Mixing
Alternative proof that $U(n^2-1)$ is not cyclic for $n>2$.
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.84 ibid. and I want to solve it using only the tools available in the textbook so far. (A free copy of the book is available online.)
Notation: The group $$({ainBbb Z_mmid gcd (a, m)=1}, times_m)$$ of units modulo $m$ under multiplication $times_m$ (or concatenation) modulo $m$ is denoted $U(m)$.
The Question:
For every integer $n$ greater than $2$, prove that the group $U(n^2-1)$ is not cyclic.
Thoughts:
I'm aware that $n^2-1=(n-1)(n+1)$ as a difference of two squares.
That $U(n^2-1)$ is cyclic for $n=2$ is clear by direct computation.
External methods (e.g., ideas of proofs that rely on, say, anachronistic techniques):
The result that $U(m)$ is cyclic iff $m$ is $2, 4, p^k, 2p^k$ for prime $p>2$ is not yet established in the text (I think).
(What I suspect is) the lemma that if a group $G$ contains at least four distinct elements $xin G$ such that $x^2=e$, then $G$ is not cyclic, is not clear to me; it's not established in the textbook so far and, yeah, I can sort of see why it's true (as the Klein four group is, intuitively, a (non-cyclic) subgroup of $G$). Please feel free to prove this lemma. It'd be sufficient, for me to understand the problem at hand.
The Chinese Remainder Theorem is not established yet (but it is on page 347; I'm up to page 92).
The theorem that $U(m)$ is cyclic iff $varphi(m)=lambda(m)$ is not established yet. (Here $varphi$ is Euler's totient function and $lambda$ is the Charmichael function.) In fact, I don't think it's mentioned at all (but I haven't looked very hard).
Please help :)
group-theory modular-arithmetic alternative-proof cyclic-groups
asked Jan 21 at 13:33
ShaunShaun
9,380113684
$endgroup$
add a comment |
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.84 ibid. and I want to solve it using only the tools available in the textbook so far. (A free copy of the book is available online.)
Notation: The group $$({ainBbb Z_mmid gcd (a, m)=1}, times_m)$$ of units modulo $m$ under multiplication $times_m$ (or concatenation) modulo $m$ is denoted $U(m)$.
The Question:
For every integer $n$ greater than $2$, prove that the group $U(n^2-1)$ is not cyclic.
Thoughts:
I'm aware that $n^2-1=(n-1)(n+1)$ as a difference of two squares.
That $U(n^2-1)$ is cyclic for $n=2$ is clear by direct computation.
External methods (e.g., ideas of proofs that rely on, say, anachronistic techniques):
The result that $U(m)$ is cyclic iff $m$ is $2, 4, p^k, 2p^k$ for prime $p>2$ is not yet established in the text (I think).
(What I suspect is) the lemma that if a group $G$ contains at least four distinct elements $xin G$ such that $x^2=e$, then $G$ is not cyclic, is not clear to me; it's not established in the textbook so far and, yeah, I can sort of see why it's true (as the Klein four group is, intuitively, a (non-cyclic) subgroup of $G$). Please feel free to prove this lemma. It'd be sufficient, for me to understand the problem at hand.
The Chinese Remainder Theorem is not established yet (but it is on page 347; I'm up to page 92).
The theorem that $U(m)$ is cyclic iff $varphi(m)=lambda(m)$ is not established yet. (Here $varphi$ is Euler's totient function and $lambda$ is the Charmichael function.) In fact, I don't think it's mentioned at all (but I haven't looked very hard).
Please help :)
group-theory modular-arithmetic alternative-proof cyclic-groups
asked Jan 21 at 13:33
ShaunShaun
9,380113684
$endgroup$
add a comment |
$begingroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.84 ibid. and I want to solve it using only the tools available in the textbook so far. (A free copy of the book is available online.)
Notation: The group $$({ainBbb Z_mmid gcd (a, m)=1}, times_m)$$ of units modulo $m$ under multiplication $times_m$ (or concatenation) modulo $m$ is denoted $U(m)$.
The Question:
For every integer $n$ greater than $2$, prove that the group $U(n^2-1)$ is not cyclic.
Thoughts:
I'm aware that $n^2-1=(n-1)(n+1)$ as a difference of two squares.
That $U(n^2-1)$ is cyclic for $n=2$ is clear by direct computation.
External methods (e.g., ideas of proofs that rely on, say, anachronistic techniques):
The result that $U(m)$ is cyclic iff $m$ is $2, 4, p^k, 2p^k$ for prime $p>2$ is not yet established in the text (I think).
(What I suspect is) the lemma that if a group $G$ contains at least four distinct elements $xin G$ such that $x^2=e$, then $G$ is not cyclic, is not clear to me; it's not established in the textbook so far and, yeah, I can sort of see why it's true (as the Klein four group is, intuitively, a (non-cyclic) subgroup of $G$). Please feel free to prove this lemma. It'd be sufficient, for me to understand the problem at hand.
The Chinese Remainder Theorem is not established yet (but it is on page 347; I'm up to page 92).
The theorem that $U(m)$ is cyclic iff $varphi(m)=lambda(m)$ is not established yet. (Here $varphi$ is Euler's totient function and $lambda$ is the Charmichael function.) In fact, I don't think it's mentioned at all (but I haven't looked very hard).
Please help :)
group-theory modular-arithmetic alternative-proof cyclic-groups
asked Jan 21 at 13:33
ShaunShaun
9,380113684
$endgroup$
I'm reading "Contemporary Abstract Algebra," by Gallian.
This is Exercise 4.84 ibid. and I want to solve it using only the tools available in the textbook so far. (A free copy of the book is available online.)
Notation: The group $$({ainBbb Z_mmid gcd (a, m)=1}, times_m)$$ of units modulo $m$ under multiplication $times_m$ (or concatenation) modulo $m$ is denoted $U(m)$.
The Question:
For every integer $n$ greater than $2$, prove that the group $U(n^2-1)$ is not cyclic.
Thoughts:
I'm aware that $n^2-1=(n-1)(n+1)$ as a difference of two squares.
That $U(n^2-1)$ is cyclic for $n=2$ is clear by direct computation.
External methods (e.g., ideas of proofs that rely on, say, anachronistic techniques):
The result that $U(m)$ is cyclic iff $m$ is $2, 4, p^k, 2p^k$ for prime $p>2$ is not yet established in the text (I think).
(What I suspect is) the lemma that if a group $G$ contains at least four distinct elements $xin G$ such that $x^2=e$, then $G$ is not cyclic, is not clear to me; it's not established in the textbook so far and, yeah, I can sort of see why it's true (as the Klein four group is, intuitively, a (non-cyclic) subgroup of $G$). Please feel free to prove this lemma. It'd be sufficient, for me to understand the problem at hand.
The Chinese Remainder Theorem is not established yet (but it is on page 347; I'm up to page 92).
The theorem that $U(m)$ is cyclic iff $varphi(m)=lambda(m)$ is not established yet. (Here $varphi$ is Euler's totient function and $lambda$ is the Charmichael function.) In fact, I don't think it's mentioned at all (but I haven't looked very hard).
Please help :)
group-theory modular-arithmetic alternative-proof cyclic-groups
group-theory modular-arithmetic alternative-proof cyclic-groups
asked Jan 21 at 13:33
ShaunShaun
9,380113684
asked Jan 21 at 13:33
ShaunShaun
9,380113684
edited Jan 21 at 20:46
Shaun
asked Jan 21 at 13:33
ShaunShaun
9,380113684
asked Jan 21 at 13:33
ShaunShaun
9,380113684
asked Jan 21 at 13:33
ShaunShaun
9,380113684
9,380113684
add a comment |
add a comment |
2 Answers
2
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oldest
votes
$begingroup$
Using Theorem 4.4, one sees that if $2$ divides $phi(n^2-1)$ (which is the order of $U(n^2-1)$, an even number) and this group is cyclic then there must be exactly $phi(2) = 1$ elements of order 2.
The order of $n^2-2$ is $2$ (it is congruent to -1 modulo $n^2-1$) and the order of $n$ is clearly $2$ as well. This contradicts Theorem 4.4 if we assume that the group is cyclic.
answered Jan 21 at 14:33
user328442user328442
1,8851516
$endgroup$
add a comment |
$begingroup$
Your second approach is right on: if $a=pm 1, pm n$, then $a^2 equiv 1 bmod n^2-1$.
The argument relies on this fact:
If $G$ is a cyclic group of order $m$ and $d$ divides $m$, then there is exactly one subgroup of $G$ of order $d$; it is the set of solutions of $x^d=1$.
For a proof, see here.
answered Jan 21 at 14:30
lhflhf
166k10171396
$endgroup$
$begingroup$
Well, no, it's not; it doesn't use any previous results from the textbook nor do I understand the lemma. Thank you anyway.
$endgroup$
– Shaun
Jan 21 at 14:31
$begingroup$
After this edit of yours, I've removed my downvote, but I can't bring myself to upvote this answer yet as, in the textbook, I can't find the fact you added here.
$endgroup$
– Shaun
Jan 21 at 15:47
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Using Theorem 4.4, one sees that if $2$ divides $phi(n^2-1)$ (which is the order of $U(n^2-1)$, an even number) and this group is cyclic then there must be exactly $phi(2) = 1$ elements of order 2.
The order of $n^2-2$ is $2$ (it is congruent to -1 modulo $n^2-1$) and the order of $n$ is clearly $2$ as well. This contradicts Theorem 4.4 if we assume that the group is cyclic.
answered Jan 21 at 14:33
user328442user328442
1,8851516
$endgroup$
add a comment |
$begingroup$
Using Theorem 4.4, one sees that if $2$ divides $phi(n^2-1)$ (which is the order of $U(n^2-1)$, an even number) and this group is cyclic then there must be exactly $phi(2) = 1$ elements of order 2.
The order of $n^2-2$ is $2$ (it is congruent to -1 modulo $n^2-1$) and the order of $n$ is clearly $2$ as well. This contradicts Theorem 4.4 if we assume that the group is cyclic.
answered Jan 21 at 14:33
user328442user328442
1,8851516
$endgroup$
add a comment |
$begingroup$
Using Theorem 4.4, one sees that if $2$ divides $phi(n^2-1)$ (which is the order of $U(n^2-1)$, an even number) and this group is cyclic then there must be exactly $phi(2) = 1$ elements of order 2.
The order of $n^2-2$ is $2$ (it is congruent to -1 modulo $n^2-1$) and the order of $n$ is clearly $2$ as well. This contradicts Theorem 4.4 if we assume that the group is cyclic.
answered Jan 21 at 14:33
user328442user328442
1,8851516
$endgroup$
Using Theorem 4.4, one sees that if $2$ divides $phi(n^2-1)$ (which is the order of $U(n^2-1)$, an even number) and this group is cyclic then there must be exactly $phi(2) = 1$ elements of order 2.
The order of $n^2-2$ is $2$ (it is congruent to -1 modulo $n^2-1$) and the order of $n$ is clearly $2$ as well. This contradicts Theorem 4.4 if we assume that the group is cyclic.
answered Jan 21 at 14:33
user328442user328442
1,8851516
answered Jan 21 at 14:33
user328442user328442
1,8851516
answered Jan 21 at 14:33
user328442user328442
1,8851516
answered Jan 21 at 14:33
user328442user328442
1,8851516
1,8851516
add a comment |
add a comment |
$begingroup$
Your second approach is right on: if $a=pm 1, pm n$, then $a^2 equiv 1 bmod n^2-1$.
The argument relies on this fact:
If $G$ is a cyclic group of order $m$ and $d$ divides $m$, then there is exactly one subgroup of $G$ of order $d$; it is the set of solutions of $x^d=1$.
For a proof, see here.
answered Jan 21 at 14:30
lhflhf
166k10171396
$endgroup$
$begingroup$
Well, no, it's not; it doesn't use any previous results from the textbook nor do I understand the lemma. Thank you anyway.
$endgroup$
– Shaun
Jan 21 at 14:31
$begingroup$
After this edit of yours, I've removed my downvote, but I can't bring myself to upvote this answer yet as, in the textbook, I can't find the fact you added here.
$endgroup$
– Shaun
Jan 21 at 15:47
add a comment |
$begingroup$
Your second approach is right on: if $a=pm 1, pm n$, then $a^2 equiv 1 bmod n^2-1$.
The argument relies on this fact:
If $G$ is a cyclic group of order $m$ and $d$ divides $m$, then there is exactly one subgroup of $G$ of order $d$; it is the set of solutions of $x^d=1$.
For a proof, see here.
answered Jan 21 at 14:30
lhflhf
166k10171396
$endgroup$
$begingroup$
Well, no, it's not; it doesn't use any previous results from the textbook nor do I understand the lemma. Thank you anyway.
$endgroup$
– Shaun
Jan 21 at 14:31
$begingroup$
After this edit of yours, I've removed my downvote, but I can't bring myself to upvote this answer yet as, in the textbook, I can't find the fact you added here.
$endgroup$
– Shaun
Jan 21 at 15:47
add a comment |
$begingroup$
Your second approach is right on: if $a=pm 1, pm n$, then $a^2 equiv 1 bmod n^2-1$.
The argument relies on this fact:
If $G$ is a cyclic group of order $m$ and $d$ divides $m$, then there is exactly one subgroup of $G$ of order $d$; it is the set of solutions of $x^d=1$.
For a proof, see here.
answered Jan 21 at 14:30
lhflhf
166k10171396
$endgroup$
Your second approach is right on: if $a=pm 1, pm n$, then $a^2 equiv 1 bmod n^2-1$.
The argument relies on this fact:
If $G$ is a cyclic group of order $m$ and $d$ divides $m$, then there is exactly one subgroup of $G$ of order $d$; it is the set of solutions of $x^d=1$.
For a proof, see here.
answered Jan 21 at 14:30
lhflhf
166k10171396
edited Jan 21 at 18:36
answered Jan 21 at 14:30
lhflhf
166k10171396
answered Jan 21 at 14:30
lhflhf
166k10171396
answered Jan 21 at 14:30
lhflhf
166k10171396
166k10171396
$begingroup$
Well, no, it's not; it doesn't use any previous results from the textbook nor do I understand the lemma. Thank you anyway.
$endgroup$
– Shaun
Jan 21 at 14:31
$begingroup$
After this edit of yours, I've removed my downvote, but I can't bring myself to upvote this answer yet as, in the textbook, I can't find the fact you added here.
$endgroup$
– Shaun
Jan 21 at 15:47
add a comment |
$begingroup$
Well, no, it's not; it doesn't use any previous results from the textbook nor do I understand the lemma. Thank you anyway.
$endgroup$
– Shaun
Jan 21 at 14:31
$begingroup$
After this edit of yours, I've removed my downvote, but I can't bring myself to upvote this answer yet as, in the textbook, I can't find the fact you added here.
$endgroup$
– Shaun
Jan 21 at 15:47
$begingroup$
Well, no, it's not; it doesn't use any previous results from the textbook nor do I understand the lemma. Thank you anyway.
$endgroup$
– Shaun
Jan 21 at 14:31
$begingroup$
Well, no, it's not; it doesn't use any previous results from the textbook nor do I understand the lemma. Thank you anyway.
$endgroup$
– Shaun
Jan 21 at 14:31
$begingroup$
After this edit of yours, I've removed my downvote, but I can't bring myself to upvote this answer yet as, in the textbook, I can't find the fact you added here.
$endgroup$
– Shaun
Jan 21 at 15:47
$begingroup$
After this edit of yours, I've removed my downvote, but I can't bring myself to upvote this answer yet as, in the textbook, I can't find the fact you added here.
$endgroup$
– Shaun
Jan 21 at 15:47
add a comment |
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