Semidirect product of two cyclic groups
9
$begingroup$
Describe all semidirect products of $C_n$ by $C_m$ (ie $C_n rtimes C_m$) where $m,n in mathbb{N_+}$
Note: For the first attempt one needs to find all homomorphisms from $C_m to U(n)$, but the situation differs a lot for different pairs of $n ,m$, is there a better way to find all structures of $C_n rtimes C_m$?
Wikipedia provided a general presentation of this product which I do not know how it was worked out.
group-theory cyclic-groups group-homomorphism semidirect-product
edited Nov 30 '18 at 3:28
Shaun
8,893113681
asked May 31 '12 at 4:25
user31899user31899
1,6091033
$endgroup$
add a comment |
9
$begingroup$
Describe all semidirect products of $C_n$ by $C_m$ (ie $C_n rtimes C_m$) where $m,n in mathbb{N_+}$
Note: For the first attempt one needs to find all homomorphisms from $C_m to U(n)$, but the situation differs a lot for different pairs of $n ,m$, is there a better way to find all structures of $C_n rtimes C_m$?
Wikipedia provided a general presentation of this product which I do not know how it was worked out.
group-theory cyclic-groups group-homomorphism semidirect-product
edited Nov 30 '18 at 3:28
Shaun
8,893113681
asked May 31 '12 at 4:25
user31899user31899
1,6091033
$endgroup$
2
$begingroup$
$U(n)$ has order $phi(n)$, so there are nontrivial homomorphisms if and only if $gcd(m,phi(n))gt 1$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:26
$begingroup$
That's true, but if there are non-trivial homomorphisms then how to get general presentations of the semidirect product?
$endgroup$
– user31899
May 31 '12 at 4:47
3
$begingroup$
The presentation of the semidirect product is $$langle x,ymid x^n = y^m=1, x^y = x^{varphi(y)}rangle,$$ where $varphicolon C_mto U(n)$ is the chosen homomorphism, and $y$ generates $C_m$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:51
2
$begingroup$
The semidirect product is completely determined by how $y$ (the generator of $C_m$) acts on $x$ (the generator of $C_n$). The action is given by conjugation, so you just need to say what $x^y = y^{-1}xy$ is; it must be a power of $x$ (since $langle xrangle$ is normal), so it is of the form $y^{-1}xy = x^k$; the order of $x^k$ must be prime to $n$, and $k^mequiv 1pmod{phi(n)}$ must hold. This gives you the multiplication table, so that's a presentation of the group.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:54
$begingroup$
Thanks! It's clear to me now. the phi(y) is in U(n) thus coprime to n which is equal to the symbol "k" in the presentaion given by the wiki site above.
$endgroup$
– user31899
May 31 '12 at 4:57
add a comment |
9
9
9
6
$begingroup$
Describe all semidirect products of $C_n$ by $C_m$ (ie $C_n rtimes C_m$) where $m,n in mathbb{N_+}$
Note: For the first attempt one needs to find all homomorphisms from $C_m to U(n)$, but the situation differs a lot for different pairs of $n ,m$, is there a better way to find all structures of $C_n rtimes C_m$?
Wikipedia provided a general presentation of this product which I do not know how it was worked out.
group-theory cyclic-groups group-homomorphism semidirect-product
edited Nov 30 '18 at 3:28
Shaun
8,893113681
asked May 31 '12 at 4:25
user31899user31899
1,6091033
$endgroup$
Describe all semidirect products of $C_n$ by $C_m$ (ie $C_n rtimes C_m$) where $m,n in mathbb{N_+}$
Note: For the first attempt one needs to find all homomorphisms from $C_m to U(n)$, but the situation differs a lot for different pairs of $n ,m$, is there a better way to find all structures of $C_n rtimes C_m$?
Wikipedia provided a general presentation of this product which I do not know how it was worked out.
group-theory cyclic-groups group-homomorphism semidirect-product
group-theory cyclic-groups group-homomorphism semidirect-product
edited Nov 30 '18 at 3:28
Shaun
8,893113681
asked May 31 '12 at 4:25
user31899user31899
1,6091033
edited Nov 30 '18 at 3:28
Shaun
8,893113681
asked May 31 '12 at 4:25
user31899user31899
1,6091033
edited Nov 30 '18 at 3:28
Shaun
8,893113681
edited Nov 30 '18 at 3:28
Shaun
8,893113681
edited Nov 30 '18 at 3:28
Shaun
8,893113681
8,893113681
asked May 31 '12 at 4:25
user31899user31899
1,6091033
asked May 31 '12 at 4:25
user31899user31899
1,6091033
asked May 31 '12 at 4:25
user31899user31899
1,6091033
1,6091033
2
$begingroup$
$U(n)$ has order $phi(n)$, so there are nontrivial homomorphisms if and only if $gcd(m,phi(n))gt 1$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:26
$begingroup$
That's true, but if there are non-trivial homomorphisms then how to get general presentations of the semidirect product?
$endgroup$
– user31899
May 31 '12 at 4:47
3
$begingroup$
The presentation of the semidirect product is $$langle x,ymid x^n = y^m=1, x^y = x^{varphi(y)}rangle,$$ where $varphicolon C_mto U(n)$ is the chosen homomorphism, and $y$ generates $C_m$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:51
2
$begingroup$
The semidirect product is completely determined by how $y$ (the generator of $C_m$) acts on $x$ (the generator of $C_n$). The action is given by conjugation, so you just need to say what $x^y = y^{-1}xy$ is; it must be a power of $x$ (since $langle xrangle$ is normal), so it is of the form $y^{-1}xy = x^k$; the order of $x^k$ must be prime to $n$, and $k^mequiv 1pmod{phi(n)}$ must hold. This gives you the multiplication table, so that's a presentation of the group.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:54
$begingroup$
Thanks! It's clear to me now. the phi(y) is in U(n) thus coprime to n which is equal to the symbol "k" in the presentaion given by the wiki site above.
$endgroup$
– user31899
May 31 '12 at 4:57
add a comment |
2
$begingroup$
$U(n)$ has order $phi(n)$, so there are nontrivial homomorphisms if and only if $gcd(m,phi(n))gt 1$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:26
$begingroup$
That's true, but if there are non-trivial homomorphisms then how to get general presentations of the semidirect product?
$endgroup$
– user31899
May 31 '12 at 4:47
3
$begingroup$
The presentation of the semidirect product is $$langle x,ymid x^n = y^m=1, x^y = x^{varphi(y)}rangle,$$ where $varphicolon C_mto U(n)$ is the chosen homomorphism, and $y$ generates $C_m$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:51
2
$begingroup$
The semidirect product is completely determined by how $y$ (the generator of $C_m$) acts on $x$ (the generator of $C_n$). The action is given by conjugation, so you just need to say what $x^y = y^{-1}xy$ is; it must be a power of $x$ (since $langle xrangle$ is normal), so it is of the form $y^{-1}xy = x^k$; the order of $x^k$ must be prime to $n$, and $k^mequiv 1pmod{phi(n)}$ must hold. This gives you the multiplication table, so that's a presentation of the group.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:54
$begingroup$
Thanks! It's clear to me now. the phi(y) is in U(n) thus coprime to n which is equal to the symbol "k" in the presentaion given by the wiki site above.
$endgroup$
– user31899
May 31 '12 at 4:57
2
2
$begingroup$
$U(n)$ has order $phi(n)$, so there are nontrivial homomorphisms if and only if $gcd(m,phi(n))gt 1$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:26
$begingroup$
$U(n)$ has order $phi(n)$, so there are nontrivial homomorphisms if and only if $gcd(m,phi(n))gt 1$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:26
$begingroup$
That's true, but if there are non-trivial homomorphisms then how to get general presentations of the semidirect product?
$endgroup$
– user31899
May 31 '12 at 4:47
$begingroup$
That's true, but if there are non-trivial homomorphisms then how to get general presentations of the semidirect product?
$endgroup$
– user31899
May 31 '12 at 4:47
3
3
$begingroup$
The presentation of the semidirect product is $$langle x,ymid x^n = y^m=1, x^y = x^{varphi(y)}rangle,$$ where $varphicolon C_mto U(n)$ is the chosen homomorphism, and $y$ generates $C_m$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:51
$begingroup$
The presentation of the semidirect product is $$langle x,ymid x^n = y^m=1, x^y = x^{varphi(y)}rangle,$$ where $varphicolon C_mto U(n)$ is the chosen homomorphism, and $y$ generates $C_m$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:51
2
2
$begingroup$
The semidirect product is completely determined by how $y$ (the generator of $C_m$) acts on $x$ (the generator of $C_n$). The action is given by conjugation, so you just need to say what $x^y = y^{-1}xy$ is; it must be a power of $x$ (since $langle xrangle$ is normal), so it is of the form $y^{-1}xy = x^k$; the order of $x^k$ must be prime to $n$, and $k^mequiv 1pmod{phi(n)}$ must hold. This gives you the multiplication table, so that's a presentation of the group.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:54
$begingroup$
The semidirect product is completely determined by how $y$ (the generator of $C_m$) acts on $x$ (the generator of $C_n$). The action is given by conjugation, so you just need to say what $x^y = y^{-1}xy$ is; it must be a power of $x$ (since $langle xrangle$ is normal), so it is of the form $y^{-1}xy = x^k$; the order of $x^k$ must be prime to $n$, and $k^mequiv 1pmod{phi(n)}$ must hold. This gives you the multiplication table, so that's a presentation of the group.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:54
$begingroup$
Thanks! It's clear to me now. the phi(y) is in U(n) thus coprime to n which is equal to the symbol "k" in the presentaion given by the wiki site above.
$endgroup$
– user31899
May 31 '12 at 4:57
$begingroup$
Thanks! It's clear to me now. the phi(y) is in U(n) thus coprime to n which is equal to the symbol "k" in the presentaion given by the wiki site above.
$endgroup$
– user31899
May 31 '12 at 4:57
add a comment |
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2
$begingroup$
$U(n)$ has order $phi(n)$, so there are nontrivial homomorphisms if and only if $gcd(m,phi(n))gt 1$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:26
$begingroup$
That's true, but if there are non-trivial homomorphisms then how to get general presentations of the semidirect product?
$endgroup$
– user31899
May 31 '12 at 4:47
3
$begingroup$
The presentation of the semidirect product is $$langle x,ymid x^n = y^m=1, x^y = x^{varphi(y)}rangle,$$ where $varphicolon C_mto U(n)$ is the chosen homomorphism, and $y$ generates $C_m$.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:51
2
$begingroup$
The semidirect product is completely determined by how $y$ (the generator of $C_m$) acts on $x$ (the generator of $C_n$). The action is given by conjugation, so you just need to say what $x^y = y^{-1}xy$ is; it must be a power of $x$ (since $langle xrangle$ is normal), so it is of the form $y^{-1}xy = x^k$; the order of $x^k$ must be prime to $n$, and $k^mequiv 1pmod{phi(n)}$ must hold. This gives you the multiplication table, so that's a presentation of the group.
$endgroup$
– Arturo Magidin
May 31 '12 at 4:54
$begingroup$
Thanks! It's clear to me now. the phi(y) is in U(n) thus coprime to n which is equal to the symbol "k" in the presentaion given by the wiki site above.
$endgroup$
– user31899
May 31 '12 at 4:57