The conditions $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$ make the group $G$ abelian?
$begingroup$
A group $G$ satisfies $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$. Prove that the group is abelian.
Also please tell me whether there is any standard approach in proving commutativity of groups like this problem.
abstract-algebra group-theory
edited Oct 25 '13 at 9:35
mrs
1
asked Oct 25 '13 at 7:20
sauviksauvik
993
$endgroup$
add a comment |
$begingroup$
A group $G$ satisfies $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$. Prove that the group is abelian.
Also please tell me whether there is any standard approach in proving commutativity of groups like this problem.
abstract-algebra group-theory
edited Oct 25 '13 at 9:35
mrs
1
asked Oct 25 '13 at 7:20
sauviksauvik
993
$endgroup$
$begingroup$
Can you share your thoughts on the problem, and explain what you've tried and what's giving you difficulty?
$endgroup$
– user61527
Oct 25 '13 at 7:21
$begingroup$
The standard approach is to prove that ab = ba, using whatever are given.
$endgroup$
– Mai
Oct 25 '13 at 7:28
add a comment |
$begingroup$
A group $G$ satisfies $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$. Prove that the group is abelian.
Also please tell me whether there is any standard approach in proving commutativity of groups like this problem.
abstract-algebra group-theory
edited Oct 25 '13 at 9:35
mrs
1
asked Oct 25 '13 at 7:20
sauviksauvik
993
$endgroup$
A group $G$ satisfies $a^2b^2=b^2a^2$ and $a^3b^3=b^3a^3$. Prove that the group is abelian.
Also please tell me whether there is any standard approach in proving commutativity of groups like this problem.
abstract-algebra group-theory
abstract-algebra group-theory
edited Oct 25 '13 at 9:35
mrs
1
asked Oct 25 '13 at 7:20
sauviksauvik
993
edited Oct 25 '13 at 9:35
mrs
1
asked Oct 25 '13 at 7:20
sauviksauvik
993
edited Oct 25 '13 at 9:35
mrs
1
edited Oct 25 '13 at 9:35
mrs
1
edited Oct 25 '13 at 9:35
mrs
1
1
asked Oct 25 '13 at 7:20
sauviksauvik
993
asked Oct 25 '13 at 7:20
sauviksauvik
993
asked Oct 25 '13 at 7:20
sauviksauvik
993
993
$begingroup$
Can you share your thoughts on the problem, and explain what you've tried and what's giving you difficulty?
$endgroup$
– user61527
Oct 25 '13 at 7:21
$begingroup$
The standard approach is to prove that ab = ba, using whatever are given.
$endgroup$
– Mai
Oct 25 '13 at 7:28
add a comment |
$begingroup$
Can you share your thoughts on the problem, and explain what you've tried and what's giving you difficulty?
$endgroup$
– user61527
Oct 25 '13 at 7:21
$begingroup$
The standard approach is to prove that ab = ba, using whatever are given.
$endgroup$
– Mai
Oct 25 '13 at 7:28
$begingroup$
Can you share your thoughts on the problem, and explain what you've tried and what's giving you difficulty?
$endgroup$
– user61527
Oct 25 '13 at 7:21
$begingroup$
Can you share your thoughts on the problem, and explain what you've tried and what's giving you difficulty?
$endgroup$
– user61527
Oct 25 '13 at 7:21
$begingroup$
The standard approach is to prove that ab = ba, using whatever are given.
$endgroup$
– Mai
Oct 25 '13 at 7:28
$begingroup$
The standard approach is to prove that ab = ba, using whatever are given.
$endgroup$
– Mai
Oct 25 '13 at 7:28
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
G does not have to be finite. Let $M subset G$ be the subgroup generated by all squares and let $N subset G$ be the subgroup generated by all cubes. These subgroups are clearly abelian normal subgroups. Since 2 and 3 are coprime $G = MN$ (use Bézout's Theorem!) and $M cap N$ is contained in the center $Z(G)$ of $G$.
To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] subseteq (M cap N)$. Let $x in M$ and $y in N$.
Then $[x, y] = x^{−1}y^{−1}xy in M cap N$. Hence $[x, y] = z$ with $z in Z(G)$. Hence $y^{−1}xy = zx$, whence $y^{−1}x^3y=z^3x^3$. Since $x^3 in N$ it commutes with $y$, so $z^3=1$. Similarly $z^2=1$. Since $2$ and $3$ are relatively prime, we conclude $z=1$.
answered Oct 25 '13 at 19:59
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
$begingroup$
Why is $Mcap N$ contained in $Z(G)$ ?
$endgroup$
– Ewan Delanoy
Oct 25 '13 at 20:05
$begingroup$
Take $g in G$, then $g=mn$, with $m in M$ and $n in N$. If $x in M cap N$, then $xg=(xm)n=(mx)n$ because $M$ is abelian and $x in M$. And $(mx)n=m(xn)=m(nx)$, because $N$ is abelian and $x in N$. Hence $xg=m(nx)=(mn)x=gx$ and so $x in Z(G)$.
$endgroup$
– Nicky Hekster
Oct 25 '13 at 21:18
$begingroup$
Right. So your argument works in any case when $a^p$ commutes with $b^p$ and $a^q$ commutes with $b^q$ for any coprime pair of integers $(p,q)$, not just $(p,q)=(2,3)$.
$endgroup$
– Ewan Delanoy
Oct 26 '13 at 4:30
$begingroup$
@Ewan that is correct, it works for any pair of non-zero coprime integers!
$endgroup$
– Nicky Hekster
Oct 26 '13 at 11:26
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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active
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votes
$begingroup$
G does not have to be finite. Let $M subset G$ be the subgroup generated by all squares and let $N subset G$ be the subgroup generated by all cubes. These subgroups are clearly abelian normal subgroups. Since 2 and 3 are coprime $G = MN$ (use Bézout's Theorem!) and $M cap N$ is contained in the center $Z(G)$ of $G$.
To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] subseteq (M cap N)$. Let $x in M$ and $y in N$.
Then $[x, y] = x^{−1}y^{−1}xy in M cap N$. Hence $[x, y] = z$ with $z in Z(G)$. Hence $y^{−1}xy = zx$, whence $y^{−1}x^3y=z^3x^3$. Since $x^3 in N$ it commutes with $y$, so $z^3=1$. Similarly $z^2=1$. Since $2$ and $3$ are relatively prime, we conclude $z=1$.
answered Oct 25 '13 at 19:59
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
$begingroup$
Why is $Mcap N$ contained in $Z(G)$ ?
$endgroup$
– Ewan Delanoy
Oct 25 '13 at 20:05
$begingroup$
Take $g in G$, then $g=mn$, with $m in M$ and $n in N$. If $x in M cap N$, then $xg=(xm)n=(mx)n$ because $M$ is abelian and $x in M$. And $(mx)n=m(xn)=m(nx)$, because $N$ is abelian and $x in N$. Hence $xg=m(nx)=(mn)x=gx$ and so $x in Z(G)$.
$endgroup$
– Nicky Hekster
Oct 25 '13 at 21:18
$begingroup$
Right. So your argument works in any case when $a^p$ commutes with $b^p$ and $a^q$ commutes with $b^q$ for any coprime pair of integers $(p,q)$, not just $(p,q)=(2,3)$.
$endgroup$
– Ewan Delanoy
Oct 26 '13 at 4:30
$begingroup$
@Ewan that is correct, it works for any pair of non-zero coprime integers!
$endgroup$
– Nicky Hekster
Oct 26 '13 at 11:26
add a comment |
$begingroup$
G does not have to be finite. Let $M subset G$ be the subgroup generated by all squares and let $N subset G$ be the subgroup generated by all cubes. These subgroups are clearly abelian normal subgroups. Since 2 and 3 are coprime $G = MN$ (use Bézout's Theorem!) and $M cap N$ is contained in the center $Z(G)$ of $G$.
To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] subseteq (M cap N)$. Let $x in M$ and $y in N$.
Then $[x, y] = x^{−1}y^{−1}xy in M cap N$. Hence $[x, y] = z$ with $z in Z(G)$. Hence $y^{−1}xy = zx$, whence $y^{−1}x^3y=z^3x^3$. Since $x^3 in N$ it commutes with $y$, so $z^3=1$. Similarly $z^2=1$. Since $2$ and $3$ are relatively prime, we conclude $z=1$.
answered Oct 25 '13 at 19:59
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
$begingroup$
Why is $Mcap N$ contained in $Z(G)$ ?
$endgroup$
– Ewan Delanoy
Oct 25 '13 at 20:05
$begingroup$
Take $g in G$, then $g=mn$, with $m in M$ and $n in N$. If $x in M cap N$, then $xg=(xm)n=(mx)n$ because $M$ is abelian and $x in M$. And $(mx)n=m(xn)=m(nx)$, because $N$ is abelian and $x in N$. Hence $xg=m(nx)=(mn)x=gx$ and so $x in Z(G)$.
$endgroup$
– Nicky Hekster
Oct 25 '13 at 21:18
$begingroup$
Right. So your argument works in any case when $a^p$ commutes with $b^p$ and $a^q$ commutes with $b^q$ for any coprime pair of integers $(p,q)$, not just $(p,q)=(2,3)$.
$endgroup$
– Ewan Delanoy
Oct 26 '13 at 4:30
$begingroup$
@Ewan that is correct, it works for any pair of non-zero coprime integers!
$endgroup$
– Nicky Hekster
Oct 26 '13 at 11:26
add a comment |
$begingroup$
G does not have to be finite. Let $M subset G$ be the subgroup generated by all squares and let $N subset G$ be the subgroup generated by all cubes. These subgroups are clearly abelian normal subgroups. Since 2 and 3 are coprime $G = MN$ (use Bézout's Theorem!) and $M cap N$ is contained in the center $Z(G)$ of $G$.
To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] subseteq (M cap N)$. Let $x in M$ and $y in N$.
Then $[x, y] = x^{−1}y^{−1}xy in M cap N$. Hence $[x, y] = z$ with $z in Z(G)$. Hence $y^{−1}xy = zx$, whence $y^{−1}x^3y=z^3x^3$. Since $x^3 in N$ it commutes with $y$, so $z^3=1$. Similarly $z^2=1$. Since $2$ and $3$ are relatively prime, we conclude $z=1$.
answered Oct 25 '13 at 19:59
Nicky HeksterNicky Hekster
28.4k53456
$endgroup$
G does not have to be finite. Let $M subset G$ be the subgroup generated by all squares and let $N subset G$ be the subgroup generated by all cubes. These subgroups are clearly abelian normal subgroups. Since 2 and 3 are coprime $G = MN$ (use Bézout's Theorem!) and $M cap N$ is contained in the center $Z(G)$ of $G$.
To prove that $G$ is abelian it suffices to show that $M$ and $N$ commute, that is $[M,N]=1$. Note that $[M,N] subseteq (M cap N)$. Let $x in M$ and $y in N$.
Then $[x, y] = x^{−1}y^{−1}xy in M cap N$. Hence $[x, y] = z$ with $z in Z(G)$. Hence $y^{−1}xy = zx$, whence $y^{−1}x^3y=z^3x^3$. Since $x^3 in N$ it commutes with $y$, so $z^3=1$. Similarly $z^2=1$. Since $2$ and $3$ are relatively prime, we conclude $z=1$.
answered Oct 25 '13 at 19:59
Nicky HeksterNicky Hekster
28.4k53456
edited Oct 25 '13 at 21:40
answered Oct 25 '13 at 19:59
Nicky HeksterNicky Hekster
28.4k53456
answered Oct 25 '13 at 19:59
Nicky HeksterNicky Hekster
28.4k53456
answered Oct 25 '13 at 19:59
Nicky HeksterNicky Hekster
28.4k53456
28.4k53456
$begingroup$
Why is $Mcap N$ contained in $Z(G)$ ?
$endgroup$
– Ewan Delanoy
Oct 25 '13 at 20:05
$begingroup$
Take $g in G$, then $g=mn$, with $m in M$ and $n in N$. If $x in M cap N$, then $xg=(xm)n=(mx)n$ because $M$ is abelian and $x in M$. And $(mx)n=m(xn)=m(nx)$, because $N$ is abelian and $x in N$. Hence $xg=m(nx)=(mn)x=gx$ and so $x in Z(G)$.
$endgroup$
– Nicky Hekster
Oct 25 '13 at 21:18
$begingroup$
Right. So your argument works in any case when $a^p$ commutes with $b^p$ and $a^q$ commutes with $b^q$ for any coprime pair of integers $(p,q)$, not just $(p,q)=(2,3)$.
$endgroup$
– Ewan Delanoy
Oct 26 '13 at 4:30
$begingroup$
@Ewan that is correct, it works for any pair of non-zero coprime integers!
$endgroup$
– Nicky Hekster
Oct 26 '13 at 11:26
add a comment |
$begingroup$
Why is $Mcap N$ contained in $Z(G)$ ?
$endgroup$
– Ewan Delanoy
Oct 25 '13 at 20:05
$begingroup$
Take $g in G$, then $g=mn$, with $m in M$ and $n in N$. If $x in M cap N$, then $xg=(xm)n=(mx)n$ because $M$ is abelian and $x in M$. And $(mx)n=m(xn)=m(nx)$, because $N$ is abelian and $x in N$. Hence $xg=m(nx)=(mn)x=gx$ and so $x in Z(G)$.
$endgroup$
– Nicky Hekster
Oct 25 '13 at 21:18
$begingroup$
Right. So your argument works in any case when $a^p$ commutes with $b^p$ and $a^q$ commutes with $b^q$ for any coprime pair of integers $(p,q)$, not just $(p,q)=(2,3)$.
$endgroup$
– Ewan Delanoy
Oct 26 '13 at 4:30
$begingroup$
@Ewan that is correct, it works for any pair of non-zero coprime integers!
$endgroup$
– Nicky Hekster
Oct 26 '13 at 11:26
$begingroup$
Why is $Mcap N$ contained in $Z(G)$ ?
$endgroup$
– Ewan Delanoy
Oct 25 '13 at 20:05
$begingroup$
Why is $Mcap N$ contained in $Z(G)$ ?
$endgroup$
– Ewan Delanoy
Oct 25 '13 at 20:05
$begingroup$
Take $g in G$, then $g=mn$, with $m in M$ and $n in N$. If $x in M cap N$, then $xg=(xm)n=(mx)n$ because $M$ is abelian and $x in M$. And $(mx)n=m(xn)=m(nx)$, because $N$ is abelian and $x in N$. Hence $xg=m(nx)=(mn)x=gx$ and so $x in Z(G)$.
$endgroup$
– Nicky Hekster
Oct 25 '13 at 21:18
$begingroup$
Take $g in G$, then $g=mn$, with $m in M$ and $n in N$. If $x in M cap N$, then $xg=(xm)n=(mx)n$ because $M$ is abelian and $x in M$. And $(mx)n=m(xn)=m(nx)$, because $N$ is abelian and $x in N$. Hence $xg=m(nx)=(mn)x=gx$ and so $x in Z(G)$.
$endgroup$
– Nicky Hekster
Oct 25 '13 at 21:18
$begingroup$
Right. So your argument works in any case when $a^p$ commutes with $b^p$ and $a^q$ commutes with $b^q$ for any coprime pair of integers $(p,q)$, not just $(p,q)=(2,3)$.
$endgroup$
– Ewan Delanoy
Oct 26 '13 at 4:30
$begingroup$
Right. So your argument works in any case when $a^p$ commutes with $b^p$ and $a^q$ commutes with $b^q$ for any coprime pair of integers $(p,q)$, not just $(p,q)=(2,3)$.
$endgroup$
– Ewan Delanoy
Oct 26 '13 at 4:30
$begingroup$
@Ewan that is correct, it works for any pair of non-zero coprime integers!
$endgroup$
– Nicky Hekster
Oct 26 '13 at 11:26
$begingroup$
@Ewan that is correct, it works for any pair of non-zero coprime integers!
$endgroup$
– Nicky Hekster
Oct 26 '13 at 11:26
add a comment |
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$begingroup$
Can you share your thoughts on the problem, and explain what you've tried and what's giving you difficulty?
$endgroup$
– user61527
Oct 25 '13 at 7:21
$begingroup$
The standard approach is to prove that ab = ba, using whatever are given.
$endgroup$
– Mai
Oct 25 '13 at 7:28