2-generated abelian group
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Given a non-cyclic 2-generated abelian group $G=langle a,brangle$, it seems straightforward from fundamental theorem of abelian groups that
$$Gconglangle xrangletimeslangle yrangle$$
for some $x,yin G$.
However, if you look closely, there will be still some tedious argument to show such a result. So I am looking for a slick proof that does NOT use fundamental theorem of abelian groups.
group-theory abelian-groups finitely-generated
edited Dec 21 '18 at 16:29
Shaun
9,759113684
asked Dec 20 '18 at 10:53
EasyEasy
3,535917
$endgroup$
add a comment |
$begingroup$
Given a non-cyclic 2-generated abelian group $G=langle a,brangle$, it seems straightforward from fundamental theorem of abelian groups that
$$Gconglangle xrangletimeslangle yrangle$$
for some $x,yin G$.
However, if you look closely, there will be still some tedious argument to show such a result. So I am looking for a slick proof that does NOT use fundamental theorem of abelian groups.
group-theory abelian-groups finitely-generated
edited Dec 21 '18 at 16:29
Shaun
9,759113684
asked Dec 20 '18 at 10:53
EasyEasy
3,535917
$endgroup$
$begingroup$
Well, it all depends on how you use words and definitions. For example, the "two generated" group $Bbb Z=langle,1,,,-1,rangle;$ doesn't fulfill $;Bbb Z=langle 1rangleopluslangle -1rangle;$ ...So what are you assuming about $;a,b;$ ?
$endgroup$
– DonAntonio
Dec 20 '18 at 11:06
1
$begingroup$
@DonAntonio Sorry, I forgot to mention it is non-cyclic.
$endgroup$
– Easy
Dec 20 '18 at 11:08
$begingroup$
But I don't think you need assume that it is noncyclic, because you could just choose $y$ to be the identity.
$endgroup$
– Derek Holt
Dec 20 '18 at 11:16
$begingroup$
@DerekHolt You can put it that way, but that would be not of interest in that case.
$endgroup$
– Easy
Dec 20 '18 at 11:42
add a comment |
$begingroup$
Given a non-cyclic 2-generated abelian group $G=langle a,brangle$, it seems straightforward from fundamental theorem of abelian groups that
$$Gconglangle xrangletimeslangle yrangle$$
for some $x,yin G$.
However, if you look closely, there will be still some tedious argument to show such a result. So I am looking for a slick proof that does NOT use fundamental theorem of abelian groups.
group-theory abelian-groups finitely-generated
edited Dec 21 '18 at 16:29
Shaun
9,759113684
asked Dec 20 '18 at 10:53
EasyEasy
3,535917
$endgroup$
Given a non-cyclic 2-generated abelian group $G=langle a,brangle$, it seems straightforward from fundamental theorem of abelian groups that
$$Gconglangle xrangletimeslangle yrangle$$
for some $x,yin G$.
However, if you look closely, there will be still some tedious argument to show such a result. So I am looking for a slick proof that does NOT use fundamental theorem of abelian groups.
group-theory abelian-groups finitely-generated
group-theory abelian-groups finitely-generated
edited Dec 21 '18 at 16:29
Shaun
9,759113684
asked Dec 20 '18 at 10:53
EasyEasy
3,535917
edited Dec 21 '18 at 16:29
Shaun
9,759113684
asked Dec 20 '18 at 10:53
EasyEasy
3,535917
edited Dec 21 '18 at 16:29
Shaun
9,759113684
edited Dec 21 '18 at 16:29
Shaun
9,759113684
edited Dec 21 '18 at 16:29
Shaun
9,759113684
9,759113684
asked Dec 20 '18 at 10:53
EasyEasy
3,535917
asked Dec 20 '18 at 10:53
EasyEasy
3,535917
asked Dec 20 '18 at 10:53
EasyEasy
3,535917
3,535917
$begingroup$
Well, it all depends on how you use words and definitions. For example, the "two generated" group $Bbb Z=langle,1,,,-1,rangle;$ doesn't fulfill $;Bbb Z=langle 1rangleopluslangle -1rangle;$ ...So what are you assuming about $;a,b;$ ?
$endgroup$
– DonAntonio
Dec 20 '18 at 11:06
1
$begingroup$
@DonAntonio Sorry, I forgot to mention it is non-cyclic.
$endgroup$
– Easy
Dec 20 '18 at 11:08
$begingroup$
But I don't think you need assume that it is noncyclic, because you could just choose $y$ to be the identity.
$endgroup$
– Derek Holt
Dec 20 '18 at 11:16
$begingroup$
@DerekHolt You can put it that way, but that would be not of interest in that case.
$endgroup$
– Easy
Dec 20 '18 at 11:42
add a comment |
$begingroup$
Well, it all depends on how you use words and definitions. For example, the "two generated" group $Bbb Z=langle,1,,,-1,rangle;$ doesn't fulfill $;Bbb Z=langle 1rangleopluslangle -1rangle;$ ...So what are you assuming about $;a,b;$ ?
$endgroup$
– DonAntonio
Dec 20 '18 at 11:06
1
$begingroup$
@DonAntonio Sorry, I forgot to mention it is non-cyclic.
$endgroup$
– Easy
Dec 20 '18 at 11:08
$begingroup$
But I don't think you need assume that it is noncyclic, because you could just choose $y$ to be the identity.
$endgroup$
– Derek Holt
Dec 20 '18 at 11:16
$begingroup$
@DerekHolt You can put it that way, but that would be not of interest in that case.
$endgroup$
– Easy
Dec 20 '18 at 11:42
$begingroup$
Well, it all depends on how you use words and definitions. For example, the "two generated" group $Bbb Z=langle,1,,,-1,rangle;$ doesn't fulfill $;Bbb Z=langle 1rangleopluslangle -1rangle;$ ...So what are you assuming about $;a,b;$ ?
$endgroup$
– DonAntonio
Dec 20 '18 at 11:06
$begingroup$
Well, it all depends on how you use words and definitions. For example, the "two generated" group $Bbb Z=langle,1,,,-1,rangle;$ doesn't fulfill $;Bbb Z=langle 1rangleopluslangle -1rangle;$ ...So what are you assuming about $;a,b;$ ?
$endgroup$
– DonAntonio
Dec 20 '18 at 11:06
1
1
$begingroup$
@DonAntonio Sorry, I forgot to mention it is non-cyclic.
$endgroup$
– Easy
Dec 20 '18 at 11:08
$begingroup$
@DonAntonio Sorry, I forgot to mention it is non-cyclic.
$endgroup$
– Easy
Dec 20 '18 at 11:08
$begingroup$
But I don't think you need assume that it is noncyclic, because you could just choose $y$ to be the identity.
$endgroup$
– Derek Holt
Dec 20 '18 at 11:16
$begingroup$
But I don't think you need assume that it is noncyclic, because you could just choose $y$ to be the identity.
$endgroup$
– Derek Holt
Dec 20 '18 at 11:16
$begingroup$
@DerekHolt You can put it that way, but that would be not of interest in that case.
$endgroup$
– Easy
Dec 20 '18 at 11:42
$begingroup$
@DerekHolt You can put it that way, but that would be not of interest in that case.
$endgroup$
– Easy
Dec 20 '18 at 11:42
add a comment |
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$begingroup$
Well, it all depends on how you use words and definitions. For example, the "two generated" group $Bbb Z=langle,1,,,-1,rangle;$ doesn't fulfill $;Bbb Z=langle 1rangleopluslangle -1rangle;$ ...So what are you assuming about $;a,b;$ ?
$endgroup$
– DonAntonio
Dec 20 '18 at 11:06
1
$begingroup$
@DonAntonio Sorry, I forgot to mention it is non-cyclic.
$endgroup$
– Easy
Dec 20 '18 at 11:08
$begingroup$
But I don't think you need assume that it is noncyclic, because you could just choose $y$ to be the identity.
$endgroup$
– Derek Holt
Dec 20 '18 at 11:16
$begingroup$
@DerekHolt You can put it that way, but that would be not of interest in that case.
$endgroup$
– Easy
Dec 20 '18 at 11:42