Suppose H⩽G, prove that if (H,G′)=⟨e⟩, then (H′,G)=⟨e⟩.
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I know this is a duplicate of Suppose $ Hleqslant G $, prove that if $ (H, G')=langle e rangle $, then $ (H', G)=langle e rangle $. but the only answer which does not use the Three Subgroups Lemma (which, at this point, Hungerford has not covered) did not make sense to me, and since my reputation is still low, I can't leave a comment asking for clarification. The given solution is as follows:
"By the previous exercise in the book, for $h,k∈H$, and $g∈G$, we have $[hk,g]=h[k,g]h^{−1}[h,g]=[k,g][h,g],$ because $(H′,G)=1$ (I am writing 1 for ⟨e⟩ and also for e.)
We have to prove that $[[h,k],g]=1$. We have $$[[h,k],g]=[hkh^{−1}k^{−1},g]=[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g].$$
Now, using $(H′,G)=1$ again, we have $h^{−1}[k,g]h=[k,g]$ and $hg^{−1}[k,g]gh^{−1}=g^{−1}[k,g]g,$ and so
$$[k,g]^{−1}[h,g]^{−1}[k,g][h,g]=[k,g]^{−1}ghg^{−1}h^{−1}[k,g]hgh^{−1}g^{−1}=[k,g]^{−1}[k,g]=1$$
as required."
The problem I have with this proof is that they seem to be using $(H',G)=1$, which is what we're trying to prove; we have that $(H,G')=1$ and we're trying to prove $(H',G)=1$, so it seems to me we shouldn't assume that in our proof. Am I missing something? If not, and this proof is invalid, would anyone be able to provide a valid proof of this (without referring to the Three Subgroups Lemma)?
abstract-algebra group-theory
asked Nov 18 at 2:19
John
356
add a comment |
up vote
2
down vote
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I know this is a duplicate of Suppose $ Hleqslant G $, prove that if $ (H, G')=langle e rangle $, then $ (H', G)=langle e rangle $. but the only answer which does not use the Three Subgroups Lemma (which, at this point, Hungerford has not covered) did not make sense to me, and since my reputation is still low, I can't leave a comment asking for clarification. The given solution is as follows:
"By the previous exercise in the book, for $h,k∈H$, and $g∈G$, we have $[hk,g]=h[k,g]h^{−1}[h,g]=[k,g][h,g],$ because $(H′,G)=1$ (I am writing 1 for ⟨e⟩ and also for e.)
We have to prove that $[[h,k],g]=1$. We have $$[[h,k],g]=[hkh^{−1}k^{−1},g]=[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g].$$
Now, using $(H′,G)=1$ again, we have $h^{−1}[k,g]h=[k,g]$ and $hg^{−1}[k,g]gh^{−1}=g^{−1}[k,g]g,$ and so
$$[k,g]^{−1}[h,g]^{−1}[k,g][h,g]=[k,g]^{−1}ghg^{−1}h^{−1}[k,g]hgh^{−1}g^{−1}=[k,g]^{−1}[k,g]=1$$
as required."
The problem I have with this proof is that they seem to be using $(H',G)=1$, which is what we're trying to prove; we have that $(H,G')=1$ and we're trying to prove $(H',G)=1$, so it seems to me we shouldn't assume that in our proof. Am I missing something? If not, and this proof is invalid, would anyone be able to provide a valid proof of this (without referring to the Three Subgroups Lemma)?
abstract-algebra group-theory
asked Nov 18 at 2:19
John
356
1
Sorry, the proof was mine, and when I wrote "using $(H',G)=1$", I meant "using $(H,G')=1$", which we are assuming. So it was just a typo, both times. I have edited the solution now to remove the mistake.
– Derek Holt
Nov 18 at 11:59
Gotcha! Thanks for the edit, I am able to follow most of the proof now. I do have one more question, though; How did you get that $[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g]$? I wrote it out and it doesn't seem to be equal at first glance; were you using something to get this equivalence?
– John
Nov 18 at 19:30
$[k^{-1},g]=[k,g]^{-1}$ follows from $[hk,g]=[k,g][h,g]$, which we have just proved.
– Derek Holt
Nov 18 at 22:18
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I know this is a duplicate of Suppose $ Hleqslant G $, prove that if $ (H, G')=langle e rangle $, then $ (H', G)=langle e rangle $. but the only answer which does not use the Three Subgroups Lemma (which, at this point, Hungerford has not covered) did not make sense to me, and since my reputation is still low, I can't leave a comment asking for clarification. The given solution is as follows:
"By the previous exercise in the book, for $h,k∈H$, and $g∈G$, we have $[hk,g]=h[k,g]h^{−1}[h,g]=[k,g][h,g],$ because $(H′,G)=1$ (I am writing 1 for ⟨e⟩ and also for e.)
We have to prove that $[[h,k],g]=1$. We have $$[[h,k],g]=[hkh^{−1}k^{−1},g]=[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g].$$
Now, using $(H′,G)=1$ again, we have $h^{−1}[k,g]h=[k,g]$ and $hg^{−1}[k,g]gh^{−1}=g^{−1}[k,g]g,$ and so
$$[k,g]^{−1}[h,g]^{−1}[k,g][h,g]=[k,g]^{−1}ghg^{−1}h^{−1}[k,g]hgh^{−1}g^{−1}=[k,g]^{−1}[k,g]=1$$
as required."
The problem I have with this proof is that they seem to be using $(H',G)=1$, which is what we're trying to prove; we have that $(H,G')=1$ and we're trying to prove $(H',G)=1$, so it seems to me we shouldn't assume that in our proof. Am I missing something? If not, and this proof is invalid, would anyone be able to provide a valid proof of this (without referring to the Three Subgroups Lemma)?
abstract-algebra group-theory
asked Nov 18 at 2:19
John
356
I know this is a duplicate of Suppose $ Hleqslant G $, prove that if $ (H, G')=langle e rangle $, then $ (H', G)=langle e rangle $. but the only answer which does not use the Three Subgroups Lemma (which, at this point, Hungerford has not covered) did not make sense to me, and since my reputation is still low, I can't leave a comment asking for clarification. The given solution is as follows:
"By the previous exercise in the book, for $h,k∈H$, and $g∈G$, we have $[hk,g]=h[k,g]h^{−1}[h,g]=[k,g][h,g],$ because $(H′,G)=1$ (I am writing 1 for ⟨e⟩ and also for e.)
We have to prove that $[[h,k],g]=1$. We have $$[[h,k],g]=[hkh^{−1}k^{−1},g]=[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g].$$
Now, using $(H′,G)=1$ again, we have $h^{−1}[k,g]h=[k,g]$ and $hg^{−1}[k,g]gh^{−1}=g^{−1}[k,g]g,$ and so
$$[k,g]^{−1}[h,g]^{−1}[k,g][h,g]=[k,g]^{−1}ghg^{−1}h^{−1}[k,g]hgh^{−1}g^{−1}=[k,g]^{−1}[k,g]=1$$
as required."
The problem I have with this proof is that they seem to be using $(H',G)=1$, which is what we're trying to prove; we have that $(H,G')=1$ and we're trying to prove $(H',G)=1$, so it seems to me we shouldn't assume that in our proof. Am I missing something? If not, and this proof is invalid, would anyone be able to provide a valid proof of this (without referring to the Three Subgroups Lemma)?
abstract-algebra group-theory
abstract-algebra group-theory
asked Nov 18 at 2:19
John
356
asked Nov 18 at 2:19
John
356
asked Nov 18 at 2:19
John
356
asked Nov 18 at 2:19
John
356
asked Nov 18 at 2:19
John
356
356
1
Sorry, the proof was mine, and when I wrote "using $(H',G)=1$", I meant "using $(H,G')=1$", which we are assuming. So it was just a typo, both times. I have edited the solution now to remove the mistake.
– Derek Holt
Nov 18 at 11:59
Gotcha! Thanks for the edit, I am able to follow most of the proof now. I do have one more question, though; How did you get that $[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g]$? I wrote it out and it doesn't seem to be equal at first glance; were you using something to get this equivalence?
– John
Nov 18 at 19:30
$[k^{-1},g]=[k,g]^{-1}$ follows from $[hk,g]=[k,g][h,g]$, which we have just proved.
– Derek Holt
Nov 18 at 22:18
add a comment |
1
Sorry, the proof was mine, and when I wrote "using $(H',G)=1$", I meant "using $(H,G')=1$", which we are assuming. So it was just a typo, both times. I have edited the solution now to remove the mistake.
– Derek Holt
Nov 18 at 11:59
Gotcha! Thanks for the edit, I am able to follow most of the proof now. I do have one more question, though; How did you get that $[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g]$? I wrote it out and it doesn't seem to be equal at first glance; were you using something to get this equivalence?
– John
Nov 18 at 19:30
$[k^{-1},g]=[k,g]^{-1}$ follows from $[hk,g]=[k,g][h,g]$, which we have just proved.
– Derek Holt
Nov 18 at 22:18
1
1
Sorry, the proof was mine, and when I wrote "using $(H',G)=1$", I meant "using $(H,G')=1$", which we are assuming. So it was just a typo, both times. I have edited the solution now to remove the mistake.
– Derek Holt
Nov 18 at 11:59
Sorry, the proof was mine, and when I wrote "using $(H',G)=1$", I meant "using $(H,G')=1$", which we are assuming. So it was just a typo, both times. I have edited the solution now to remove the mistake.
– Derek Holt
Nov 18 at 11:59
Gotcha! Thanks for the edit, I am able to follow most of the proof now. I do have one more question, though; How did you get that $[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g]$? I wrote it out and it doesn't seem to be equal at first glance; were you using something to get this equivalence?
– John
Nov 18 at 19:30
Gotcha! Thanks for the edit, I am able to follow most of the proof now. I do have one more question, though; How did you get that $[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g]$? I wrote it out and it doesn't seem to be equal at first glance; were you using something to get this equivalence?
– John
Nov 18 at 19:30
$[k^{-1},g]=[k,g]^{-1}$ follows from $[hk,g]=[k,g][h,g]$, which we have just proved.
– Derek Holt
Nov 18 at 22:18
$[k^{-1},g]=[k,g]^{-1}$ follows from $[hk,g]=[k,g][h,g]$, which we have just proved.
– Derek Holt
Nov 18 at 22:18
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Sorry, the proof was mine, and when I wrote "using $(H',G)=1$", I meant "using $(H,G')=1$", which we are assuming. So it was just a typo, both times. I have edited the solution now to remove the mistake.
– Derek Holt
Nov 18 at 11:59
Gotcha! Thanks for the edit, I am able to follow most of the proof now. I do have one more question, though; How did you get that $[k^{−1},g][h^{−1},g][k,g][h,g]=[k,g]^{−1}[h,g]^{−1}[k,g][h,g]$? I wrote it out and it doesn't seem to be equal at first glance; were you using something to get this equivalence?
– John
Nov 18 at 19:30
$[k^{-1},g]=[k,g]^{-1}$ follows from $[hk,g]=[k,g][h,g]$, which we have just proved.
– Derek Holt
Nov 18 at 22:18