,G]=a^n_1b^m_1...a^n_kb^m_k[G,G]=a^n_1+...+n_kb^m_1+...+m_k[G,G]=[G,G]$
So $Kerphileq[G,G]$ and there is the result.
answered 15 mins ago
Mark
1,983111
add a comment |Â
up vote
2
down vote
up vote
2
down vote
First of all by the first isomorphism theorem you know that $G/Kerphi$ is isomorphic to $Imphi$ which is an abelian group. Hence $[G,G]leq Kerphi$.
The other direction is a bit harder. Every element in $G$ can be written as the product of powers of $a$ and powers of $b$. $mathbbZtimesmathbbZ$ is abelian and hence every element in $Imphi$ can be written in the form $Tphi(a)+Sphi(b)$ when $T,SinmathbbZ$. From here you get that $phi(a),phi(b)$ generate $mathbbZtimesmathbbZ$ (as $phi$ is surjective), and because they are two elements we can conclude that they form a basis. So $Tphi(a)+Sphi(b)=0$ iff $T=S=0$. So $Kerphi$ is the set of all elements of the form $a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. It is left to show that every such element is in $[G,G]$.
So let $g=a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. We know that the quotient group $G/[G,G]$ is abelian, and hence:
$g[G,G]=a^n_1b^m_1...a^n_kb^m_k[G,G]=a^n_1+...+n_kb^m_1+...+m_k[G,G]=[G,G]$
So $Kerphileq[G,G]$ and there is the result.
answered 15 mins ago
Mark
1,983111
First of all by the first isomorphism theorem you know that $G/Kerphi$ is isomorphic to $Imphi$ which is an abelian group. Hence $[G,G]leq Kerphi$.
The other direction is a bit harder. Every element in $G$ can be written as the product of powers of $a$ and powers of $b$. $mathbbZtimesmathbbZ$ is abelian and hence every element in $Imphi$ can be written in the form $Tphi(a)+Sphi(b)$ when $T,SinmathbbZ$. From here you get that $phi(a),phi(b)$ generate $mathbbZtimesmathbbZ$ (as $phi$ is surjective), and because they are two elements we can conclude that they form a basis. So $Tphi(a)+Sphi(b)=0$ iff $T=S=0$. So $Kerphi$ is the set of all elements of the form $a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. It is left to show that every such element is in $[G,G]$.
So let $g=a^n_1b^m_1...a^n_kb^m_k$ where $n_1+...+n_k=m_1+...+m_k=0$. We know that the quotient group $G/[G,G]$ is abelian, and hence:
$g[G,G]=a^n_1b^m_1...a^n_kb^m_k[G,G]=a^n_1+...+n_kb^m_1+...+m_k[G,G]=[G,G]$
So $Kerphileq[G,G]$ and there is the result.
answered 15 mins ago
Mark
1,983111
answered 15 mins ago
Mark
1,983111
answered 15 mins ago
Mark
1,983111
answered 15 mins ago
Mark
1,983111
1,983111
add a comment |Â
add a comment |Â
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What is the definition of the commutator subgroup $[G,G]$?
– crispypizza
1 hour ago
@crispypizza it's the group generated by the elements [g,h]:=g^-1h^-1gh, with g,h in G
– John N.
1 hour ago