G is a group and $(ab)^3=a^3b^3$ for all $a,b in G$. Prove (or disprove with a counterexample) that if...
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G is a group and $(ab)^3=a^3b^3$ for all $a,b in G$. Prove (or disprove with a counterexample) that if $(ab)^3=(ba)^3$, then $ab=ba$.
The Next CEO of Stack OverflowA criterion for a group to be abelianAre these exactly the abelian groups (2)?Prove that a group with exponent 3 is abelian.Let $H triangleleft G$ such that $[G:H]=n$. Show that $g^n in H$ for all $g in G$.A fact regarding the interaction of powers and commutators in a groupProve that if $G$ is a group with $(ab)^{2}=a^{2}b^{2}$ for all $a,bin G$, then $G$ is abelian.If a group isomorphism is an involution with only a trivial fixed point, then the group is abelianIs it only the generator of the group that commutes with all the other elements?If $G$ has inverse of all elements, then $G$ is a group. $($ true / false$) ?$All the $m$th powers commute with each other and all the $n$th powers commute with each other, $m$ and $n$ relative prime, is abelian.
5
$begingroup$
Proposition. Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a,b in G$. If $(ab)^3=(ba)^3$, then $ab=ba$.
Is it true or false? So far I've only been able to prove that powers of $a$ commute with $b^3$ and powers of $b$ with $a^3$.
abstract-algebra group-theory abelian-groups
asked Mar 18 at 0:42
The FootprintThe Footprint
877
$endgroup$
add a comment |
5
$begingroup$
Proposition. Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a,b in G$. If $(ab)^3=(ba)^3$, then $ab=ba$.
Is it true or false? So far I've only been able to prove that powers of $a$ commute with $b^3$ and powers of $b$ with $a^3$.
abstract-algebra group-theory abelian-groups
asked Mar 18 at 0:42
The FootprintThe Footprint
877
$endgroup$
add a comment |
5
5
5
2
$begingroup$
Proposition. Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a,b in G$. If $(ab)^3=(ba)^3$, then $ab=ba$.
Is it true or false? So far I've only been able to prove that powers of $a$ commute with $b^3$ and powers of $b$ with $a^3$.
abstract-algebra group-theory abelian-groups
asked Mar 18 at 0:42
The FootprintThe Footprint
877
$endgroup$
Proposition. Let $G$ be a group such that $(ab)^3=a^3b^3$ for all $a,b in G$. If $(ab)^3=(ba)^3$, then $ab=ba$.
Is it true or false? So far I've only been able to prove that powers of $a$ commute with $b^3$ and powers of $b$ with $a^3$.
abstract-algebra group-theory abelian-groups
abstract-algebra group-theory abelian-groups
asked Mar 18 at 0:42
The FootprintThe Footprint
877
asked Mar 18 at 0:42
The FootprintThe Footprint
877
edited Mar 18 at 12:01
The Footprint
asked Mar 18 at 0:42
The FootprintThe Footprint
877
asked Mar 18 at 0:42
The FootprintThe Footprint
877
asked Mar 18 at 0:42
The FootprintThe Footprint
877
877
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
5
$begingroup$
Hint:
Let $U(3, mathbb F_3)$ be the group of all $3 times 3$ upper triangular matrices with all diagonal entries $1$, over the field $mathbb F_3$. Thus, the elements are all the matrices of the form $$begin{bmatrix}1 & a & b\0 & 1 & c\0 & 0 & 1end{bmatrix}$$ where $a, b, c in mathbb F_3 = {0,1,2}$, the field of order $3$.
- What is the exponent of the group [the least positive $n$ such that $g^n = 1$ for all group elements $g$]? Or: Determine the order of each element.
- Is the group Abelian?
- What do Points 1 and 2 imply about the status of the Proposition in this group?
answered Mar 18 at 2:09
M. VinayM. Vinay
7,22822135
$endgroup$
1
$begingroup$
@TheFootprint That's an odd imposition (doesn't really make sense as written), but let's say that you insist that the group should not have exponent $3$, so there should be at least one element $g$ such that $g^3 /ne 1$. Fine, let $H = U(3, mathbb F_3)$ be the group described above, and let $K$ be your favourite Abelian group with exponent not equal to $3$ (say it $|K|$ is even, e.g.). Let $G = H times K$ (direct product). Now $[(h_1, k_1)(h_2, k_2)]^3 = [(h_2, k_2)(h_1, k_1)]^3$ (Because the components get cubed and multiplied independently, and in each respective group the cubes commute!)
$endgroup$
– M. Vinay
Mar 18 at 12:19
1
$begingroup$
So this ↑ $G$ also satisfies the condition of the Proposition, but it's still not Abelian (Because the direct factor $H$ is not). However its exponent is not $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:20
1
$begingroup$
@TheFootprint I guess it's a somewhat standard example. Personally I know it from studying power graphs of groups, where it is an important counterexample (for almost the same reason). The exponent of $G$ being the least positive integer $n$ (if any) such that $g^n = 1 forall g in G$, it must be divisible by the order of every element of the group. Any finite group of even order has an element of order $2$ [this is a good basic exercise!], so its exponent cannot be $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:40
1
$begingroup$
@TheFootprint Just to be clear: The exponent is not always the "maximum" order. For example, the maximum order of an element of $S_3$ is $3$, but the exponent of $S_3$ is $6$ (because there are some elements of order $2$). In other words, the exponent need not be the order of any element of the group. However, that is in a non-Abelian group. In an Abelian group of finite exponent, there is always an element of order equal to the exponent!
$endgroup$
– M. Vinay
Mar 18 at 13:38
1
$begingroup$
@TheFootprint Yes, that makes for a more involved exercise. You may first need to make a couple of observations about how element orders work in Abelian groups, then prove this. The proof of this itself needs some thinking, maybe some small construction — so in short, it's the kind of proof that is very common in group theory (especially finite group theory).
$endgroup$
– M. Vinay
Mar 19 at 1:38
|
show 3 more comments
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1 Answer
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1 Answer
1
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oldest
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oldest
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5
$begingroup$
Hint:
Let $U(3, mathbb F_3)$ be the group of all $3 times 3$ upper triangular matrices with all diagonal entries $1$, over the field $mathbb F_3$. Thus, the elements are all the matrices of the form $$begin{bmatrix}1 & a & b\0 & 1 & c\0 & 0 & 1end{bmatrix}$$ where $a, b, c in mathbb F_3 = {0,1,2}$, the field of order $3$.
- What is the exponent of the group [the least positive $n$ such that $g^n = 1$ for all group elements $g$]? Or: Determine the order of each element.
- Is the group Abelian?
- What do Points 1 and 2 imply about the status of the Proposition in this group?
answered Mar 18 at 2:09
M. VinayM. Vinay
7,22822135
$endgroup$
1
$begingroup$
@TheFootprint That's an odd imposition (doesn't really make sense as written), but let's say that you insist that the group should not have exponent $3$, so there should be at least one element $g$ such that $g^3 /ne 1$. Fine, let $H = U(3, mathbb F_3)$ be the group described above, and let $K$ be your favourite Abelian group with exponent not equal to $3$ (say it $|K|$ is even, e.g.). Let $G = H times K$ (direct product). Now $[(h_1, k_1)(h_2, k_2)]^3 = [(h_2, k_2)(h_1, k_1)]^3$ (Because the components get cubed and multiplied independently, and in each respective group the cubes commute!)
$endgroup$
– M. Vinay
Mar 18 at 12:19
1
$begingroup$
So this ↑ $G$ also satisfies the condition of the Proposition, but it's still not Abelian (Because the direct factor $H$ is not). However its exponent is not $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:20
1
$begingroup$
@TheFootprint I guess it's a somewhat standard example. Personally I know it from studying power graphs of groups, where it is an important counterexample (for almost the same reason). The exponent of $G$ being the least positive integer $n$ (if any) such that $g^n = 1 forall g in G$, it must be divisible by the order of every element of the group. Any finite group of even order has an element of order $2$ [this is a good basic exercise!], so its exponent cannot be $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:40
1
$begingroup$
@TheFootprint Just to be clear: The exponent is not always the "maximum" order. For example, the maximum order of an element of $S_3$ is $3$, but the exponent of $S_3$ is $6$ (because there are some elements of order $2$). In other words, the exponent need not be the order of any element of the group. However, that is in a non-Abelian group. In an Abelian group of finite exponent, there is always an element of order equal to the exponent!
$endgroup$
– M. Vinay
Mar 18 at 13:38
1
$begingroup$
@TheFootprint Yes, that makes for a more involved exercise. You may first need to make a couple of observations about how element orders work in Abelian groups, then prove this. The proof of this itself needs some thinking, maybe some small construction — so in short, it's the kind of proof that is very common in group theory (especially finite group theory).
$endgroup$
– M. Vinay
Mar 19 at 1:38
|
show 3 more comments
5
$begingroup$
Hint:
Let $U(3, mathbb F_3)$ be the group of all $3 times 3$ upper triangular matrices with all diagonal entries $1$, over the field $mathbb F_3$. Thus, the elements are all the matrices of the form $$begin{bmatrix}1 & a & b\0 & 1 & c\0 & 0 & 1end{bmatrix}$$ where $a, b, c in mathbb F_3 = {0,1,2}$, the field of order $3$.
- What is the exponent of the group [the least positive $n$ such that $g^n = 1$ for all group elements $g$]? Or: Determine the order of each element.
- Is the group Abelian?
- What do Points 1 and 2 imply about the status of the Proposition in this group?
answered Mar 18 at 2:09
M. VinayM. Vinay
7,22822135
$endgroup$
1
$begingroup$
@TheFootprint That's an odd imposition (doesn't really make sense as written), but let's say that you insist that the group should not have exponent $3$, so there should be at least one element $g$ such that $g^3 /ne 1$. Fine, let $H = U(3, mathbb F_3)$ be the group described above, and let $K$ be your favourite Abelian group with exponent not equal to $3$ (say it $|K|$ is even, e.g.). Let $G = H times K$ (direct product). Now $[(h_1, k_1)(h_2, k_2)]^3 = [(h_2, k_2)(h_1, k_1)]^3$ (Because the components get cubed and multiplied independently, and in each respective group the cubes commute!)
$endgroup$
– M. Vinay
Mar 18 at 12:19
1
$begingroup$
So this ↑ $G$ also satisfies the condition of the Proposition, but it's still not Abelian (Because the direct factor $H$ is not). However its exponent is not $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:20
1
$begingroup$
@TheFootprint I guess it's a somewhat standard example. Personally I know it from studying power graphs of groups, where it is an important counterexample (for almost the same reason). The exponent of $G$ being the least positive integer $n$ (if any) such that $g^n = 1 forall g in G$, it must be divisible by the order of every element of the group. Any finite group of even order has an element of order $2$ [this is a good basic exercise!], so its exponent cannot be $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:40
1
$begingroup$
@TheFootprint Just to be clear: The exponent is not always the "maximum" order. For example, the maximum order of an element of $S_3$ is $3$, but the exponent of $S_3$ is $6$ (because there are some elements of order $2$). In other words, the exponent need not be the order of any element of the group. However, that is in a non-Abelian group. In an Abelian group of finite exponent, there is always an element of order equal to the exponent!
$endgroup$
– M. Vinay
Mar 18 at 13:38
1
$begingroup$
@TheFootprint Yes, that makes for a more involved exercise. You may first need to make a couple of observations about how element orders work in Abelian groups, then prove this. The proof of this itself needs some thinking, maybe some small construction — so in short, it's the kind of proof that is very common in group theory (especially finite group theory).
$endgroup$
– M. Vinay
Mar 19 at 1:38
|
show 3 more comments
5
5
5
$begingroup$
Hint:
Let $U(3, mathbb F_3)$ be the group of all $3 times 3$ upper triangular matrices with all diagonal entries $1$, over the field $mathbb F_3$. Thus, the elements are all the matrices of the form $$begin{bmatrix}1 & a & b\0 & 1 & c\0 & 0 & 1end{bmatrix}$$ where $a, b, c in mathbb F_3 = {0,1,2}$, the field of order $3$.
- What is the exponent of the group [the least positive $n$ such that $g^n = 1$ for all group elements $g$]? Or: Determine the order of each element.
- Is the group Abelian?
- What do Points 1 and 2 imply about the status of the Proposition in this group?
answered Mar 18 at 2:09
M. VinayM. Vinay
7,22822135
$endgroup$
Hint:
Let $U(3, mathbb F_3)$ be the group of all $3 times 3$ upper triangular matrices with all diagonal entries $1$, over the field $mathbb F_3$. Thus, the elements are all the matrices of the form $$begin{bmatrix}1 & a & b\0 & 1 & c\0 & 0 & 1end{bmatrix}$$ where $a, b, c in mathbb F_3 = {0,1,2}$, the field of order $3$.
- What is the exponent of the group [the least positive $n$ such that $g^n = 1$ for all group elements $g$]? Or: Determine the order of each element.
- Is the group Abelian?
- What do Points 1 and 2 imply about the status of the Proposition in this group?
answered Mar 18 at 2:09
M. VinayM. Vinay
7,22822135
edited Mar 18 at 2:48
answered Mar 18 at 2:09
M. VinayM. Vinay
7,22822135
answered Mar 18 at 2:09
M. VinayM. Vinay
7,22822135
answered Mar 18 at 2:09
M. VinayM. Vinay
7,22822135
7,22822135
1
$begingroup$
@TheFootprint That's an odd imposition (doesn't really make sense as written), but let's say that you insist that the group should not have exponent $3$, so there should be at least one element $g$ such that $g^3 /ne 1$. Fine, let $H = U(3, mathbb F_3)$ be the group described above, and let $K$ be your favourite Abelian group with exponent not equal to $3$ (say it $|K|$ is even, e.g.). Let $G = H times K$ (direct product). Now $[(h_1, k_1)(h_2, k_2)]^3 = [(h_2, k_2)(h_1, k_1)]^3$ (Because the components get cubed and multiplied independently, and in each respective group the cubes commute!)
$endgroup$
– M. Vinay
Mar 18 at 12:19
1
$begingroup$
So this ↑ $G$ also satisfies the condition of the Proposition, but it's still not Abelian (Because the direct factor $H$ is not). However its exponent is not $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:20
1
$begingroup$
@TheFootprint I guess it's a somewhat standard example. Personally I know it from studying power graphs of groups, where it is an important counterexample (for almost the same reason). The exponent of $G$ being the least positive integer $n$ (if any) such that $g^n = 1 forall g in G$, it must be divisible by the order of every element of the group. Any finite group of even order has an element of order $2$ [this is a good basic exercise!], so its exponent cannot be $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:40
1
$begingroup$
@TheFootprint Just to be clear: The exponent is not always the "maximum" order. For example, the maximum order of an element of $S_3$ is $3$, but the exponent of $S_3$ is $6$ (because there are some elements of order $2$). In other words, the exponent need not be the order of any element of the group. However, that is in a non-Abelian group. In an Abelian group of finite exponent, there is always an element of order equal to the exponent!
$endgroup$
– M. Vinay
Mar 18 at 13:38
1
$begingroup$
@TheFootprint Yes, that makes for a more involved exercise. You may first need to make a couple of observations about how element orders work in Abelian groups, then prove this. The proof of this itself needs some thinking, maybe some small construction — so in short, it's the kind of proof that is very common in group theory (especially finite group theory).
$endgroup$
– M. Vinay
Mar 19 at 1:38
|
show 3 more comments
1
$begingroup$
@TheFootprint That's an odd imposition (doesn't really make sense as written), but let's say that you insist that the group should not have exponent $3$, so there should be at least one element $g$ such that $g^3 /ne 1$. Fine, let $H = U(3, mathbb F_3)$ be the group described above, and let $K$ be your favourite Abelian group with exponent not equal to $3$ (say it $|K|$ is even, e.g.). Let $G = H times K$ (direct product). Now $[(h_1, k_1)(h_2, k_2)]^3 = [(h_2, k_2)(h_1, k_1)]^3$ (Because the components get cubed and multiplied independently, and in each respective group the cubes commute!)
$endgroup$
– M. Vinay
Mar 18 at 12:19
1
$begingroup$
So this ↑ $G$ also satisfies the condition of the Proposition, but it's still not Abelian (Because the direct factor $H$ is not). However its exponent is not $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:20
1
$begingroup$
@TheFootprint I guess it's a somewhat standard example. Personally I know it from studying power graphs of groups, where it is an important counterexample (for almost the same reason). The exponent of $G$ being the least positive integer $n$ (if any) such that $g^n = 1 forall g in G$, it must be divisible by the order of every element of the group. Any finite group of even order has an element of order $2$ [this is a good basic exercise!], so its exponent cannot be $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:40
1
$begingroup$
@TheFootprint Just to be clear: The exponent is not always the "maximum" order. For example, the maximum order of an element of $S_3$ is $3$, but the exponent of $S_3$ is $6$ (because there are some elements of order $2$). In other words, the exponent need not be the order of any element of the group. However, that is in a non-Abelian group. In an Abelian group of finite exponent, there is always an element of order equal to the exponent!
$endgroup$
– M. Vinay
Mar 18 at 13:38
1
$begingroup$
@TheFootprint Yes, that makes for a more involved exercise. You may first need to make a couple of observations about how element orders work in Abelian groups, then prove this. The proof of this itself needs some thinking, maybe some small construction — so in short, it's the kind of proof that is very common in group theory (especially finite group theory).
$endgroup$
– M. Vinay
Mar 19 at 1:38
1
1
$begingroup$
@TheFootprint That's an odd imposition (doesn't really make sense as written), but let's say that you insist that the group should not have exponent $3$, so there should be at least one element $g$ such that $g^3 /ne 1$. Fine, let $H = U(3, mathbb F_3)$ be the group described above, and let $K$ be your favourite Abelian group with exponent not equal to $3$ (say it $|K|$ is even, e.g.). Let $G = H times K$ (direct product). Now $[(h_1, k_1)(h_2, k_2)]^3 = [(h_2, k_2)(h_1, k_1)]^3$ (Because the components get cubed and multiplied independently, and in each respective group the cubes commute!)
$endgroup$
– M. Vinay
Mar 18 at 12:19
$begingroup$
@TheFootprint That's an odd imposition (doesn't really make sense as written), but let's say that you insist that the group should not have exponent $3$, so there should be at least one element $g$ such that $g^3 /ne 1$. Fine, let $H = U(3, mathbb F_3)$ be the group described above, and let $K$ be your favourite Abelian group with exponent not equal to $3$ (say it $|K|$ is even, e.g.). Let $G = H times K$ (direct product). Now $[(h_1, k_1)(h_2, k_2)]^3 = [(h_2, k_2)(h_1, k_1)]^3$ (Because the components get cubed and multiplied independently, and in each respective group the cubes commute!)
$endgroup$
– M. Vinay
Mar 18 at 12:19
1
1
$begingroup$
So this ↑ $G$ also satisfies the condition of the Proposition, but it's still not Abelian (Because the direct factor $H$ is not). However its exponent is not $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:20
$begingroup$
So this ↑ $G$ also satisfies the condition of the Proposition, but it's still not Abelian (Because the direct factor $H$ is not). However its exponent is not $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:20
1
1
$begingroup$
@TheFootprint I guess it's a somewhat standard example. Personally I know it from studying power graphs of groups, where it is an important counterexample (for almost the same reason). The exponent of $G$ being the least positive integer $n$ (if any) such that $g^n = 1 forall g in G$, it must be divisible by the order of every element of the group. Any finite group of even order has an element of order $2$ [this is a good basic exercise!], so its exponent cannot be $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:40
$begingroup$
@TheFootprint I guess it's a somewhat standard example. Personally I know it from studying power graphs of groups, where it is an important counterexample (for almost the same reason). The exponent of $G$ being the least positive integer $n$ (if any) such that $g^n = 1 forall g in G$, it must be divisible by the order of every element of the group. Any finite group of even order has an element of order $2$ [this is a good basic exercise!], so its exponent cannot be $3$.
$endgroup$
– M. Vinay
Mar 18 at 12:40
1
1
$begingroup$
@TheFootprint Just to be clear: The exponent is not always the "maximum" order. For example, the maximum order of an element of $S_3$ is $3$, but the exponent of $S_3$ is $6$ (because there are some elements of order $2$). In other words, the exponent need not be the order of any element of the group. However, that is in a non-Abelian group. In an Abelian group of finite exponent, there is always an element of order equal to the exponent!
$endgroup$
– M. Vinay
Mar 18 at 13:38
$begingroup$
@TheFootprint Just to be clear: The exponent is not always the "maximum" order. For example, the maximum order of an element of $S_3$ is $3$, but the exponent of $S_3$ is $6$ (because there are some elements of order $2$). In other words, the exponent need not be the order of any element of the group. However, that is in a non-Abelian group. In an Abelian group of finite exponent, there is always an element of order equal to the exponent!
$endgroup$
– M. Vinay
Mar 18 at 13:38
1
1
$begingroup$
@TheFootprint Yes, that makes for a more involved exercise. You may first need to make a couple of observations about how element orders work in Abelian groups, then prove this. The proof of this itself needs some thinking, maybe some small construction — so in short, it's the kind of proof that is very common in group theory (especially finite group theory).
$endgroup$
– M. Vinay
Mar 19 at 1:38
$begingroup$
@TheFootprint Yes, that makes for a more involved exercise. You may first need to make a couple of observations about how element orders work in Abelian groups, then prove this. The proof of this itself needs some thinking, maybe some small construction — so in short, it's the kind of proof that is very common in group theory (especially finite group theory).
$endgroup$
– M. Vinay
Mar 19 at 1:38
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