Mixing
If $operatorname{class}(G) = 2$ and $exp(G) = 4$ then $exp(G') = 2$?
$begingroup$
Let $G$ be a finite $p$-group. I'd like to prove (or disprove) that if the nilpotency class of $G$ equals two (i.e., $1 neq G' le Z$, where $Z$ is the center of $G$) and the exponent of $G$ equals four (i.e., $g^4 = 1$ for all $g in G$), then the exponent of $G'$ equals two.
Note: By a formula $[gh, k] = [g, k]^h[h, k]$, we have $[gh, k] = [g, k][h, k]$ because $G' le Z$. So $[g^2, h] = [g, h]^2$ in particular and our goal is equivalent to $G^2 le Z$, where $G^{p^i} := langle, g^{p^i} mid g in G ,rangle$.
I tried to prove it by showing $(G')^2 le G^4$ but I couldn't prove it so far. As $G' le G^2$ is true by a trick $[g, h] = (g^{-1})^2(gh^{-1})^2h^2$, I've been expecting something similar does the job. I also checked that the statement is true for $p$-groups of order dividing $2^6$ by a computer.
group-theory finite-groups p-groups nilpotent-groups
edited Jan 23 at 12:07
Shaun
9,366113684
asked Jan 23 at 7:13
OratOrat
2,88021231
$endgroup$
add a comment |
$begingroup$
Let $G$ be a finite $p$-group. I'd like to prove (or disprove) that if the nilpotency class of $G$ equals two (i.e., $1 neq G' le Z$, where $Z$ is the center of $G$) and the exponent of $G$ equals four (i.e., $g^4 = 1$ for all $g in G$), then the exponent of $G'$ equals two.
Note: By a formula $[gh, k] = [g, k]^h[h, k]$, we have $[gh, k] = [g, k][h, k]$ because $G' le Z$. So $[g^2, h] = [g, h]^2$ in particular and our goal is equivalent to $G^2 le Z$, where $G^{p^i} := langle, g^{p^i} mid g in G ,rangle$.
I tried to prove it by showing $(G')^2 le G^4$ but I couldn't prove it so far. As $G' le G^2$ is true by a trick $[g, h] = (g^{-1})^2(gh^{-1})^2h^2$, I've been expecting something similar does the job. I also checked that the statement is true for $p$-groups of order dividing $2^6$ by a computer.
group-theory finite-groups p-groups nilpotent-groups
edited Jan 23 at 12:07
Shaun
9,366113684
asked Jan 23 at 7:13
OratOrat
2,88021231
$endgroup$
$begingroup$
What code & software did you use to check the statement?
$endgroup$
– Shaun
Jan 23 at 12:07
1
$begingroup$
@Shaun I used GAP. A code is a fairly simple for-loop test usingAllSmallGroups.
$endgroup$
– Orat
Jan 23 at 13:24
add a comment |
$begingroup$
Let $G$ be a finite $p$-group. I'd like to prove (or disprove) that if the nilpotency class of $G$ equals two (i.e., $1 neq G' le Z$, where $Z$ is the center of $G$) and the exponent of $G$ equals four (i.e., $g^4 = 1$ for all $g in G$), then the exponent of $G'$ equals two.
Note: By a formula $[gh, k] = [g, k]^h[h, k]$, we have $[gh, k] = [g, k][h, k]$ because $G' le Z$. So $[g^2, h] = [g, h]^2$ in particular and our goal is equivalent to $G^2 le Z$, where $G^{p^i} := langle, g^{p^i} mid g in G ,rangle$.
I tried to prove it by showing $(G')^2 le G^4$ but I couldn't prove it so far. As $G' le G^2$ is true by a trick $[g, h] = (g^{-1})^2(gh^{-1})^2h^2$, I've been expecting something similar does the job. I also checked that the statement is true for $p$-groups of order dividing $2^6$ by a computer.
group-theory finite-groups p-groups nilpotent-groups
edited Jan 23 at 12:07
Shaun
9,366113684
asked Jan 23 at 7:13
OratOrat
2,88021231
$endgroup$
Let $G$ be a finite $p$-group. I'd like to prove (or disprove) that if the nilpotency class of $G$ equals two (i.e., $1 neq G' le Z$, where $Z$ is the center of $G$) and the exponent of $G$ equals four (i.e., $g^4 = 1$ for all $g in G$), then the exponent of $G'$ equals two.
Note: By a formula $[gh, k] = [g, k]^h[h, k]$, we have $[gh, k] = [g, k][h, k]$ because $G' le Z$. So $[g^2, h] = [g, h]^2$ in particular and our goal is equivalent to $G^2 le Z$, where $G^{p^i} := langle, g^{p^i} mid g in G ,rangle$.
I tried to prove it by showing $(G')^2 le G^4$ but I couldn't prove it so far. As $G' le G^2$ is true by a trick $[g, h] = (g^{-1})^2(gh^{-1})^2h^2$, I've been expecting something similar does the job. I also checked that the statement is true for $p$-groups of order dividing $2^6$ by a computer.
group-theory finite-groups p-groups nilpotent-groups
group-theory finite-groups p-groups nilpotent-groups
edited Jan 23 at 12:07
Shaun
9,366113684
asked Jan 23 at 7:13
OratOrat
2,88021231
edited Jan 23 at 12:07
Shaun
9,366113684
asked Jan 23 at 7:13
OratOrat
2,88021231
edited Jan 23 at 12:07
Shaun
9,366113684
edited Jan 23 at 12:07
Shaun
9,366113684
edited Jan 23 at 12:07
Shaun
9,366113684
9,366113684
asked Jan 23 at 7:13
OratOrat
2,88021231
asked Jan 23 at 7:13
OratOrat
2,88021231
asked Jan 23 at 7:13
OratOrat
2,88021231
2,88021231
$begingroup$
What code & software did you use to check the statement?
$endgroup$
– Shaun
Jan 23 at 12:07
1
$begingroup$
@Shaun I used GAP. A code is a fairly simple for-loop test usingAllSmallGroups.
$endgroup$
– Orat
Jan 23 at 13:24
add a comment |
$begingroup$
What code & software did you use to check the statement?
$endgroup$
– Shaun
Jan 23 at 12:07
1
$begingroup$
@Shaun I used GAP. A code is a fairly simple for-loop test usingAllSmallGroups.
$endgroup$
– Orat
Jan 23 at 13:24
$begingroup$
What code & software did you use to check the statement?
$endgroup$
– Shaun
Jan 23 at 12:07
$begingroup$
What code & software did you use to check the statement?
$endgroup$
– Shaun
Jan 23 at 12:07
1
1
$begingroup$
@Shaun I used GAP. A code is a fairly simple for-loop test using
AllSmallGroups.$endgroup$
– Orat
Jan 23 at 13:24
$begingroup$
@Shaun I used GAP. A code is a fairly simple for-loop test using
AllSmallGroups.$endgroup$
– Orat
Jan 23 at 13:24
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have $$[g,h]^2 = (g^{-1}h^{-1}gh)^2 = (g^{-2}gh^{-2}hgh)^2.$$
Now collecting the terms $g^{-2}$ and $h^{-2}$ and the resulting commutators (which are central) to the left gives $$g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4 = g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}].$$
But commutators are bilinear in groups of class $2$, and $[g^2,h^2]=[g,h]^4=1$, so this simplifies to $g^{-4}h^{-4}[g^4,h^{-2}][h^2,g^{-2}]=1$.
answered Jan 23 at 8:44
Derek HoltDerek Holt
54.2k53571
$endgroup$
$begingroup$
Thank you for your answer. But I'm afraid I don't understand how $(g^{-2}gh^{-2}hgh)^2$ leads to $g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4$. Could you elaborate it a little more, please?
$endgroup$
– Orat
Jan 23 at 9:39
2
$begingroup$
This is a standard process known as collection to the left. It's just repeated use of the identity $ba=ab[b,a]$, so for example, the collection process starts on the left of the word with $g^{-2}gh^{-2}hcdots = g^{-2}h^{-2}g[g,h^{-2}]hcdots = g^{-2}h^{-2}[g,h^{-2}]ghcdots$.
$endgroup$
– Derek Holt
Jan 23 at 9:55
$begingroup$
Ah! I should've noticed it by myself. Thank you so much for your help.
$endgroup$
– Orat
Jan 23 at 10:02
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
We have $$[g,h]^2 = (g^{-1}h^{-1}gh)^2 = (g^{-2}gh^{-2}hgh)^2.$$
Now collecting the terms $g^{-2}$ and $h^{-2}$ and the resulting commutators (which are central) to the left gives $$g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4 = g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}].$$
But commutators are bilinear in groups of class $2$, and $[g^2,h^2]=[g,h]^4=1$, so this simplifies to $g^{-4}h^{-4}[g^4,h^{-2}][h^2,g^{-2}]=1$.
answered Jan 23 at 8:44
Derek HoltDerek Holt
54.2k53571
$endgroup$
$begingroup$
Thank you for your answer. But I'm afraid I don't understand how $(g^{-2}gh^{-2}hgh)^2$ leads to $g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4$. Could you elaborate it a little more, please?
$endgroup$
– Orat
Jan 23 at 9:39
2
$begingroup$
This is a standard process known as collection to the left. It's just repeated use of the identity $ba=ab[b,a]$, so for example, the collection process starts on the left of the word with $g^{-2}gh^{-2}hcdots = g^{-2}h^{-2}g[g,h^{-2}]hcdots = g^{-2}h^{-2}[g,h^{-2}]ghcdots$.
$endgroup$
– Derek Holt
Jan 23 at 9:55
$begingroup$
Ah! I should've noticed it by myself. Thank you so much for your help.
$endgroup$
– Orat
Jan 23 at 10:02
add a comment |
$begingroup$
We have $$[g,h]^2 = (g^{-1}h^{-1}gh)^2 = (g^{-2}gh^{-2}hgh)^2.$$
Now collecting the terms $g^{-2}$ and $h^{-2}$ and the resulting commutators (which are central) to the left gives $$g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4 = g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}].$$
But commutators are bilinear in groups of class $2$, and $[g^2,h^2]=[g,h]^4=1$, so this simplifies to $g^{-4}h^{-4}[g^4,h^{-2}][h^2,g^{-2}]=1$.
answered Jan 23 at 8:44
Derek HoltDerek Holt
54.2k53571
$endgroup$
$begingroup$
Thank you for your answer. But I'm afraid I don't understand how $(g^{-2}gh^{-2}hgh)^2$ leads to $g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4$. Could you elaborate it a little more, please?
$endgroup$
– Orat
Jan 23 at 9:39
2
$begingroup$
This is a standard process known as collection to the left. It's just repeated use of the identity $ba=ab[b,a]$, so for example, the collection process starts on the left of the word with $g^{-2}gh^{-2}hcdots = g^{-2}h^{-2}g[g,h^{-2}]hcdots = g^{-2}h^{-2}[g,h^{-2}]ghcdots$.
$endgroup$
– Derek Holt
Jan 23 at 9:55
$begingroup$
Ah! I should've noticed it by myself. Thank you so much for your help.
$endgroup$
– Orat
Jan 23 at 10:02
add a comment |
$begingroup$
We have $$[g,h]^2 = (g^{-1}h^{-1}gh)^2 = (g^{-2}gh^{-2}hgh)^2.$$
Now collecting the terms $g^{-2}$ and $h^{-2}$ and the resulting commutators (which are central) to the left gives $$g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4 = g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}].$$
But commutators are bilinear in groups of class $2$, and $[g^2,h^2]=[g,h]^4=1$, so this simplifies to $g^{-4}h^{-4}[g^4,h^{-2}][h^2,g^{-2}]=1$.
answered Jan 23 at 8:44
Derek HoltDerek Holt
54.2k53571
$endgroup$
We have $$[g,h]^2 = (g^{-1}h^{-1}gh)^2 = (g^{-2}gh^{-2}hgh)^2.$$
Now collecting the terms $g^{-2}$ and $h^{-2}$ and the resulting commutators (which are central) to the left gives $$g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4 = g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}].$$
But commutators are bilinear in groups of class $2$, and $[g^2,h^2]=[g,h]^4=1$, so this simplifies to $g^{-4}h^{-4}[g^4,h^{-2}][h^2,g^{-2}]=1$.
answered Jan 23 at 8:44
Derek HoltDerek Holt
54.2k53571
answered Jan 23 at 8:44
Derek HoltDerek Holt
54.2k53571
answered Jan 23 at 8:44
Derek HoltDerek Holt
54.2k53571
answered Jan 23 at 8:44
Derek HoltDerek Holt
54.2k53571
54.2k53571
$begingroup$
Thank you for your answer. But I'm afraid I don't understand how $(g^{-2}gh^{-2}hgh)^2$ leads to $g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4$. Could you elaborate it a little more, please?
$endgroup$
– Orat
Jan 23 at 9:39
2
$begingroup$
This is a standard process known as collection to the left. It's just repeated use of the identity $ba=ab[b,a]$, so for example, the collection process starts on the left of the word with $g^{-2}gh^{-2}hcdots = g^{-2}h^{-2}g[g,h^{-2}]hcdots = g^{-2}h^{-2}[g,h^{-2}]ghcdots$.
$endgroup$
– Derek Holt
Jan 23 at 9:55
$begingroup$
Ah! I should've noticed it by myself. Thank you so much for your help.
$endgroup$
– Orat
Jan 23 at 10:02
add a comment |
$begingroup$
Thank you for your answer. But I'm afraid I don't understand how $(g^{-2}gh^{-2}hgh)^2$ leads to $g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4$. Could you elaborate it a little more, please?
$endgroup$
– Orat
Jan 23 at 9:39
2
$begingroup$
This is a standard process known as collection to the left. It's just repeated use of the identity $ba=ab[b,a]$, so for example, the collection process starts on the left of the word with $g^{-2}gh^{-2}hcdots = g^{-2}h^{-2}g[g,h^{-2}]hcdots = g^{-2}h^{-2}[g,h^{-2}]ghcdots$.
$endgroup$
– Derek Holt
Jan 23 at 9:55
$begingroup$
Ah! I should've noticed it by myself. Thank you so much for your help.
$endgroup$
– Orat
Jan 23 at 10:02
$begingroup$
Thank you for your answer. But I'm afraid I don't understand how $(g^{-2}gh^{-2}hgh)^2$ leads to $g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4$. Could you elaborate it a little more, please?
$endgroup$
– Orat
Jan 23 at 9:39
$begingroup$
Thank you for your answer. But I'm afraid I don't understand how $(g^{-2}gh^{-2}hgh)^2$ leads to $g^{-2}h^{-2}g^{-2}h^{-2}[g,h^{-2}][h^2,g^{-2}][g^3,h^{-2}](gh)^4$. Could you elaborate it a little more, please?
$endgroup$
– Orat
Jan 23 at 9:39
2
2
$begingroup$
This is a standard process known as collection to the left. It's just repeated use of the identity $ba=ab[b,a]$, so for example, the collection process starts on the left of the word with $g^{-2}gh^{-2}hcdots = g^{-2}h^{-2}g[g,h^{-2}]hcdots = g^{-2}h^{-2}[g,h^{-2}]ghcdots$.
$endgroup$
– Derek Holt
Jan 23 at 9:55
$begingroup$
This is a standard process known as collection to the left. It's just repeated use of the identity $ba=ab[b,a]$, so for example, the collection process starts on the left of the word with $g^{-2}gh^{-2}hcdots = g^{-2}h^{-2}g[g,h^{-2}]hcdots = g^{-2}h^{-2}[g,h^{-2}]ghcdots$.
$endgroup$
– Derek Holt
Jan 23 at 9:55
$begingroup$
Ah! I should've noticed it by myself. Thank you so much for your help.
$endgroup$
– Orat
Jan 23 at 10:02
$begingroup$
Ah! I should've noticed it by myself. Thank you so much for your help.
$endgroup$
– Orat
Jan 23 at 10:02
add a comment |
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$begingroup$
What code & software did you use to check the statement?
$endgroup$
– Shaun
Jan 23 at 12:07
1
$begingroup$
@Shaun I used GAP. A code is a fairly simple for-loop test using
AllSmallGroups.$endgroup$
– Orat
Jan 23 at 13:24