How to derive the relation between $k$ and $l$ given $langle g^k rangle = langle g^l rangle$ in a cyclic...
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How to derive the relation between $k$ and $l$ given $langle g^k rangle = langle g^l rangle$ in a cyclic group $C_n = langle g rangle$?
The Next CEO of Stack OverflowProve that the order of the cyclic subgroup $langle g^krangle $ is $n/{operatorname{gcd}(n,k)}$Is $mathbb{Z}/langle nrangle$ cyclic and of order $n$?If an element $c$ of order $d$ belongs to both $langle arangle$ and $langle brangle$, then $langle arangle=langle brangle=langle crangle$?Cyclic group, automorphism, and isomorphism$G$ cyclic group of order $24$, and $H=langle x^6rangle$, find the order of each element of $G/H$Prove that the presentation of a cyclic group of infinite order is $langle amid rangle$Show that ${C_infty }/leftlangle {{c^n}} rightrangle simeq {C_n}$Efficient solution finding quotient group and is there generator in quotient group $mathbb{Z}_{15}^*/langle 11 rangle$??Prove that the order of the cyclic subgroup $langle g^krangle $ is $n/{operatorname{gcd}(n,k)}$What is the difference between cyclic subgroup $left langle aright rangle$ and $left langle a^{k} right rangle$?Prove $langle k rangle / langle km rangle simeq mathbb {Z}_m,$ where $langle k rangle$ is the subgroup generated by $k$ and $kin mathbb{Z}$
$begingroup$
It is known that
For a cyclic group $C_n = langle g rangle$ of order $n$, we have $langle g^k rangle = langle g^{(k, n)} rangle$, where $k in mathbb{Z}$.
I am able to verify this result.
Now, I try to derive it by showing something like
If $langle g^k rangle = langle g^l rangle$, then we have $l = (k, n)$ (in terms of "mod n").
$langle g^k rangle = langle g^l rangle$ means that $g^k in langle g^l rangle$ and $g^l in langle g^k rangle$.
That is, $exists t in mathbb{Z}: g^k = g^{lt}$ and $exists s in mathbb{Z}: g^l = g^{ks}$.
Therefore, $k equiv lt ; (text{mod}; n)$ and $l equiv ks ; (text{mod}; n)$.
How to proceed with this argument?
abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups
asked Mar 16 at 8:05
hengxinhengxin
1,6221429
$endgroup$
add a comment |
$begingroup$
It is known that
For a cyclic group $C_n = langle g rangle$ of order $n$, we have $langle g^k rangle = langle g^{(k, n)} rangle$, where $k in mathbb{Z}$.
I am able to verify this result.
Now, I try to derive it by showing something like
If $langle g^k rangle = langle g^l rangle$, then we have $l = (k, n)$ (in terms of "mod n").
$langle g^k rangle = langle g^l rangle$ means that $g^k in langle g^l rangle$ and $g^l in langle g^k rangle$.
That is, $exists t in mathbb{Z}: g^k = g^{lt}$ and $exists s in mathbb{Z}: g^l = g^{ks}$.
Therefore, $k equiv lt ; (text{mod}; n)$ and $l equiv ks ; (text{mod}; n)$.
How to proceed with this argument?
abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups
asked Mar 16 at 8:05
hengxinhengxin
1,6221429
$endgroup$
$begingroup$
You can proceed like here.
$endgroup$
– Dietrich Burde
Mar 16 at 9:09
$begingroup$
@DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
$endgroup$
– hengxin
Mar 16 at 9:41
add a comment |
$begingroup$
It is known that
For a cyclic group $C_n = langle g rangle$ of order $n$, we have $langle g^k rangle = langle g^{(k, n)} rangle$, where $k in mathbb{Z}$.
I am able to verify this result.
Now, I try to derive it by showing something like
If $langle g^k rangle = langle g^l rangle$, then we have $l = (k, n)$ (in terms of "mod n").
$langle g^k rangle = langle g^l rangle$ means that $g^k in langle g^l rangle$ and $g^l in langle g^k rangle$.
That is, $exists t in mathbb{Z}: g^k = g^{lt}$ and $exists s in mathbb{Z}: g^l = g^{ks}$.
Therefore, $k equiv lt ; (text{mod}; n)$ and $l equiv ks ; (text{mod}; n)$.
How to proceed with this argument?
abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups
asked Mar 16 at 8:05
hengxinhengxin
1,6221429
$endgroup$
It is known that
For a cyclic group $C_n = langle g rangle$ of order $n$, we have $langle g^k rangle = langle g^{(k, n)} rangle$, where $k in mathbb{Z}$.
I am able to verify this result.
Now, I try to derive it by showing something like
If $langle g^k rangle = langle g^l rangle$, then we have $l = (k, n)$ (in terms of "mod n").
$langle g^k rangle = langle g^l rangle$ means that $g^k in langle g^l rangle$ and $g^l in langle g^k rangle$.
That is, $exists t in mathbb{Z}: g^k = g^{lt}$ and $exists s in mathbb{Z}: g^l = g^{ks}$.
Therefore, $k equiv lt ; (text{mod}; n)$ and $l equiv ks ; (text{mod}; n)$.
How to proceed with this argument?
abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups
abstract-algebra group-theory modular-arithmetic greatest-common-divisor cyclic-groups
asked Mar 16 at 8:05
hengxinhengxin
1,6221429
asked Mar 16 at 8:05
hengxinhengxin
1,6221429
edited Mar 16 at 11:53
hengxin
asked Mar 16 at 8:05
hengxinhengxin
1,6221429
asked Mar 16 at 8:05
hengxinhengxin
1,6221429
asked Mar 16 at 8:05
hengxinhengxin
1,6221429
1,6221429
$begingroup$
You can proceed like here.
$endgroup$
– Dietrich Burde
Mar 16 at 9:09
$begingroup$
@DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
$endgroup$
– hengxin
Mar 16 at 9:41
add a comment |
$begingroup$
You can proceed like here.
$endgroup$
– Dietrich Burde
Mar 16 at 9:09
$begingroup$
@DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
$endgroup$
– hengxin
Mar 16 at 9:41
$begingroup$
You can proceed like here.
$endgroup$
– Dietrich Burde
Mar 16 at 9:09
$begingroup$
You can proceed like here.
$endgroup$
– Dietrich Burde
Mar 16 at 9:09
$begingroup$
@DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
$endgroup$
– hengxin
Mar 16 at 9:41
$begingroup$
@DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
$endgroup$
– hengxin
Mar 16 at 9:41
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.
answered Mar 16 at 17:16
W-t-PW-t-P
1,579612
$endgroup$
add a comment |
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$begingroup$
If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.
answered Mar 16 at 17:16
W-t-PW-t-P
1,579612
$endgroup$
add a comment |
$begingroup$
If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.
answered Mar 16 at 17:16
W-t-PW-t-P
1,579612
$endgroup$
add a comment |
$begingroup$
If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.
answered Mar 16 at 17:16
W-t-PW-t-P
1,579612
$endgroup$
If $dmid n$, then the order of $langle g^drangle$ is $n/d$. Thus, the order of $langle g^krangle=langle g^{(n,k)}rangle$ is $n/(n,k)$, and similarly, the order of $langle g^lrangle=langle g^{(n,l)}rangle$ is $n/(n,l)$. However, a cyclic group has just one subgroup of every possible order. Therefore, $langle g^krangle=langle g^lrangle$ if and only if $n/(k,n)=n/(l,n)$; that is, $(k,n)=(l,n)$.
answered Mar 16 at 17:16
W-t-PW-t-P
1,579612
answered Mar 16 at 17:16
W-t-PW-t-P
1,579612
answered Mar 16 at 17:16
W-t-PW-t-P
1,579612
answered Mar 16 at 17:16
W-t-PW-t-P
1,579612
1,579612
add a comment |
add a comment |
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$begingroup$
You can proceed like here.
$endgroup$
– Dietrich Burde
Mar 16 at 9:09
$begingroup$
@DietrichBurde Thanks. Because $frac{n}{(k,n)} = frac{n}{((k, n), n)}$, we can guess that $l = (k, n)$. However, how to derive that from $langle g^k rangle = langle g^l rangle$? Could you please show me more hints?
$endgroup$
– hengxin
Mar 16 at 9:41