Does the additive group modulo $n$ have proper subgroups if and only if $n$ is not prime?
$begingroup$
After checking a few examples is seems as if it is true and I would have thought the elements of the group would be the multiples of the divisors of $n$ up until $n$, and how would I attempt to prove this?
group-theory modular-arithmetic
edited Dec 13 '18 at 22:27
Tianlalu
3,08621038
asked Dec 13 '18 at 21:27
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
406
$endgroup$
add a comment |
$begingroup$
After checking a few examples is seems as if it is true and I would have thought the elements of the group would be the multiples of the divisors of $n$ up until $n$, and how would I attempt to prove this?
group-theory modular-arithmetic
edited Dec 13 '18 at 22:27
Tianlalu
3,08621038
asked Dec 13 '18 at 21:27
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
406
$endgroup$
1
$begingroup$
Would lagrange's theorem be used?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 13 '18 at 21:29
$begingroup$
Proper subgroups, you mean? I presume yes.
$endgroup$
– Sean Roberson
Dec 13 '18 at 21:31
add a comment |
$begingroup$
After checking a few examples is seems as if it is true and I would have thought the elements of the group would be the multiples of the divisors of $n$ up until $n$, and how would I attempt to prove this?
group-theory modular-arithmetic
edited Dec 13 '18 at 22:27
Tianlalu
3,08621038
asked Dec 13 '18 at 21:27
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
406
$endgroup$
After checking a few examples is seems as if it is true and I would have thought the elements of the group would be the multiples of the divisors of $n$ up until $n$, and how would I attempt to prove this?
group-theory modular-arithmetic
group-theory modular-arithmetic
edited Dec 13 '18 at 22:27
Tianlalu
3,08621038
asked Dec 13 '18 at 21:27
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
406
edited Dec 13 '18 at 22:27
Tianlalu
3,08621038
asked Dec 13 '18 at 21:27
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
406
edited Dec 13 '18 at 22:27
Tianlalu
3,08621038
edited Dec 13 '18 at 22:27
Tianlalu
3,08621038
edited Dec 13 '18 at 22:27
Tianlalu
3,08621038
3,08621038
asked Dec 13 '18 at 21:27
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
406
asked Dec 13 '18 at 21:27
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
406
asked Dec 13 '18 at 21:27
4M4D3U5 M0Z4RT4M4D3U5 M0Z4RT
406
406
1
$begingroup$
Would lagrange's theorem be used?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 13 '18 at 21:29
$begingroup$
Proper subgroups, you mean? I presume yes.
$endgroup$
– Sean Roberson
Dec 13 '18 at 21:31
add a comment |
1
$begingroup$
Would lagrange's theorem be used?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 13 '18 at 21:29
$begingroup$
Proper subgroups, you mean? I presume yes.
$endgroup$
– Sean Roberson
Dec 13 '18 at 21:31
1
1
$begingroup$
Would lagrange's theorem be used?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 13 '18 at 21:29
$begingroup$
Would lagrange's theorem be used?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 13 '18 at 21:29
$begingroup$
Proper subgroups, you mean? I presume yes.
$endgroup$
– Sean Roberson
Dec 13 '18 at 21:31
$begingroup$
Proper subgroups, you mean? I presume yes.
$endgroup$
– Sean Roberson
Dec 13 '18 at 21:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You have that $(mathbb{Z}_n,+)$ is cyclic, generated by $1$ (modulo $n$, course), so $1$ has order $n$ in $(mathbb{Z}_n,+)$, then you can show that for every $d|n$, $d cdot 1$ has order $n/d$, and the subgroups generated by $d cdot 1, d | n$ are the only subgroups of $(mathbb{Z}_n,+)$.
You can show that $d cdot 1$ has order $n/d$ in the following way: let $o$ be the order of $d cdot 1$. You have that $displaystyle frac{n}{d} cdot left(d cdot 1right) = 0,$ so $o | displaystyle frac{n}{d}.$
On the other hand, you also have that $displaystyle left(o cdot d right)cdot 1 = d cdot (o cdot 1) = 0,$ so $n | (o cdot d) implies left( displaystyle d cdot frac{n}{d} right) | (o cdot d) implies displaystyle frac{n}{d} | o,$ so $o = displaystyle frac{n}{d}$.
If $n$ is not prime, you have that there is some $d in mathbb{N}, d|n$ such that the subgroup generetaed by $d cdot 1$ is proper.
If $(mathbb{Z}_n, +)$ has proper subgroups, then assume that $n$ is prime. Then, from Lagrange's theorem, you would have that the order of every subgroup would be a divisor of $n$, but since $n$ is prime, you would have no proper subgroups, which is a contradiction.
answered Dec 13 '18 at 21:48
CosminCosmin
1,4441527
$endgroup$
$begingroup$
A remark not important enough to warrant an Answer, since it falls out from your argument: It’s a Theorem that in a cyclic group of order $n$, there is one and only one subgroup of each order dividing $n$.
$endgroup$
– Lubin
Dec 13 '18 at 23:02
$begingroup$
I do not see the point of your comment. I know that it's a theorem, but the OP asked how should he prove it.
$endgroup$
– Cosmin
Dec 13 '18 at 23:10
add a comment |
Your Answer
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1 Answer
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$begingroup$
You have that $(mathbb{Z}_n,+)$ is cyclic, generated by $1$ (modulo $n$, course), so $1$ has order $n$ in $(mathbb{Z}_n,+)$, then you can show that for every $d|n$, $d cdot 1$ has order $n/d$, and the subgroups generated by $d cdot 1, d | n$ are the only subgroups of $(mathbb{Z}_n,+)$.
You can show that $d cdot 1$ has order $n/d$ in the following way: let $o$ be the order of $d cdot 1$. You have that $displaystyle frac{n}{d} cdot left(d cdot 1right) = 0,$ so $o | displaystyle frac{n}{d}.$
On the other hand, you also have that $displaystyle left(o cdot d right)cdot 1 = d cdot (o cdot 1) = 0,$ so $n | (o cdot d) implies left( displaystyle d cdot frac{n}{d} right) | (o cdot d) implies displaystyle frac{n}{d} | o,$ so $o = displaystyle frac{n}{d}$.
If $n$ is not prime, you have that there is some $d in mathbb{N}, d|n$ such that the subgroup generetaed by $d cdot 1$ is proper.
If $(mathbb{Z}_n, +)$ has proper subgroups, then assume that $n$ is prime. Then, from Lagrange's theorem, you would have that the order of every subgroup would be a divisor of $n$, but since $n$ is prime, you would have no proper subgroups, which is a contradiction.
answered Dec 13 '18 at 21:48
CosminCosmin
1,4441527
$endgroup$
$begingroup$
A remark not important enough to warrant an Answer, since it falls out from your argument: It’s a Theorem that in a cyclic group of order $n$, there is one and only one subgroup of each order dividing $n$.
$endgroup$
– Lubin
Dec 13 '18 at 23:02
$begingroup$
I do not see the point of your comment. I know that it's a theorem, but the OP asked how should he prove it.
$endgroup$
– Cosmin
Dec 13 '18 at 23:10
add a comment |
$begingroup$
You have that $(mathbb{Z}_n,+)$ is cyclic, generated by $1$ (modulo $n$, course), so $1$ has order $n$ in $(mathbb{Z}_n,+)$, then you can show that for every $d|n$, $d cdot 1$ has order $n/d$, and the subgroups generated by $d cdot 1, d | n$ are the only subgroups of $(mathbb{Z}_n,+)$.
You can show that $d cdot 1$ has order $n/d$ in the following way: let $o$ be the order of $d cdot 1$. You have that $displaystyle frac{n}{d} cdot left(d cdot 1right) = 0,$ so $o | displaystyle frac{n}{d}.$
On the other hand, you also have that $displaystyle left(o cdot d right)cdot 1 = d cdot (o cdot 1) = 0,$ so $n | (o cdot d) implies left( displaystyle d cdot frac{n}{d} right) | (o cdot d) implies displaystyle frac{n}{d} | o,$ so $o = displaystyle frac{n}{d}$.
If $n$ is not prime, you have that there is some $d in mathbb{N}, d|n$ such that the subgroup generetaed by $d cdot 1$ is proper.
If $(mathbb{Z}_n, +)$ has proper subgroups, then assume that $n$ is prime. Then, from Lagrange's theorem, you would have that the order of every subgroup would be a divisor of $n$, but since $n$ is prime, you would have no proper subgroups, which is a contradiction.
answered Dec 13 '18 at 21:48
CosminCosmin
1,4441527
$endgroup$
$begingroup$
A remark not important enough to warrant an Answer, since it falls out from your argument: It’s a Theorem that in a cyclic group of order $n$, there is one and only one subgroup of each order dividing $n$.
$endgroup$
– Lubin
Dec 13 '18 at 23:02
$begingroup$
I do not see the point of your comment. I know that it's a theorem, but the OP asked how should he prove it.
$endgroup$
– Cosmin
Dec 13 '18 at 23:10
add a comment |
$begingroup$
You have that $(mathbb{Z}_n,+)$ is cyclic, generated by $1$ (modulo $n$, course), so $1$ has order $n$ in $(mathbb{Z}_n,+)$, then you can show that for every $d|n$, $d cdot 1$ has order $n/d$, and the subgroups generated by $d cdot 1, d | n$ are the only subgroups of $(mathbb{Z}_n,+)$.
You can show that $d cdot 1$ has order $n/d$ in the following way: let $o$ be the order of $d cdot 1$. You have that $displaystyle frac{n}{d} cdot left(d cdot 1right) = 0,$ so $o | displaystyle frac{n}{d}.$
On the other hand, you also have that $displaystyle left(o cdot d right)cdot 1 = d cdot (o cdot 1) = 0,$ so $n | (o cdot d) implies left( displaystyle d cdot frac{n}{d} right) | (o cdot d) implies displaystyle frac{n}{d} | o,$ so $o = displaystyle frac{n}{d}$.
If $n$ is not prime, you have that there is some $d in mathbb{N}, d|n$ such that the subgroup generetaed by $d cdot 1$ is proper.
If $(mathbb{Z}_n, +)$ has proper subgroups, then assume that $n$ is prime. Then, from Lagrange's theorem, you would have that the order of every subgroup would be a divisor of $n$, but since $n$ is prime, you would have no proper subgroups, which is a contradiction.
answered Dec 13 '18 at 21:48
CosminCosmin
1,4441527
$endgroup$
You have that $(mathbb{Z}_n,+)$ is cyclic, generated by $1$ (modulo $n$, course), so $1$ has order $n$ in $(mathbb{Z}_n,+)$, then you can show that for every $d|n$, $d cdot 1$ has order $n/d$, and the subgroups generated by $d cdot 1, d | n$ are the only subgroups of $(mathbb{Z}_n,+)$.
You can show that $d cdot 1$ has order $n/d$ in the following way: let $o$ be the order of $d cdot 1$. You have that $displaystyle frac{n}{d} cdot left(d cdot 1right) = 0,$ so $o | displaystyle frac{n}{d}.$
On the other hand, you also have that $displaystyle left(o cdot d right)cdot 1 = d cdot (o cdot 1) = 0,$ so $n | (o cdot d) implies left( displaystyle d cdot frac{n}{d} right) | (o cdot d) implies displaystyle frac{n}{d} | o,$ so $o = displaystyle frac{n}{d}$.
If $n$ is not prime, you have that there is some $d in mathbb{N}, d|n$ such that the subgroup generetaed by $d cdot 1$ is proper.
If $(mathbb{Z}_n, +)$ has proper subgroups, then assume that $n$ is prime. Then, from Lagrange's theorem, you would have that the order of every subgroup would be a divisor of $n$, but since $n$ is prime, you would have no proper subgroups, which is a contradiction.
answered Dec 13 '18 at 21:48
CosminCosmin
1,4441527
edited Dec 13 '18 at 22:03
answered Dec 13 '18 at 21:48
CosminCosmin
1,4441527
answered Dec 13 '18 at 21:48
CosminCosmin
1,4441527
answered Dec 13 '18 at 21:48
CosminCosmin
1,4441527
1,4441527
$begingroup$
A remark not important enough to warrant an Answer, since it falls out from your argument: It’s a Theorem that in a cyclic group of order $n$, there is one and only one subgroup of each order dividing $n$.
$endgroup$
– Lubin
Dec 13 '18 at 23:02
$begingroup$
I do not see the point of your comment. I know that it's a theorem, but the OP asked how should he prove it.
$endgroup$
– Cosmin
Dec 13 '18 at 23:10
add a comment |
$begingroup$
A remark not important enough to warrant an Answer, since it falls out from your argument: It’s a Theorem that in a cyclic group of order $n$, there is one and only one subgroup of each order dividing $n$.
$endgroup$
– Lubin
Dec 13 '18 at 23:02
$begingroup$
I do not see the point of your comment. I know that it's a theorem, but the OP asked how should he prove it.
$endgroup$
– Cosmin
Dec 13 '18 at 23:10
$begingroup$
A remark not important enough to warrant an Answer, since it falls out from your argument: It’s a Theorem that in a cyclic group of order $n$, there is one and only one subgroup of each order dividing $n$.
$endgroup$
– Lubin
Dec 13 '18 at 23:02
$begingroup$
A remark not important enough to warrant an Answer, since it falls out from your argument: It’s a Theorem that in a cyclic group of order $n$, there is one and only one subgroup of each order dividing $n$.
$endgroup$
– Lubin
Dec 13 '18 at 23:02
$begingroup$
I do not see the point of your comment. I know that it's a theorem, but the OP asked how should he prove it.
$endgroup$
– Cosmin
Dec 13 '18 at 23:10
$begingroup$
I do not see the point of your comment. I know that it's a theorem, but the OP asked how should he prove it.
$endgroup$
– Cosmin
Dec 13 '18 at 23:10
add a comment |
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1
$begingroup$
Would lagrange's theorem be used?
$endgroup$
– 4M4D3U5 M0Z4RT
Dec 13 '18 at 21:29
$begingroup$
Proper subgroups, you mean? I presume yes.
$endgroup$
– Sean Roberson
Dec 13 '18 at 21:31