Mixing
An attempted proof of Cauchy's theorem for abelian groups using composition series.
$begingroup$
I came up with this proof for the abelian version of Cauchy's Theorem (if a prime $p$ divides the order of an abelian group then it has a subgroup of order $p$). I'm hoping someone could please check it's correct, and then answer a few questions about it. I hope this kind of question is appropriate here.
If $G$ is cyclic then the result is obvious, so assume otherwise and proceed by induction on $lvert G rvert$. Since $G$ is finite we can take a composition series $$G supset G_1 supset G_2 supset ldots supset {e},$$ and since $lvert G rvert$ is the product of the orders of its composition factors there must be a factor $frac{G_i}{G_{i+1}}$ with order divisible by $p$. But then $lvert G_i rvert = lvert frac{G_i}{G_{i+1}} rvert lvert G_{i+1} rvert $ is also divisible by $p$. If $G_i neq G$ then the result follows via induction, so assume $G_i = G$, and pick $x in G setminus G_{1}$. By simplicity of the first composition factor, $G_{1}$ is a maximal proper subgroup, so it must be that $langle x rangle G_{1} = G$, so $lvert G rvert = frac{lvert x rvert lvert G_{1} rvert}{lvert langle x rangle cap G_{1} rvert }$. Then $lvert x rvert = lvert langle x rangle cap G_{1} rvert lvert frac{G}{G_{1}} rvert$, so $p$ divides $lvert x rvert$ which completes the proof.
Questions:
Is the proof correct? It feels too easy which makes me worried. If it is correct, have I added any unnecessary complications that could be removed to simplify it? If it's not correct can it be salvaged?
Is there any way to make this work for non-abelian groups? I'm pretty sure I only used commutativity in the second last sentence, in the non-commutative case $G_1$ is a maximal normal subgroup and $langle x rangle$ isn't necessarily normal so the proof doesn't go through. Is there some way to work around this? All I can think of is replacing $langle x rangle$ with the normal subgroup generated by $x$, but I don't know how to show that it's proper (if it even is).
abstract-algebra group-theory proof-verification proof-writing finite-groups
asked Jan 22 at 17:45
EscadaraEscadara
1106
$endgroup$
add a comment |
$begingroup$
I came up with this proof for the abelian version of Cauchy's Theorem (if a prime $p$ divides the order of an abelian group then it has a subgroup of order $p$). I'm hoping someone could please check it's correct, and then answer a few questions about it. I hope this kind of question is appropriate here.
If $G$ is cyclic then the result is obvious, so assume otherwise and proceed by induction on $lvert G rvert$. Since $G$ is finite we can take a composition series $$G supset G_1 supset G_2 supset ldots supset {e},$$ and since $lvert G rvert$ is the product of the orders of its composition factors there must be a factor $frac{G_i}{G_{i+1}}$ with order divisible by $p$. But then $lvert G_i rvert = lvert frac{G_i}{G_{i+1}} rvert lvert G_{i+1} rvert $ is also divisible by $p$. If $G_i neq G$ then the result follows via induction, so assume $G_i = G$, and pick $x in G setminus G_{1}$. By simplicity of the first composition factor, $G_{1}$ is a maximal proper subgroup, so it must be that $langle x rangle G_{1} = G$, so $lvert G rvert = frac{lvert x rvert lvert G_{1} rvert}{lvert langle x rangle cap G_{1} rvert }$. Then $lvert x rvert = lvert langle x rangle cap G_{1} rvert lvert frac{G}{G_{1}} rvert$, so $p$ divides $lvert x rvert$ which completes the proof.
Questions:
Is the proof correct? It feels too easy which makes me worried. If it is correct, have I added any unnecessary complications that could be removed to simplify it? If it's not correct can it be salvaged?
Is there any way to make this work for non-abelian groups? I'm pretty sure I only used commutativity in the second last sentence, in the non-commutative case $G_1$ is a maximal normal subgroup and $langle x rangle$ isn't necessarily normal so the proof doesn't go through. Is there some way to work around this? All I can think of is replacing $langle x rangle$ with the normal subgroup generated by $x$, but I don't know how to show that it's proper (if it even is).
abstract-algebra group-theory proof-verification proof-writing finite-groups
asked Jan 22 at 17:45
EscadaraEscadara
1106
$endgroup$
1
$begingroup$
You can compare your proof with a "book proof" here of Cauchy's Theorem for abelian groups, which is may be a bit easier than yours.
$endgroup$
– Dietrich Burde
Jan 22 at 17:51
1
$begingroup$
You've used the dreaded word "obvious". Is it really obvious (to you or to the intended reader) or just familiar? Remember that there's a plethora of would-be theorems out there that seem obvious but turn out to be wrong.
$endgroup$
– Shaun
Jan 22 at 18:58
add a comment |
$begingroup$
I came up with this proof for the abelian version of Cauchy's Theorem (if a prime $p$ divides the order of an abelian group then it has a subgroup of order $p$). I'm hoping someone could please check it's correct, and then answer a few questions about it. I hope this kind of question is appropriate here.
If $G$ is cyclic then the result is obvious, so assume otherwise and proceed by induction on $lvert G rvert$. Since $G$ is finite we can take a composition series $$G supset G_1 supset G_2 supset ldots supset {e},$$ and since $lvert G rvert$ is the product of the orders of its composition factors there must be a factor $frac{G_i}{G_{i+1}}$ with order divisible by $p$. But then $lvert G_i rvert = lvert frac{G_i}{G_{i+1}} rvert lvert G_{i+1} rvert $ is also divisible by $p$. If $G_i neq G$ then the result follows via induction, so assume $G_i = G$, and pick $x in G setminus G_{1}$. By simplicity of the first composition factor, $G_{1}$ is a maximal proper subgroup, so it must be that $langle x rangle G_{1} = G$, so $lvert G rvert = frac{lvert x rvert lvert G_{1} rvert}{lvert langle x rangle cap G_{1} rvert }$. Then $lvert x rvert = lvert langle x rangle cap G_{1} rvert lvert frac{G}{G_{1}} rvert$, so $p$ divides $lvert x rvert$ which completes the proof.
Questions:
Is the proof correct? It feels too easy which makes me worried. If it is correct, have I added any unnecessary complications that could be removed to simplify it? If it's not correct can it be salvaged?
Is there any way to make this work for non-abelian groups? I'm pretty sure I only used commutativity in the second last sentence, in the non-commutative case $G_1$ is a maximal normal subgroup and $langle x rangle$ isn't necessarily normal so the proof doesn't go through. Is there some way to work around this? All I can think of is replacing $langle x rangle$ with the normal subgroup generated by $x$, but I don't know how to show that it's proper (if it even is).
abstract-algebra group-theory proof-verification proof-writing finite-groups
asked Jan 22 at 17:45
EscadaraEscadara
1106
$endgroup$
I came up with this proof for the abelian version of Cauchy's Theorem (if a prime $p$ divides the order of an abelian group then it has a subgroup of order $p$). I'm hoping someone could please check it's correct, and then answer a few questions about it. I hope this kind of question is appropriate here.
If $G$ is cyclic then the result is obvious, so assume otherwise and proceed by induction on $lvert G rvert$. Since $G$ is finite we can take a composition series $$G supset G_1 supset G_2 supset ldots supset {e},$$ and since $lvert G rvert$ is the product of the orders of its composition factors there must be a factor $frac{G_i}{G_{i+1}}$ with order divisible by $p$. But then $lvert G_i rvert = lvert frac{G_i}{G_{i+1}} rvert lvert G_{i+1} rvert $ is also divisible by $p$. If $G_i neq G$ then the result follows via induction, so assume $G_i = G$, and pick $x in G setminus G_{1}$. By simplicity of the first composition factor, $G_{1}$ is a maximal proper subgroup, so it must be that $langle x rangle G_{1} = G$, so $lvert G rvert = frac{lvert x rvert lvert G_{1} rvert}{lvert langle x rangle cap G_{1} rvert }$. Then $lvert x rvert = lvert langle x rangle cap G_{1} rvert lvert frac{G}{G_{1}} rvert$, so $p$ divides $lvert x rvert$ which completes the proof.
Questions:
Is the proof correct? It feels too easy which makes me worried. If it is correct, have I added any unnecessary complications that could be removed to simplify it? If it's not correct can it be salvaged?
Is there any way to make this work for non-abelian groups? I'm pretty sure I only used commutativity in the second last sentence, in the non-commutative case $G_1$ is a maximal normal subgroup and $langle x rangle$ isn't necessarily normal so the proof doesn't go through. Is there some way to work around this? All I can think of is replacing $langle x rangle$ with the normal subgroup generated by $x$, but I don't know how to show that it's proper (if it even is).
abstract-algebra group-theory proof-verification proof-writing finite-groups
abstract-algebra group-theory proof-verification proof-writing finite-groups
asked Jan 22 at 17:45
EscadaraEscadara
1106
asked Jan 22 at 17:45
EscadaraEscadara
1106
asked Jan 22 at 17:45
EscadaraEscadara
1106
asked Jan 22 at 17:45
EscadaraEscadara
1106
asked Jan 22 at 17:45
EscadaraEscadara
1106
1106
1
$begingroup$
You can compare your proof with a "book proof" here of Cauchy's Theorem for abelian groups, which is may be a bit easier than yours.
$endgroup$
– Dietrich Burde
Jan 22 at 17:51
1
$begingroup$
You've used the dreaded word "obvious". Is it really obvious (to you or to the intended reader) or just familiar? Remember that there's a plethora of would-be theorems out there that seem obvious but turn out to be wrong.
$endgroup$
– Shaun
Jan 22 at 18:58
add a comment |
1
$begingroup$
You can compare your proof with a "book proof" here of Cauchy's Theorem for abelian groups, which is may be a bit easier than yours.
$endgroup$
– Dietrich Burde
Jan 22 at 17:51
1
$begingroup$
You've used the dreaded word "obvious". Is it really obvious (to you or to the intended reader) or just familiar? Remember that there's a plethora of would-be theorems out there that seem obvious but turn out to be wrong.
$endgroup$
– Shaun
Jan 22 at 18:58
1
1
$begingroup$
You can compare your proof with a "book proof" here of Cauchy's Theorem for abelian groups, which is may be a bit easier than yours.
$endgroup$
– Dietrich Burde
Jan 22 at 17:51
$begingroup$
You can compare your proof with a "book proof" here of Cauchy's Theorem for abelian groups, which is may be a bit easier than yours.
$endgroup$
– Dietrich Burde
Jan 22 at 17:51
1
1
$begingroup$
You've used the dreaded word "obvious". Is it really obvious (to you or to the intended reader) or just familiar? Remember that there's a plethora of would-be theorems out there that seem obvious but turn out to be wrong.
$endgroup$
– Shaun
Jan 22 at 18:58
$begingroup$
You've used the dreaded word "obvious". Is it really obvious (to you or to the intended reader) or just familiar? Remember that there's a plethora of would-be theorems out there that seem obvious but turn out to be wrong.
$endgroup$
– Shaun
Jan 22 at 18:58
add a comment |
2 Answers
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oldest
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$begingroup$
Your proof is fine. You can actually simplify it a bit since you don't even need to use induction at all once you have your composition series. Indeed, let $i$ be such that $p$ divides $G_i/G_{i+1}$. Then there is an element $xin G_i/G_{i+1}$ of order $p$ (since $G_i/G_{i+1}$ is simple and in particular cyclic). Now pick $yin G_i$ whose image in $G_i/G_{i+1}$ is $x$. The order of $y$ is a multiple of the order of $x$, and so the order of $y$ is divisible by $p$ and we're done.
If you try to make a similar argument for nonabelian groups, it would reduce to proving Cauchy's theorem for simple groups. This is a nontrivial reduction, but I don't think it actually really helps: I don't see how it would be any easier to prove Cauchy's theorem for simple groups than for arbitrary groups, except in that it handles the abelian case for proofs of Cauchy's theorem that treat the abelian and nonabelian cases separately.
answered Jan 22 at 18:04
Eric WofseyEric Wofsey
189k14216347
$endgroup$
add a comment |
$begingroup$
A different way of reaching the conclusion is as follows.
Let $phi$ be the Frobenius map $x rightarrow x^{p}$. As $G$ is abelian it is easy to see that $phi$ is a homomorphism. Clearly $im(phi) subseteq G_{1}$. Hence $ker(phi) neq 1$. Since the non-trivial elements in $ker(phi)$ are elements of order $p$, the result follows.
answered Jan 22 at 21:53
user515430user515430
17613
$endgroup$
add a comment |
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2 Answers
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$begingroup$
Your proof is fine. You can actually simplify it a bit since you don't even need to use induction at all once you have your composition series. Indeed, let $i$ be such that $p$ divides $G_i/G_{i+1}$. Then there is an element $xin G_i/G_{i+1}$ of order $p$ (since $G_i/G_{i+1}$ is simple and in particular cyclic). Now pick $yin G_i$ whose image in $G_i/G_{i+1}$ is $x$. The order of $y$ is a multiple of the order of $x$, and so the order of $y$ is divisible by $p$ and we're done.
If you try to make a similar argument for nonabelian groups, it would reduce to proving Cauchy's theorem for simple groups. This is a nontrivial reduction, but I don't think it actually really helps: I don't see how it would be any easier to prove Cauchy's theorem for simple groups than for arbitrary groups, except in that it handles the abelian case for proofs of Cauchy's theorem that treat the abelian and nonabelian cases separately.
answered Jan 22 at 18:04
Eric WofseyEric Wofsey
189k14216347
$endgroup$
add a comment |
$begingroup$
Your proof is fine. You can actually simplify it a bit since you don't even need to use induction at all once you have your composition series. Indeed, let $i$ be such that $p$ divides $G_i/G_{i+1}$. Then there is an element $xin G_i/G_{i+1}$ of order $p$ (since $G_i/G_{i+1}$ is simple and in particular cyclic). Now pick $yin G_i$ whose image in $G_i/G_{i+1}$ is $x$. The order of $y$ is a multiple of the order of $x$, and so the order of $y$ is divisible by $p$ and we're done.
If you try to make a similar argument for nonabelian groups, it would reduce to proving Cauchy's theorem for simple groups. This is a nontrivial reduction, but I don't think it actually really helps: I don't see how it would be any easier to prove Cauchy's theorem for simple groups than for arbitrary groups, except in that it handles the abelian case for proofs of Cauchy's theorem that treat the abelian and nonabelian cases separately.
answered Jan 22 at 18:04
Eric WofseyEric Wofsey
189k14216347
$endgroup$
add a comment |
$begingroup$
Your proof is fine. You can actually simplify it a bit since you don't even need to use induction at all once you have your composition series. Indeed, let $i$ be such that $p$ divides $G_i/G_{i+1}$. Then there is an element $xin G_i/G_{i+1}$ of order $p$ (since $G_i/G_{i+1}$ is simple and in particular cyclic). Now pick $yin G_i$ whose image in $G_i/G_{i+1}$ is $x$. The order of $y$ is a multiple of the order of $x$, and so the order of $y$ is divisible by $p$ and we're done.
If you try to make a similar argument for nonabelian groups, it would reduce to proving Cauchy's theorem for simple groups. This is a nontrivial reduction, but I don't think it actually really helps: I don't see how it would be any easier to prove Cauchy's theorem for simple groups than for arbitrary groups, except in that it handles the abelian case for proofs of Cauchy's theorem that treat the abelian and nonabelian cases separately.
answered Jan 22 at 18:04
Eric WofseyEric Wofsey
189k14216347
$endgroup$
Your proof is fine. You can actually simplify it a bit since you don't even need to use induction at all once you have your composition series. Indeed, let $i$ be such that $p$ divides $G_i/G_{i+1}$. Then there is an element $xin G_i/G_{i+1}$ of order $p$ (since $G_i/G_{i+1}$ is simple and in particular cyclic). Now pick $yin G_i$ whose image in $G_i/G_{i+1}$ is $x$. The order of $y$ is a multiple of the order of $x$, and so the order of $y$ is divisible by $p$ and we're done.
If you try to make a similar argument for nonabelian groups, it would reduce to proving Cauchy's theorem for simple groups. This is a nontrivial reduction, but I don't think it actually really helps: I don't see how it would be any easier to prove Cauchy's theorem for simple groups than for arbitrary groups, except in that it handles the abelian case for proofs of Cauchy's theorem that treat the abelian and nonabelian cases separately.
answered Jan 22 at 18:04
Eric WofseyEric Wofsey
189k14216347
answered Jan 22 at 18:04
Eric WofseyEric Wofsey
189k14216347
answered Jan 22 at 18:04
Eric WofseyEric Wofsey
189k14216347
answered Jan 22 at 18:04
Eric WofseyEric Wofsey
189k14216347
189k14216347
add a comment |
add a comment |
$begingroup$
A different way of reaching the conclusion is as follows.
Let $phi$ be the Frobenius map $x rightarrow x^{p}$. As $G$ is abelian it is easy to see that $phi$ is a homomorphism. Clearly $im(phi) subseteq G_{1}$. Hence $ker(phi) neq 1$. Since the non-trivial elements in $ker(phi)$ are elements of order $p$, the result follows.
answered Jan 22 at 21:53
user515430user515430
17613
$endgroup$
add a comment |
$begingroup$
A different way of reaching the conclusion is as follows.
Let $phi$ be the Frobenius map $x rightarrow x^{p}$. As $G$ is abelian it is easy to see that $phi$ is a homomorphism. Clearly $im(phi) subseteq G_{1}$. Hence $ker(phi) neq 1$. Since the non-trivial elements in $ker(phi)$ are elements of order $p$, the result follows.
answered Jan 22 at 21:53
user515430user515430
17613
$endgroup$
add a comment |
$begingroup$
A different way of reaching the conclusion is as follows.
Let $phi$ be the Frobenius map $x rightarrow x^{p}$. As $G$ is abelian it is easy to see that $phi$ is a homomorphism. Clearly $im(phi) subseteq G_{1}$. Hence $ker(phi) neq 1$. Since the non-trivial elements in $ker(phi)$ are elements of order $p$, the result follows.
answered Jan 22 at 21:53
user515430user515430
17613
$endgroup$
A different way of reaching the conclusion is as follows.
Let $phi$ be the Frobenius map $x rightarrow x^{p}$. As $G$ is abelian it is easy to see that $phi$ is a homomorphism. Clearly $im(phi) subseteq G_{1}$. Hence $ker(phi) neq 1$. Since the non-trivial elements in $ker(phi)$ are elements of order $p$, the result follows.
answered Jan 22 at 21:53
user515430user515430
17613
answered Jan 22 at 21:53
user515430user515430
17613
answered Jan 22 at 21:53
user515430user515430
17613
answered Jan 22 at 21:53
user515430user515430
17613
17613
add a comment |
add a comment |
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$begingroup$
You can compare your proof with a "book proof" here of Cauchy's Theorem for abelian groups, which is may be a bit easier than yours.
$endgroup$
– Dietrich Burde
Jan 22 at 17:51
1
$begingroup$
You've used the dreaded word "obvious". Is it really obvious (to you or to the intended reader) or just familiar? Remember that there's a plethora of would-be theorems out there that seem obvious but turn out to be wrong.
$endgroup$
– Shaun
Jan 22 at 18:58