Mixing
Generator of $mathbb Z/mmathbb Ztimesmathbb Z/nmathbb Z$ for coprime $n$ and $m$.
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
For coprime $n$ and $m$ we have that $mathbb Z/mmathbb Ztimesmathbb Z/nmathbb Zcongmathbb Z/mnmathbb Z$. An isomorphism is given by $$a+mnmathbb Zmapsto (a+mmathbb Z,a+nmathbb Z)$$
Question: Since the cyclic group is generated by (for example) $bar 1$ and an isomorphism has to map generators to generators to maintain injective, does this mean that $(bar 1,bar 1)$ is a generator of the product?
abstract-algebra
asked Jul 27 at 8:47
Buh
59427
add a comment |Â
up vote
3
down vote
favorite
For coprime $n$ and $m$ we have that $mathbb Z/mmathbb Ztimesmathbb Z/nmathbb Zcongmathbb Z/mnmathbb Z$. An isomorphism is given by $$a+mnmathbb Zmapsto (a+mmathbb Z,a+nmathbb Z)$$
Question: Since the cyclic group is generated by (for example) $bar 1$ and an isomorphism has to map generators to generators to maintain injective, does this mean that $(bar 1,bar 1)$ is a generator of the product?
abstract-algebra
asked Jul 27 at 8:47
Buh
59427
1
Yes, this is a consequence of $m$ and $n$ being coprime. Try choosing smallish $m$ and $n$ and look at what happens to $(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1)+(bar 1,bar 1), ...$ and you should quickly get an idea of why it works.
– Nephry
Jul 27 at 8:52
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
For coprime $n$ and $m$ we have that $mathbb Z/mmathbb Ztimesmathbb Z/nmathbb Zcongmathbb Z/mnmathbb Z$. An isomorphism is given by $$a+mnmathbb Zmapsto (a+mmathbb Z,a+nmathbb Z)$$
Question: Since the cyclic group is generated by (for example) $bar 1$ and an isomorphism has to map generators to generators to maintain injective, does this mean that $(bar 1,bar 1)$ is a generator of the product?
abstract-algebra
asked Jul 27 at 8:47
Buh
59427
For coprime $n$ and $m$ we have that $mathbb Z/mmathbb Ztimesmathbb Z/nmathbb Zcongmathbb Z/mnmathbb Z$. An isomorphism is given by $$a+mnmathbb Zmapsto (a+mmathbb Z,a+nmathbb Z)$$
Question: Since the cyclic group is generated by (for example) $bar 1$ and an isomorphism has to map generators to generators to maintain injective, does this mean that $(bar 1,bar 1)$ is a generator of the product?
abstract-algebra
asked Jul 27 at 8:47
Buh
59427
asked Jul 27 at 8:47
Buh
59427
asked Jul 27 at 8:47
Buh
59427
asked Jul 27 at 8:47
Buh
59427
59427
1
Yes, this is a consequence of $m$ and $n$ being coprime. Try choosing smallish $m$ and $n$ and look at what happens to $(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1)+(bar 1,bar 1), ...$ and you should quickly get an idea of why it works.
– Nephry
Jul 27 at 8:52
add a comment |Â
1
Yes, this is a consequence of $m$ and $n$ being coprime. Try choosing smallish $m$ and $n$ and look at what happens to $(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1)+(bar 1,bar 1), ...$ and you should quickly get an idea of why it works.
– Nephry
Jul 27 at 8:52
1
1
Yes, this is a consequence of $m$ and $n$ being coprime. Try choosing smallish $m$ and $n$ and look at what happens to $(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1)+(bar 1,bar 1), ...$ and you should quickly get an idea of why it works.
– Nephry
Jul 27 at 8:52
Yes, this is a consequence of $m$ and $n$ being coprime. Try choosing smallish $m$ and $n$ and look at what happens to $(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1)+(bar 1,bar 1), ...$ and you should quickly get an idea of why it works.
– Nephry
Jul 27 at 8:52
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
If $fcolon Gto G'$ is an isomorphism of cyclic groups, then $f(x)$ is a generator of $G'$ if and only if $x$ is a generator of $G$.
(Easy proof.)
Since, for $m$ and $n$ coprime, the map
$$
a+mnmathbbZmapsto(a+mmathbbZ,a+nmathbbZ)
$$
is (well-defined and) an isomorphism, this implies that
$$
(1+mmathbbZ,1+nmathbbZ)
$$
is a generator of $mathbbZ/mmathbbZtimesmathbbZ/nmathbbZ$.
By the way, every generator of the product is of the form $(a+mmathbbZ,a+nmathbbZ)$, where $a+mnmathbbZ$ is a generator of $mathbbZ/mnmathbbZ$ (that is, $gcd(a,mn)=1$).
answered Jul 27 at 10:30
egreg
164k1180187
add a comment |Â
up vote
1
down vote
Yes, for the reasons you wrote.
There is also a more explicit way to see this: if $d$ is a common divisor of $m$ and $n$, we can write $m = dk,$ $n = dl$ and we can see that $dkl(1,1) = (0,0)$, so the order of $(1,1)$ divides $dkl$ which is strictly lower than $mn = d^2kl$ for $d>1$.
answered Jul 27 at 8:57
Ennar
12.9k32343
So can I say that the order of any element $(g,h)in Gtimes H$ for two groups $G$ and $H$ is given by $textlcm(o(g),o(h))$ where $o(cdot )$ denotes the order? In particular, for any two generators $langle arangle=G, langle brangle H$ we have a generator $langle(a,b)rangle =Gtimes H$ if the product is cyclic?
– Buh
Jul 27 at 9:07
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $fcolon Gto G'$ is an isomorphism of cyclic groups, then $f(x)$ is a generator of $G'$ if and only if $x$ is a generator of $G$.
(Easy proof.)
Since, for $m$ and $n$ coprime, the map
$$
a+mnmathbbZmapsto(a+mmathbbZ,a+nmathbbZ)
$$
is (well-defined and) an isomorphism, this implies that
$$
(1+mmathbbZ,1+nmathbbZ)
$$
is a generator of $mathbbZ/mmathbbZtimesmathbbZ/nmathbbZ$.
By the way, every generator of the product is of the form $(a+mmathbbZ,a+nmathbbZ)$, where $a+mnmathbbZ$ is a generator of $mathbbZ/mnmathbbZ$ (that is, $gcd(a,mn)=1$).
answered Jul 27 at 10:30
egreg
164k1180187
add a comment |Â
up vote
1
down vote
accepted
If $fcolon Gto G'$ is an isomorphism of cyclic groups, then $f(x)$ is a generator of $G'$ if and only if $x$ is a generator of $G$.
(Easy proof.)
Since, for $m$ and $n$ coprime, the map
$$
a+mnmathbbZmapsto(a+mmathbbZ,a+nmathbbZ)
$$
is (well-defined and) an isomorphism, this implies that
$$
(1+mmathbbZ,1+nmathbbZ)
$$
is a generator of $mathbbZ/mmathbbZtimesmathbbZ/nmathbbZ$.
By the way, every generator of the product is of the form $(a+mmathbbZ,a+nmathbbZ)$, where $a+mnmathbbZ$ is a generator of $mathbbZ/mnmathbbZ$ (that is, $gcd(a,mn)=1$).
answered Jul 27 at 10:30
egreg
164k1180187
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $fcolon Gto G'$ is an isomorphism of cyclic groups, then $f(x)$ is a generator of $G'$ if and only if $x$ is a generator of $G$.
(Easy proof.)
Since, for $m$ and $n$ coprime, the map
$$
a+mnmathbbZmapsto(a+mmathbbZ,a+nmathbbZ)
$$
is (well-defined and) an isomorphism, this implies that
$$
(1+mmathbbZ,1+nmathbbZ)
$$
is a generator of $mathbbZ/mmathbbZtimesmathbbZ/nmathbbZ$.
By the way, every generator of the product is of the form $(a+mmathbbZ,a+nmathbbZ)$, where $a+mnmathbbZ$ is a generator of $mathbbZ/mnmathbbZ$ (that is, $gcd(a,mn)=1$).
answered Jul 27 at 10:30
egreg
164k1180187
If $fcolon Gto G'$ is an isomorphism of cyclic groups, then $f(x)$ is a generator of $G'$ if and only if $x$ is a generator of $G$.
(Easy proof.)
Since, for $m$ and $n$ coprime, the map
$$
a+mnmathbbZmapsto(a+mmathbbZ,a+nmathbbZ)
$$
is (well-defined and) an isomorphism, this implies that
$$
(1+mmathbbZ,1+nmathbbZ)
$$
is a generator of $mathbbZ/mmathbbZtimesmathbbZ/nmathbbZ$.
By the way, every generator of the product is of the form $(a+mmathbbZ,a+nmathbbZ)$, where $a+mnmathbbZ$ is a generator of $mathbbZ/mnmathbbZ$ (that is, $gcd(a,mn)=1$).
answered Jul 27 at 10:30
egreg
164k1180187
answered Jul 27 at 10:30
egreg
164k1180187
answered Jul 27 at 10:30
egreg
164k1180187
answered Jul 27 at 10:30
egreg
164k1180187
164k1180187
add a comment |Â
add a comment |Â
up vote
1
down vote
Yes, for the reasons you wrote.
There is also a more explicit way to see this: if $d$ is a common divisor of $m$ and $n$, we can write $m = dk,$ $n = dl$ and we can see that $dkl(1,1) = (0,0)$, so the order of $(1,1)$ divides $dkl$ which is strictly lower than $mn = d^2kl$ for $d>1$.
answered Jul 27 at 8:57
Ennar
12.9k32343
So can I say that the order of any element $(g,h)in Gtimes H$ for two groups $G$ and $H$ is given by $textlcm(o(g),o(h))$ where $o(cdot )$ denotes the order? In particular, for any two generators $langle arangle=G, langle brangle H$ we have a generator $langle(a,b)rangle =Gtimes H$ if the product is cyclic?
– Buh
Jul 27 at 9:07
add a comment |Â
up vote
1
down vote
Yes, for the reasons you wrote.
There is also a more explicit way to see this: if $d$ is a common divisor of $m$ and $n$, we can write $m = dk,$ $n = dl$ and we can see that $dkl(1,1) = (0,0)$, so the order of $(1,1)$ divides $dkl$ which is strictly lower than $mn = d^2kl$ for $d>1$.
answered Jul 27 at 8:57
Ennar
12.9k32343
So can I say that the order of any element $(g,h)in Gtimes H$ for two groups $G$ and $H$ is given by $textlcm(o(g),o(h))$ where $o(cdot )$ denotes the order? In particular, for any two generators $langle arangle=G, langle brangle H$ we have a generator $langle(a,b)rangle =Gtimes H$ if the product is cyclic?
– Buh
Jul 27 at 9:07
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, for the reasons you wrote.
There is also a more explicit way to see this: if $d$ is a common divisor of $m$ and $n$, we can write $m = dk,$ $n = dl$ and we can see that $dkl(1,1) = (0,0)$, so the order of $(1,1)$ divides $dkl$ which is strictly lower than $mn = d^2kl$ for $d>1$.
answered Jul 27 at 8:57
Ennar
12.9k32343
Yes, for the reasons you wrote.
There is also a more explicit way to see this: if $d$ is a common divisor of $m$ and $n$, we can write $m = dk,$ $n = dl$ and we can see that $dkl(1,1) = (0,0)$, so the order of $(1,1)$ divides $dkl$ which is strictly lower than $mn = d^2kl$ for $d>1$.
answered Jul 27 at 8:57
Ennar
12.9k32343
answered Jul 27 at 8:57
Ennar
12.9k32343
answered Jul 27 at 8:57
Ennar
12.9k32343
answered Jul 27 at 8:57
Ennar
12.9k32343
12.9k32343
So can I say that the order of any element $(g,h)in Gtimes H$ for two groups $G$ and $H$ is given by $textlcm(o(g),o(h))$ where $o(cdot )$ denotes the order? In particular, for any two generators $langle arangle=G, langle brangle H$ we have a generator $langle(a,b)rangle =Gtimes H$ if the product is cyclic?
– Buh
Jul 27 at 9:07
add a comment |Â
So can I say that the order of any element $(g,h)in Gtimes H$ for two groups $G$ and $H$ is given by $textlcm(o(g),o(h))$ where $o(cdot )$ denotes the order? In particular, for any two generators $langle arangle=G, langle brangle H$ we have a generator $langle(a,b)rangle =Gtimes H$ if the product is cyclic?
– Buh
Jul 27 at 9:07
So can I say that the order of any element $(g,h)in Gtimes H$ for two groups $G$ and $H$ is given by $textlcm(o(g),o(h))$ where $o(cdot )$ denotes the order? In particular, for any two generators $langle arangle=G, langle brangle H$ we have a generator $langle(a,b)rangle =Gtimes H$ if the product is cyclic?
– Buh
Jul 27 at 9:07
So can I say that the order of any element $(g,h)in Gtimes H$ for two groups $G$ and $H$ is given by $textlcm(o(g),o(h))$ where $o(cdot )$ denotes the order? In particular, for any two generators $langle arangle=G, langle brangle H$ we have a generator $langle(a,b)rangle =Gtimes H$ if the product is cyclic?
– Buh
Jul 27 at 9:07
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2864191%2fgenerator-of-mathbb-z-m-mathbb-z-times-mathbb-z-n-mathbb-z-for-coprime-n-an%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
Yes, this is a consequence of $m$ and $n$ being coprime. Try choosing smallish $m$ and $n$ and look at what happens to $(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1), (bar 1,bar 1)+(bar 1,bar 1)+(bar 1,bar 1), ...$ and you should quickly get an idea of why it works.
– Nephry
Jul 27 at 8:52