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Property of prime divisors in group theory
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Suppose you have a group of order $ms$ and it is $m leq p$ for every prime factor of $s$. We want to show that a subgroup $H$ with $(G:H)=m$ is a normal subgroup of $G$.
So let $H$ be a subgroup of $G$ with $vert Hvert=s$ and $K$ be another subgroup of $G$ of order $s$. Then we have
$$ vert HKvert= fracvert H vert vert K vertvert H cap K vert=fracs^2vert H cap K vert leq ms Longleftrightarrow fracsvert H cap Kvert leq m.$$
Suppose now that $fracsvert H cap Kvert>1$. Is it always true that in this case $m$ is the smallest prime divisor of the group order? I read this in another post, but I don't see an argument for this.
My attempt:
$vert H cap K vert$ is a divisor of $s$ (how to show?). Then $m$ is the smallest prime divisor of $fracs^2vert H cap Kvert$. But how do we get to the order of $G$ in this case?
Maybe there is an easier way to prove this statement...
abstract-algebra group-theory number-theory normal-subgroups
asked 16 hours ago
Jan
116111
add a comment |Â
up vote
1
down vote
favorite
Suppose you have a group of order $ms$ and it is $m leq p$ for every prime factor of $s$. We want to show that a subgroup $H$ with $(G:H)=m$ is a normal subgroup of $G$.
So let $H$ be a subgroup of $G$ with $vert Hvert=s$ and $K$ be another subgroup of $G$ of order $s$. Then we have
$$ vert HKvert= fracvert H vert vert K vertvert H cap K vert=fracs^2vert H cap K vert leq ms Longleftrightarrow fracsvert H cap Kvert leq m.$$
Suppose now that $fracsvert H cap Kvert>1$. Is it always true that in this case $m$ is the smallest prime divisor of the group order? I read this in another post, but I don't see an argument for this.
My attempt:
$vert H cap K vert$ is a divisor of $s$ (how to show?). Then $m$ is the smallest prime divisor of $fracs^2vert H cap Kvert$. But how do we get to the order of $G$ in this case?
Maybe there is an easier way to prove this statement...
abstract-algebra group-theory number-theory normal-subgroups
asked 16 hours ago
Jan
116111
Yes if $|G:H|=p$ is the smallest prime dividing $|G|$, then $H$ is normal in $G$.
– Alan Wang
16 hours ago
@AlanWang I know, but why is in the upper case $m$ the smallest prime divisor of $vert Gvert=ms$? With the explanation of this the given statement follows immediately.
– Jan
16 hours ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose you have a group of order $ms$ and it is $m leq p$ for every prime factor of $s$. We want to show that a subgroup $H$ with $(G:H)=m$ is a normal subgroup of $G$.
So let $H$ be a subgroup of $G$ with $vert Hvert=s$ and $K$ be another subgroup of $G$ of order $s$. Then we have
$$ vert HKvert= fracvert H vert vert K vertvert H cap K vert=fracs^2vert H cap K vert leq ms Longleftrightarrow fracsvert H cap Kvert leq m.$$
Suppose now that $fracsvert H cap Kvert>1$. Is it always true that in this case $m$ is the smallest prime divisor of the group order? I read this in another post, but I don't see an argument for this.
My attempt:
$vert H cap K vert$ is a divisor of $s$ (how to show?). Then $m$ is the smallest prime divisor of $fracs^2vert H cap Kvert$. But how do we get to the order of $G$ in this case?
Maybe there is an easier way to prove this statement...
abstract-algebra group-theory number-theory normal-subgroups
asked 16 hours ago
Jan
116111
Suppose you have a group of order $ms$ and it is $m leq p$ for every prime factor of $s$. We want to show that a subgroup $H$ with $(G:H)=m$ is a normal subgroup of $G$.
So let $H$ be a subgroup of $G$ with $vert Hvert=s$ and $K$ be another subgroup of $G$ of order $s$. Then we have
$$ vert HKvert= fracvert H vert vert K vertvert H cap K vert=fracs^2vert H cap K vert leq ms Longleftrightarrow fracsvert H cap Kvert leq m.$$
Suppose now that $fracsvert H cap Kvert>1$. Is it always true that in this case $m$ is the smallest prime divisor of the group order? I read this in another post, but I don't see an argument for this.
My attempt:
$vert H cap K vert$ is a divisor of $s$ (how to show?). Then $m$ is the smallest prime divisor of $fracs^2vert H cap Kvert$. But how do we get to the order of $G$ in this case?
Maybe there is an easier way to prove this statement...
abstract-algebra group-theory number-theory normal-subgroups
asked 16 hours ago
Jan
116111
asked 16 hours ago
Jan
116111
asked 16 hours ago
Jan
116111
asked 16 hours ago
Jan
116111
116111
Yes if $|G:H|=p$ is the smallest prime dividing $|G|$, then $H$ is normal in $G$.
– Alan Wang
16 hours ago
@AlanWang I know, but why is in the upper case $m$ the smallest prime divisor of $vert Gvert=ms$? With the explanation of this the given statement follows immediately.
– Jan
16 hours ago
add a comment |Â
Yes if $|G:H|=p$ is the smallest prime dividing $|G|$, then $H$ is normal in $G$.
– Alan Wang
16 hours ago
@AlanWang I know, but why is in the upper case $m$ the smallest prime divisor of $vert Gvert=ms$? With the explanation of this the given statement follows immediately.
– Jan
16 hours ago
Yes if $|G:H|=p$ is the smallest prime dividing $|G|$, then $H$ is normal in $G$.
– Alan Wang
16 hours ago
Yes if $|G:H|=p$ is the smallest prime dividing $|G|$, then $H$ is normal in $G$.
– Alan Wang
16 hours ago
@AlanWang I know, but why is in the upper case $m$ the smallest prime divisor of $vert Gvert=ms$? With the explanation of this the given statement follows immediately.
– Jan
16 hours ago
@AlanWang I know, but why is in the upper case $m$ the smallest prime divisor of $vert Gvert=ms$? With the explanation of this the given statement follows immediately.
– Jan
16 hours ago
add a comment |Â
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Yes if $|G:H|=p$ is the smallest prime dividing $|G|$, then $H$ is normal in $G$.
– Alan Wang
16 hours ago
@AlanWang I know, but why is in the upper case $m$ the smallest prime divisor of $vert Gvert=ms$? With the explanation of this the given statement follows immediately.
– Jan
16 hours ago