Mixing
If $G$ is a product of subnormal $S$ and a $p$-group, does the $p$-group need to be normal to know...
$begingroup$
I am reading I. Martin Isaacs' beautiful book Finite Group Theory. On p. 287 is a lemma that includes a condition (that $P$ is normal; notation fixed below) that I cannot find a need for in the proof.
My question: Am I missing something? How does the proof require $P$ to be normal? Or else, is this condition superfluous?
Fixing notation: given a finite group $G$ and a prime $p$, let $mathbf{O}^p(G)$ be the unique minimal subgroup of $G$ such that the quotient is a $p$-group. If $S$ is a subgroup, the relation $Striangleleft G$ means $S$ is normal in $G$, while $Strianglelefttriangleleft G$ means $S$ is subnormal in $G$, i.e. there is an ascending sequence of subgroups $S=S_1< S_2< dots< S_r = G$ where each $S_i$ is normal in $S_{i+1}$. The symbol $<$ means proper containment between subgroups.
Here is the statement and proof of the lemma from p. 287 in Isaacs' book.
9.26. Lemma. Let $G = SP$, where $G$ is a finite group, and where $Strianglelefttriangleleft G$ and $Ptriangleleft G$, and $P$ is a $p$-group for some prime $p$. Then $mathbf{O}^p(G) = mathbf{O}^p(S)$.
Proof. If $S=G$, there is nothing to prove, so we can assume $S<G$, and we choose $M$ with $Ssubseteq Mtriangleleft G$ and $M<G$. Then $M=S(Mcap P)$, and so working by induction on $|G|$ and applying the inductive hypothesis to $M$, we have $mathbf{O}^p(M) = mathbf{O}^p(S)$, and it suffices to show that $mathbf{O}^p(M) = mathbf{O}^p(G)$.
Now $mathbf{O}^p(M)triangleleft G$, and we write $overline G = G/mathbf{O}^p(M)$. Since $Ssubseteq M$, we have $G=MP$, and thus $overline G = overline M, overline P$, which is a $p$-group. It follows that $mathbf{O}^p(G)subseteq mathbf{O}^p(M)$. The reverse containment holds because $M/mathbf{O}^p(G)$ is a $p$-group, and this completes the proof.
I can't see any need for the hypothesis that $P$ is normal in this proof! The hypothesis that $S$ is subnormal is needed in order for $M$ to exist, and the weight of the first paragraph is reduction to the case that $S$ (now replaced with $M$) is normal. This is already enough to know $mathbf{O}^p(M)$ is normal in $G$ (as it is a characteristic subgroup of a normal subgroup). Then $overline G = G/mathbf{O}^p(M)$ exists. Then $overline M$, which is $M$'s image under this canonical map, is a $p$-group, by definition of $mathbf{O}^p(M)$. Meanwhile, $overline P$, the image of $P$, is obviously also a $p$-group, and the cardinality of $overline G = overline M ,overline P$ is thus a factor of a product of powers of $p$ and thus a power of $p$. Why isn't this all that's needed?
Some context: This proof follows the pattern of an earlier result, Lemma 9.15 found on p. 281 with the proof on pp. 281-2. It is the exact same statement except only assuming $P$ is nilpotent (rather than a $p$-group), and therefore with $S^infty, G^infty$ in place of $mathbf{O}^p(S), mathbf{O}^p(G)$. ($S^infty, G^infty$ are the unique minimal subgroups such that the quotients $S/S^infty$ and $G/G^infty$ are nilpotent.) The proof is also essentially the same, but in the earlier case, I do see the way the proof uses the normality of $P$: in order to conclude $overline G = overline M, overline P$ is nilpotent from the nilpotence of $overline M, overline P$, Isaacs argues that since they are both normal, they are both contained in the Fitting subgroup of $overline G$, and thus so is their product. But I don't see a similar need in the $p$-group situation -- what am I missing?
abstract-algebra group-theory finite-groups
asked Jan 13 at 22:02
Ben Blum-SmithBen Blum-Smith
10.2k23086
$endgroup$
add a comment |
$begingroup$
I am reading I. Martin Isaacs' beautiful book Finite Group Theory. On p. 287 is a lemma that includes a condition (that $P$ is normal; notation fixed below) that I cannot find a need for in the proof.
My question: Am I missing something? How does the proof require $P$ to be normal? Or else, is this condition superfluous?
Fixing notation: given a finite group $G$ and a prime $p$, let $mathbf{O}^p(G)$ be the unique minimal subgroup of $G$ such that the quotient is a $p$-group. If $S$ is a subgroup, the relation $Striangleleft G$ means $S$ is normal in $G$, while $Strianglelefttriangleleft G$ means $S$ is subnormal in $G$, i.e. there is an ascending sequence of subgroups $S=S_1< S_2< dots< S_r = G$ where each $S_i$ is normal in $S_{i+1}$. The symbol $<$ means proper containment between subgroups.
Here is the statement and proof of the lemma from p. 287 in Isaacs' book.
9.26. Lemma. Let $G = SP$, where $G$ is a finite group, and where $Strianglelefttriangleleft G$ and $Ptriangleleft G$, and $P$ is a $p$-group for some prime $p$. Then $mathbf{O}^p(G) = mathbf{O}^p(S)$.
Proof. If $S=G$, there is nothing to prove, so we can assume $S<G$, and we choose $M$ with $Ssubseteq Mtriangleleft G$ and $M<G$. Then $M=S(Mcap P)$, and so working by induction on $|G|$ and applying the inductive hypothesis to $M$, we have $mathbf{O}^p(M) = mathbf{O}^p(S)$, and it suffices to show that $mathbf{O}^p(M) = mathbf{O}^p(G)$.
Now $mathbf{O}^p(M)triangleleft G$, and we write $overline G = G/mathbf{O}^p(M)$. Since $Ssubseteq M$, we have $G=MP$, and thus $overline G = overline M, overline P$, which is a $p$-group. It follows that $mathbf{O}^p(G)subseteq mathbf{O}^p(M)$. The reverse containment holds because $M/mathbf{O}^p(G)$ is a $p$-group, and this completes the proof.
I can't see any need for the hypothesis that $P$ is normal in this proof! The hypothesis that $S$ is subnormal is needed in order for $M$ to exist, and the weight of the first paragraph is reduction to the case that $S$ (now replaced with $M$) is normal. This is already enough to know $mathbf{O}^p(M)$ is normal in $G$ (as it is a characteristic subgroup of a normal subgroup). Then $overline G = G/mathbf{O}^p(M)$ exists. Then $overline M$, which is $M$'s image under this canonical map, is a $p$-group, by definition of $mathbf{O}^p(M)$. Meanwhile, $overline P$, the image of $P$, is obviously also a $p$-group, and the cardinality of $overline G = overline M ,overline P$ is thus a factor of a product of powers of $p$ and thus a power of $p$. Why isn't this all that's needed?
Some context: This proof follows the pattern of an earlier result, Lemma 9.15 found on p. 281 with the proof on pp. 281-2. It is the exact same statement except only assuming $P$ is nilpotent (rather than a $p$-group), and therefore with $S^infty, G^infty$ in place of $mathbf{O}^p(S), mathbf{O}^p(G)$. ($S^infty, G^infty$ are the unique minimal subgroups such that the quotients $S/S^infty$ and $G/G^infty$ are nilpotent.) The proof is also essentially the same, but in the earlier case, I do see the way the proof uses the normality of $P$: in order to conclude $overline G = overline M, overline P$ is nilpotent from the nilpotence of $overline M, overline P$, Isaacs argues that since they are both normal, they are both contained in the Fitting subgroup of $overline G$, and thus so is their product. But I don't see a similar need in the $p$-group situation -- what am I missing?
abstract-algebra group-theory finite-groups
asked Jan 13 at 22:02
Ben Blum-SmithBen Blum-Smith
10.2k23086
$endgroup$
3
$begingroup$
Without having looked through the proof in detail, I'll just say that if you don't assume $P$ is normal, then there is no reason for $SP$ to be a subgroup of $G$, which makes the hypothesis $SP=G$ more difficult to check (because then it's not enough to check that a generating set of $G$ lies in $SP$). So even if it's not strictly needed, it may be that in practice, it will always be the case when we apply this result.
$endgroup$
– Ted
Jan 13 at 22:37
1
$begingroup$
I agree with you, the hypothesis $P lhd G$ is superfluous. But since this is just a lemma that forms part of a proof of a theorem rather than a result of independent interest, perhaps the author was not interested in making sure that the hypotheses were all necessary.
$endgroup$
– Derek Holt
Jan 14 at 8:42
$begingroup$
@Ted - the way the lemma is applied in the text fits the pattern of your point. Corollary 9.27 reads "Let $G$ be a finite group. Suppose that $Strianglelefttriangleleft G$, and let $Ptriangleleft G$ be a $p$-group. Then $P$ normalizes $mathbf{O}^p(S)$." The argument is to apply Lemma 9.26 in the subgroup $SP$, which is of course does not need to be a subgroup of $P$ is not normal.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
$begingroup$
@DerekHolt - thank you. Yeah, that seems to explain it.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
add a comment |
$begingroup$
I am reading I. Martin Isaacs' beautiful book Finite Group Theory. On p. 287 is a lemma that includes a condition (that $P$ is normal; notation fixed below) that I cannot find a need for in the proof.
My question: Am I missing something? How does the proof require $P$ to be normal? Or else, is this condition superfluous?
Fixing notation: given a finite group $G$ and a prime $p$, let $mathbf{O}^p(G)$ be the unique minimal subgroup of $G$ such that the quotient is a $p$-group. If $S$ is a subgroup, the relation $Striangleleft G$ means $S$ is normal in $G$, while $Strianglelefttriangleleft G$ means $S$ is subnormal in $G$, i.e. there is an ascending sequence of subgroups $S=S_1< S_2< dots< S_r = G$ where each $S_i$ is normal in $S_{i+1}$. The symbol $<$ means proper containment between subgroups.
Here is the statement and proof of the lemma from p. 287 in Isaacs' book.
9.26. Lemma. Let $G = SP$, where $G$ is a finite group, and where $Strianglelefttriangleleft G$ and $Ptriangleleft G$, and $P$ is a $p$-group for some prime $p$. Then $mathbf{O}^p(G) = mathbf{O}^p(S)$.
Proof. If $S=G$, there is nothing to prove, so we can assume $S<G$, and we choose $M$ with $Ssubseteq Mtriangleleft G$ and $M<G$. Then $M=S(Mcap P)$, and so working by induction on $|G|$ and applying the inductive hypothesis to $M$, we have $mathbf{O}^p(M) = mathbf{O}^p(S)$, and it suffices to show that $mathbf{O}^p(M) = mathbf{O}^p(G)$.
Now $mathbf{O}^p(M)triangleleft G$, and we write $overline G = G/mathbf{O}^p(M)$. Since $Ssubseteq M$, we have $G=MP$, and thus $overline G = overline M, overline P$, which is a $p$-group. It follows that $mathbf{O}^p(G)subseteq mathbf{O}^p(M)$. The reverse containment holds because $M/mathbf{O}^p(G)$ is a $p$-group, and this completes the proof.
I can't see any need for the hypothesis that $P$ is normal in this proof! The hypothesis that $S$ is subnormal is needed in order for $M$ to exist, and the weight of the first paragraph is reduction to the case that $S$ (now replaced with $M$) is normal. This is already enough to know $mathbf{O}^p(M)$ is normal in $G$ (as it is a characteristic subgroup of a normal subgroup). Then $overline G = G/mathbf{O}^p(M)$ exists. Then $overline M$, which is $M$'s image under this canonical map, is a $p$-group, by definition of $mathbf{O}^p(M)$. Meanwhile, $overline P$, the image of $P$, is obviously also a $p$-group, and the cardinality of $overline G = overline M ,overline P$ is thus a factor of a product of powers of $p$ and thus a power of $p$. Why isn't this all that's needed?
Some context: This proof follows the pattern of an earlier result, Lemma 9.15 found on p. 281 with the proof on pp. 281-2. It is the exact same statement except only assuming $P$ is nilpotent (rather than a $p$-group), and therefore with $S^infty, G^infty$ in place of $mathbf{O}^p(S), mathbf{O}^p(G)$. ($S^infty, G^infty$ are the unique minimal subgroups such that the quotients $S/S^infty$ and $G/G^infty$ are nilpotent.) The proof is also essentially the same, but in the earlier case, I do see the way the proof uses the normality of $P$: in order to conclude $overline G = overline M, overline P$ is nilpotent from the nilpotence of $overline M, overline P$, Isaacs argues that since they are both normal, they are both contained in the Fitting subgroup of $overline G$, and thus so is their product. But I don't see a similar need in the $p$-group situation -- what am I missing?
abstract-algebra group-theory finite-groups
asked Jan 13 at 22:02
Ben Blum-SmithBen Blum-Smith
10.2k23086
$endgroup$
I am reading I. Martin Isaacs' beautiful book Finite Group Theory. On p. 287 is a lemma that includes a condition (that $P$ is normal; notation fixed below) that I cannot find a need for in the proof.
My question: Am I missing something? How does the proof require $P$ to be normal? Or else, is this condition superfluous?
Fixing notation: given a finite group $G$ and a prime $p$, let $mathbf{O}^p(G)$ be the unique minimal subgroup of $G$ such that the quotient is a $p$-group. If $S$ is a subgroup, the relation $Striangleleft G$ means $S$ is normal in $G$, while $Strianglelefttriangleleft G$ means $S$ is subnormal in $G$, i.e. there is an ascending sequence of subgroups $S=S_1< S_2< dots< S_r = G$ where each $S_i$ is normal in $S_{i+1}$. The symbol $<$ means proper containment between subgroups.
Here is the statement and proof of the lemma from p. 287 in Isaacs' book.
9.26. Lemma. Let $G = SP$, where $G$ is a finite group, and where $Strianglelefttriangleleft G$ and $Ptriangleleft G$, and $P$ is a $p$-group for some prime $p$. Then $mathbf{O}^p(G) = mathbf{O}^p(S)$.
Proof. If $S=G$, there is nothing to prove, so we can assume $S<G$, and we choose $M$ with $Ssubseteq Mtriangleleft G$ and $M<G$. Then $M=S(Mcap P)$, and so working by induction on $|G|$ and applying the inductive hypothesis to $M$, we have $mathbf{O}^p(M) = mathbf{O}^p(S)$, and it suffices to show that $mathbf{O}^p(M) = mathbf{O}^p(G)$.
Now $mathbf{O}^p(M)triangleleft G$, and we write $overline G = G/mathbf{O}^p(M)$. Since $Ssubseteq M$, we have $G=MP$, and thus $overline G = overline M, overline P$, which is a $p$-group. It follows that $mathbf{O}^p(G)subseteq mathbf{O}^p(M)$. The reverse containment holds because $M/mathbf{O}^p(G)$ is a $p$-group, and this completes the proof.
I can't see any need for the hypothesis that $P$ is normal in this proof! The hypothesis that $S$ is subnormal is needed in order for $M$ to exist, and the weight of the first paragraph is reduction to the case that $S$ (now replaced with $M$) is normal. This is already enough to know $mathbf{O}^p(M)$ is normal in $G$ (as it is a characteristic subgroup of a normal subgroup). Then $overline G = G/mathbf{O}^p(M)$ exists. Then $overline M$, which is $M$'s image under this canonical map, is a $p$-group, by definition of $mathbf{O}^p(M)$. Meanwhile, $overline P$, the image of $P$, is obviously also a $p$-group, and the cardinality of $overline G = overline M ,overline P$ is thus a factor of a product of powers of $p$ and thus a power of $p$. Why isn't this all that's needed?
Some context: This proof follows the pattern of an earlier result, Lemma 9.15 found on p. 281 with the proof on pp. 281-2. It is the exact same statement except only assuming $P$ is nilpotent (rather than a $p$-group), and therefore with $S^infty, G^infty$ in place of $mathbf{O}^p(S), mathbf{O}^p(G)$. ($S^infty, G^infty$ are the unique minimal subgroups such that the quotients $S/S^infty$ and $G/G^infty$ are nilpotent.) The proof is also essentially the same, but in the earlier case, I do see the way the proof uses the normality of $P$: in order to conclude $overline G = overline M, overline P$ is nilpotent from the nilpotence of $overline M, overline P$, Isaacs argues that since they are both normal, they are both contained in the Fitting subgroup of $overline G$, and thus so is their product. But I don't see a similar need in the $p$-group situation -- what am I missing?
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked Jan 13 at 22:02
Ben Blum-SmithBen Blum-Smith
10.2k23086
asked Jan 13 at 22:02
Ben Blum-SmithBen Blum-Smith
10.2k23086
edited Jan 13 at 22:28
Ben Blum-Smith
asked Jan 13 at 22:02
Ben Blum-SmithBen Blum-Smith
10.2k23086
asked Jan 13 at 22:02
Ben Blum-SmithBen Blum-Smith
10.2k23086
asked Jan 13 at 22:02
Ben Blum-SmithBen Blum-Smith
10.2k23086
10.2k23086
3
$begingroup$
Without having looked through the proof in detail, I'll just say that if you don't assume $P$ is normal, then there is no reason for $SP$ to be a subgroup of $G$, which makes the hypothesis $SP=G$ more difficult to check (because then it's not enough to check that a generating set of $G$ lies in $SP$). So even if it's not strictly needed, it may be that in practice, it will always be the case when we apply this result.
$endgroup$
– Ted
Jan 13 at 22:37
1
$begingroup$
I agree with you, the hypothesis $P lhd G$ is superfluous. But since this is just a lemma that forms part of a proof of a theorem rather than a result of independent interest, perhaps the author was not interested in making sure that the hypotheses were all necessary.
$endgroup$
– Derek Holt
Jan 14 at 8:42
$begingroup$
@Ted - the way the lemma is applied in the text fits the pattern of your point. Corollary 9.27 reads "Let $G$ be a finite group. Suppose that $Strianglelefttriangleleft G$, and let $Ptriangleleft G$ be a $p$-group. Then $P$ normalizes $mathbf{O}^p(S)$." The argument is to apply Lemma 9.26 in the subgroup $SP$, which is of course does not need to be a subgroup of $P$ is not normal.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
$begingroup$
@DerekHolt - thank you. Yeah, that seems to explain it.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
add a comment |
3
$begingroup$
Without having looked through the proof in detail, I'll just say that if you don't assume $P$ is normal, then there is no reason for $SP$ to be a subgroup of $G$, which makes the hypothesis $SP=G$ more difficult to check (because then it's not enough to check that a generating set of $G$ lies in $SP$). So even if it's not strictly needed, it may be that in practice, it will always be the case when we apply this result.
$endgroup$
– Ted
Jan 13 at 22:37
1
$begingroup$
I agree with you, the hypothesis $P lhd G$ is superfluous. But since this is just a lemma that forms part of a proof of a theorem rather than a result of independent interest, perhaps the author was not interested in making sure that the hypotheses were all necessary.
$endgroup$
– Derek Holt
Jan 14 at 8:42
$begingroup$
@Ted - the way the lemma is applied in the text fits the pattern of your point. Corollary 9.27 reads "Let $G$ be a finite group. Suppose that $Strianglelefttriangleleft G$, and let $Ptriangleleft G$ be a $p$-group. Then $P$ normalizes $mathbf{O}^p(S)$." The argument is to apply Lemma 9.26 in the subgroup $SP$, which is of course does not need to be a subgroup of $P$ is not normal.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
$begingroup$
@DerekHolt - thank you. Yeah, that seems to explain it.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
3
3
$begingroup$
Without having looked through the proof in detail, I'll just say that if you don't assume $P$ is normal, then there is no reason for $SP$ to be a subgroup of $G$, which makes the hypothesis $SP=G$ more difficult to check (because then it's not enough to check that a generating set of $G$ lies in $SP$). So even if it's not strictly needed, it may be that in practice, it will always be the case when we apply this result.
$endgroup$
– Ted
Jan 13 at 22:37
$begingroup$
Without having looked through the proof in detail, I'll just say that if you don't assume $P$ is normal, then there is no reason for $SP$ to be a subgroup of $G$, which makes the hypothesis $SP=G$ more difficult to check (because then it's not enough to check that a generating set of $G$ lies in $SP$). So even if it's not strictly needed, it may be that in practice, it will always be the case when we apply this result.
$endgroup$
– Ted
Jan 13 at 22:37
1
1
$begingroup$
I agree with you, the hypothesis $P lhd G$ is superfluous. But since this is just a lemma that forms part of a proof of a theorem rather than a result of independent interest, perhaps the author was not interested in making sure that the hypotheses were all necessary.
$endgroup$
– Derek Holt
Jan 14 at 8:42
$begingroup$
I agree with you, the hypothesis $P lhd G$ is superfluous. But since this is just a lemma that forms part of a proof of a theorem rather than a result of independent interest, perhaps the author was not interested in making sure that the hypotheses were all necessary.
$endgroup$
– Derek Holt
Jan 14 at 8:42
$begingroup$
@Ted - the way the lemma is applied in the text fits the pattern of your point. Corollary 9.27 reads "Let $G$ be a finite group. Suppose that $Strianglelefttriangleleft G$, and let $Ptriangleleft G$ be a $p$-group. Then $P$ normalizes $mathbf{O}^p(S)$." The argument is to apply Lemma 9.26 in the subgroup $SP$, which is of course does not need to be a subgroup of $P$ is not normal.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
$begingroup$
@Ted - the way the lemma is applied in the text fits the pattern of your point. Corollary 9.27 reads "Let $G$ be a finite group. Suppose that $Strianglelefttriangleleft G$, and let $Ptriangleleft G$ be a $p$-group. Then $P$ normalizes $mathbf{O}^p(S)$." The argument is to apply Lemma 9.26 in the subgroup $SP$, which is of course does not need to be a subgroup of $P$ is not normal.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
$begingroup$
@DerekHolt - thank you. Yeah, that seems to explain it.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
$begingroup$
@DerekHolt - thank you. Yeah, that seems to explain it.
$endgroup$
– Ben Blum-Smith
Jan 15 at 14:07
add a comment |
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Without having looked through the proof in detail, I'll just say that if you don't assume $P$ is normal, then there is no reason for $SP$ to be a subgroup of $G$, which makes the hypothesis $SP=G$ more difficult to check (because then it's not enough to check that a generating set of $G$ lies in $SP$). So even if it's not strictly needed, it may be that in practice, it will always be the case when we apply this result.
$endgroup$
– Ted
Jan 13 at 22:37
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I agree with you, the hypothesis $P lhd G$ is superfluous. But since this is just a lemma that forms part of a proof of a theorem rather than a result of independent interest, perhaps the author was not interested in making sure that the hypotheses were all necessary.
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– Derek Holt
Jan 14 at 8:42
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@Ted - the way the lemma is applied in the text fits the pattern of your point. Corollary 9.27 reads "Let $G$ be a finite group. Suppose that $Strianglelefttriangleleft G$, and let $Ptriangleleft G$ be a $p$-group. Then $P$ normalizes $mathbf{O}^p(S)$." The argument is to apply Lemma 9.26 in the subgroup $SP$, which is of course does not need to be a subgroup of $P$ is not normal.
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– Ben Blum-Smith
Jan 15 at 14:07
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@DerekHolt - thank you. Yeah, that seems to explain it.
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– Ben Blum-Smith
Jan 15 at 14:07