The order of an element in a quotient group
Suppose $G$ is a finite group, that $H$ is a subgroup of $G$, and that $N$ is a normal subgroup of $G$. Suppose that $|H| = n$ and $|G| = m|N|$, where $m$ and $n$ are coprime.
Consider the quotient group $G/N$ and let $h in H$. Determine the order of the element $hN$ in the group $G/N$.
Attempt:
Then $|G/N|=m$ , $|H|=n$ , $operatorname{gcd}(m,n)=1$
Order of the element $hN$ in $G/N$ is $k$ where $k$ is the smallest positive integer such that $h^k$ belongs to $N$. $(hN)^k = (h^k)(N)=1 $ iff $h^k$ belongs to $N$.
I am not sure how to use $|H|=n$ and $gcd(m,n)=1$ to find the order of $hN$.
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 19 '18 at 0:26
J. Dawson
143
add a comment |
Suppose $G$ is a finite group, that $H$ is a subgroup of $G$, and that $N$ is a normal subgroup of $G$. Suppose that $|H| = n$ and $|G| = m|N|$, where $m$ and $n$ are coprime.
Consider the quotient group $G/N$ and let $h in H$. Determine the order of the element $hN$ in the group $G/N$.
Attempt:
Then $|G/N|=m$ , $|H|=n$ , $operatorname{gcd}(m,n)=1$
Order of the element $hN$ in $G/N$ is $k$ where $k$ is the smallest positive integer such that $h^k$ belongs to $N$. $(hN)^k = (h^k)(N)=1 $ iff $h^k$ belongs to $N$.
I am not sure how to use $|H|=n$ and $gcd(m,n)=1$ to find the order of $hN$.
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 19 '18 at 0:26
J. Dawson
143
The order of $hN$ divides the order of $h$.
– Tortoise
Nov 19 '18 at 10:16
add a comment |
Suppose $G$ is a finite group, that $H$ is a subgroup of $G$, and that $N$ is a normal subgroup of $G$. Suppose that $|H| = n$ and $|G| = m|N|$, where $m$ and $n$ are coprime.
Consider the quotient group $G/N$ and let $h in H$. Determine the order of the element $hN$ in the group $G/N$.
Attempt:
Then $|G/N|=m$ , $|H|=n$ , $operatorname{gcd}(m,n)=1$
Order of the element $hN$ in $G/N$ is $k$ where $k$ is the smallest positive integer such that $h^k$ belongs to $N$. $(hN)^k = (h^k)(N)=1 $ iff $h^k$ belongs to $N$.
I am not sure how to use $|H|=n$ and $gcd(m,n)=1$ to find the order of $hN$.
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 19 '18 at 0:26
J. Dawson
143
Suppose $G$ is a finite group, that $H$ is a subgroup of $G$, and that $N$ is a normal subgroup of $G$. Suppose that $|H| = n$ and $|G| = m|N|$, where $m$ and $n$ are coprime.
Consider the quotient group $G/N$ and let $h in H$. Determine the order of the element $hN$ in the group $G/N$.
Attempt:
Then $|G/N|=m$ , $|H|=n$ , $operatorname{gcd}(m,n)=1$
Order of the element $hN$ in $G/N$ is $k$ where $k$ is the smallest positive integer such that $h^k$ belongs to $N$. $(hN)^k = (h^k)(N)=1 $ iff $h^k$ belongs to $N$.
I am not sure how to use $|H|=n$ and $gcd(m,n)=1$ to find the order of $hN$.
abstract-algebra group-theory normal-subgroups quotient-group
abstract-algebra group-theory normal-subgroups quotient-group
asked Nov 19 '18 at 0:26
J. Dawson
143
asked Nov 19 '18 at 0:26
J. Dawson
143
edited Nov 19 '18 at 0:34
asked Nov 19 '18 at 0:26
J. Dawson
143
asked Nov 19 '18 at 0:26
J. Dawson
143
asked Nov 19 '18 at 0:26
J. Dawson
143
143
The order of $hN$ divides the order of $h$.
– Tortoise
Nov 19 '18 at 10:16
add a comment |
The order of $hN$ divides the order of $h$.
– Tortoise
Nov 19 '18 at 10:16
The order of $hN$ divides the order of $h$.
– Tortoise
Nov 19 '18 at 10:16
The order of $hN$ divides the order of $h$.
– Tortoise
Nov 19 '18 at 10:16
add a comment |
1 Answer
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I got the idea now. $(|H|,|G/N|)=1$ is key to do this problem. Notice that $|hN|||h|$ (check) and also $|h|||H|$ so that $|hN|||H|$ but $|hN||(G/N)$ since $|H|$ and $|G/N|$ are relatively prime so that $|hN|=1$. Thus $hN=N$ and then $Hsubset N.$ This finishes the proof.
answered Nov 19 '18 at 0:46
gb2017
944
add a comment |
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I got the idea now. $(|H|,|G/N|)=1$ is key to do this problem. Notice that $|hN|||h|$ (check) and also $|h|||H|$ so that $|hN|||H|$ but $|hN||(G/N)$ since $|H|$ and $|G/N|$ are relatively prime so that $|hN|=1$. Thus $hN=N$ and then $Hsubset N.$ This finishes the proof.
answered Nov 19 '18 at 0:46
gb2017
944
add a comment |
I got the idea now. $(|H|,|G/N|)=1$ is key to do this problem. Notice that $|hN|||h|$ (check) and also $|h|||H|$ so that $|hN|||H|$ but $|hN||(G/N)$ since $|H|$ and $|G/N|$ are relatively prime so that $|hN|=1$. Thus $hN=N$ and then $Hsubset N.$ This finishes the proof.
answered Nov 19 '18 at 0:46
gb2017
944
add a comment |
I got the idea now. $(|H|,|G/N|)=1$ is key to do this problem. Notice that $|hN|||h|$ (check) and also $|h|||H|$ so that $|hN|||H|$ but $|hN||(G/N)$ since $|H|$ and $|G/N|$ are relatively prime so that $|hN|=1$. Thus $hN=N$ and then $Hsubset N.$ This finishes the proof.
answered Nov 19 '18 at 0:46
gb2017
944
I got the idea now. $(|H|,|G/N|)=1$ is key to do this problem. Notice that $|hN|||h|$ (check) and also $|h|||H|$ so that $|hN|||H|$ but $|hN||(G/N)$ since $|H|$ and $|G/N|$ are relatively prime so that $|hN|=1$. Thus $hN=N$ and then $Hsubset N.$ This finishes the proof.
answered Nov 19 '18 at 0:46
gb2017
944
edited Nov 19 '18 at 2:13
answered Nov 19 '18 at 0:46
gb2017
944
answered Nov 19 '18 at 0:46
gb2017
944
answered Nov 19 '18 at 0:46
gb2017
944
944
add a comment |
add a comment |
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The order of $hN$ divides the order of $h$.
– Tortoise
Nov 19 '18 at 10:16