Mixing
$H,K lt G$ with $HK=G$
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$G$ is a group and $H lt G$ and $K lt G$, with $|G|=n$ then If $GCD( (G:H) ,(G:K))=1$ then $G=HK$
Any thoughts?
Edit by Batominovski: To prevent this thread from being closed or getting unnecessary downvotes, the OP has made an attempt to solve the problem. But the attempt was wrong, so it was removed.
group-theory finite-groups
edited Aug 3 at 20:33
Batominovski
22.6k22776
asked Aug 3 at 14:36
user528821
1406
 |Â
show 2 more comments
up vote
2
down vote
favorite
$G$ is a group and $H lt G$ and $K lt G$, with $|G|=n$ then If $GCD( (G:H) ,(G:K))=1$ then $G=HK$
Any thoughts?
Edit by Batominovski: To prevent this thread from being closed or getting unnecessary downvotes, the OP has made an attempt to solve the problem. But the attempt was wrong, so it was removed.
group-theory finite-groups
edited Aug 3 at 20:33
Batominovski
22.6k22776
asked Aug 3 at 14:36
user528821
1406
1
What is MDC? Is it GCD?
– Batominovski
Aug 3 at 14:40
@Batominovski yup, forgot tô translate, Will do
– user528821
Aug 3 at 14:41
You write LCD, but do you mean GCD? I mean if lcm$(k,l)=1$, then $k=1=l$.
– Nicky Hekster
Aug 3 at 14:48
1
I don't understand why you think the intersection of $H$ and $K$ is trivial. For example, take $G=mathbbZ_2timesmathbbZ_3times mathbbZ_5$, $H=mathbbZ_2timesmathbbZ_3times 0$, and $K=0timesmathbbZ_3timesmathbbZ_5$.
– Batominovski
Aug 3 at 14:50
1
@NickyHekster fixed it
– user528821
Aug 3 at 14:50
 |Â
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$G$ is a group and $H lt G$ and $K lt G$, with $|G|=n$ then If $GCD( (G:H) ,(G:K))=1$ then $G=HK$
Any thoughts?
Edit by Batominovski: To prevent this thread from being closed or getting unnecessary downvotes, the OP has made an attempt to solve the problem. But the attempt was wrong, so it was removed.
group-theory finite-groups
edited Aug 3 at 20:33
Batominovski
22.6k22776
asked Aug 3 at 14:36
user528821
1406
$G$ is a group and $H lt G$ and $K lt G$, with $|G|=n$ then If $GCD( (G:H) ,(G:K))=1$ then $G=HK$
Any thoughts?
Edit by Batominovski: To prevent this thread from being closed or getting unnecessary downvotes, the OP has made an attempt to solve the problem. But the attempt was wrong, so it was removed.
group-theory finite-groups
edited Aug 3 at 20:33
Batominovski
22.6k22776
asked Aug 3 at 14:36
user528821
1406
edited Aug 3 at 20:33
Batominovski
22.6k22776
edited Aug 3 at 20:33
Batominovski
22.6k22776
edited Aug 3 at 20:33
Batominovski
22.6k22776
22.6k22776
asked Aug 3 at 14:36
user528821
1406
asked Aug 3 at 14:36
user528821
1406
asked Aug 3 at 14:36
user528821
1406
1406
1
What is MDC? Is it GCD?
– Batominovski
Aug 3 at 14:40
@Batominovski yup, forgot tô translate, Will do
– user528821
Aug 3 at 14:41
You write LCD, but do you mean GCD? I mean if lcm$(k,l)=1$, then $k=1=l$.
– Nicky Hekster
Aug 3 at 14:48
1
I don't understand why you think the intersection of $H$ and $K$ is trivial. For example, take $G=mathbbZ_2timesmathbbZ_3times mathbbZ_5$, $H=mathbbZ_2timesmathbbZ_3times 0$, and $K=0timesmathbbZ_3timesmathbbZ_5$.
– Batominovski
Aug 3 at 14:50
1
@NickyHekster fixed it
– user528821
Aug 3 at 14:50
 |Â
show 2 more comments
1
What is MDC? Is it GCD?
– Batominovski
Aug 3 at 14:40
@Batominovski yup, forgot tô translate, Will do
– user528821
Aug 3 at 14:41
You write LCD, but do you mean GCD? I mean if lcm$(k,l)=1$, then $k=1=l$.
– Nicky Hekster
Aug 3 at 14:48
1
I don't understand why you think the intersection of $H$ and $K$ is trivial. For example, take $G=mathbbZ_2timesmathbbZ_3times mathbbZ_5$, $H=mathbbZ_2timesmathbbZ_3times 0$, and $K=0timesmathbbZ_3timesmathbbZ_5$.
– Batominovski
Aug 3 at 14:50
1
@NickyHekster fixed it
– user528821
Aug 3 at 14:50
1
1
What is MDC? Is it GCD?
– Batominovski
Aug 3 at 14:40
What is MDC? Is it GCD?
– Batominovski
Aug 3 at 14:40
@Batominovski yup, forgot tô translate, Will do
– user528821
Aug 3 at 14:41
@Batominovski yup, forgot tô translate, Will do
– user528821
Aug 3 at 14:41
You write LCD, but do you mean GCD? I mean if lcm$(k,l)=1$, then $k=1=l$.
– Nicky Hekster
Aug 3 at 14:48
You write LCD, but do you mean GCD? I mean if lcm$(k,l)=1$, then $k=1=l$.
– Nicky Hekster
Aug 3 at 14:48
1
1
I don't understand why you think the intersection of $H$ and $K$ is trivial. For example, take $G=mathbbZ_2timesmathbbZ_3times mathbbZ_5$, $H=mathbbZ_2timesmathbbZ_3times 0$, and $K=0timesmathbbZ_3timesmathbbZ_5$.
– Batominovski
Aug 3 at 14:50
I don't understand why you think the intersection of $H$ and $K$ is trivial. For example, take $G=mathbbZ_2timesmathbbZ_3times mathbbZ_5$, $H=mathbbZ_2timesmathbbZ_3times 0$, and $K=0timesmathbbZ_3timesmathbbZ_5$.
– Batominovski
Aug 3 at 14:50
1
1
@NickyHekster fixed it
– user528821
Aug 3 at 14:50
@NickyHekster fixed it
– user528821
Aug 3 at 14:50
 |Â
show 2 more comments
1 Answer
1
active
oldest
votes
up vote
5
down vote
accepted
I assume that $G$ is a finite group (due to the tag "finite-groups" and the quote "$|G|=n$"). Note (for example, from here) that
$$|HK|,|Hcap K|=|H|,|K|,.$$
That is,
$$frac=fracKHcap K=[K:Hcap K]inmathbbZ,.$$
Similarly,
$$fracK=fracHcap K=[H:Hcap K]inmathbbZ,.$$
That is, $|H|$ and $|K|$ both divides $|HK|$. This mean $textlcmbig(|H|,|K|big)$ divides $|HK|$, or
$$|HK|geq textlcmbig(|H|,|K|big),.$$
Consequently,
$$fracleq fractextlcmbig(=gcdleft(frac,fracKright),.$$
From the given condition
$$gcdleft(frac,fracKright)=gcdbig([G:H],[G:K]big)=1,,$$
we conclude that
$$fracleq 1text or |G|leq |HK|,.$$
Since $HKsubseteq G$ and $G$ is a finite set, we deduce that $G=HK$.
P.S. We indeed have the following proposition. See a comment by xsnl below for a sketch of a proof.
Proposition. Let $H$ and $K$ be subgroups of an arbitrary (finite or infinite) group $G$ such that $[G:H]$ and $[G:K]$ are finite. Suppose further that $gcdbig([G:H],[G:K]big)=1$. Then, $G=HK$.
answered Aug 3 at 15:43
Batominovski
22.6k22776
1
It follows from finite case. $[G, H cap K] leq [G, H][G, K]$, in particular, is finite. $H cap K$ can be not normal, but taking intersection of conjugates we have finite quotient of $G$ with kernel contained in both $H$ and $K$. Images of $H$ and $K$ generate it, so $G = HK$.
– xsnl
Aug 3 at 16:39
@xsnl Great argument! If you put that as an answer here, I will upvote it. This deserves to be an answer.
– Batominovski
Aug 3 at 16:41
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
I assume that $G$ is a finite group (due to the tag "finite-groups" and the quote "$|G|=n$"). Note (for example, from here) that
$$|HK|,|Hcap K|=|H|,|K|,.$$
That is,
$$frac=fracKHcap K=[K:Hcap K]inmathbbZ,.$$
Similarly,
$$fracK=fracHcap K=[H:Hcap K]inmathbbZ,.$$
That is, $|H|$ and $|K|$ both divides $|HK|$. This mean $textlcmbig(|H|,|K|big)$ divides $|HK|$, or
$$|HK|geq textlcmbig(|H|,|K|big),.$$
Consequently,
$$fracleq fractextlcmbig(=gcdleft(frac,fracKright),.$$
From the given condition
$$gcdleft(frac,fracKright)=gcdbig([G:H],[G:K]big)=1,,$$
we conclude that
$$fracleq 1text or |G|leq |HK|,.$$
Since $HKsubseteq G$ and $G$ is a finite set, we deduce that $G=HK$.
P.S. We indeed have the following proposition. See a comment by xsnl below for a sketch of a proof.
Proposition. Let $H$ and $K$ be subgroups of an arbitrary (finite or infinite) group $G$ such that $[G:H]$ and $[G:K]$ are finite. Suppose further that $gcdbig([G:H],[G:K]big)=1$. Then, $G=HK$.
answered Aug 3 at 15:43
Batominovski
22.6k22776
1
It follows from finite case. $[G, H cap K] leq [G, H][G, K]$, in particular, is finite. $H cap K$ can be not normal, but taking intersection of conjugates we have finite quotient of $G$ with kernel contained in both $H$ and $K$. Images of $H$ and $K$ generate it, so $G = HK$.
– xsnl
Aug 3 at 16:39
@xsnl Great argument! If you put that as an answer here, I will upvote it. This deserves to be an answer.
– Batominovski
Aug 3 at 16:41
add a comment |Â
up vote
5
down vote
accepted
I assume that $G$ is a finite group (due to the tag "finite-groups" and the quote "$|G|=n$"). Note (for example, from here) that
$$|HK|,|Hcap K|=|H|,|K|,.$$
That is,
$$frac=fracKHcap K=[K:Hcap K]inmathbbZ,.$$
Similarly,
$$fracK=fracHcap K=[H:Hcap K]inmathbbZ,.$$
That is, $|H|$ and $|K|$ both divides $|HK|$. This mean $textlcmbig(|H|,|K|big)$ divides $|HK|$, or
$$|HK|geq textlcmbig(|H|,|K|big),.$$
Consequently,
$$fracleq fractextlcmbig(=gcdleft(frac,fracKright),.$$
From the given condition
$$gcdleft(frac,fracKright)=gcdbig([G:H],[G:K]big)=1,,$$
we conclude that
$$fracleq 1text or |G|leq |HK|,.$$
Since $HKsubseteq G$ and $G$ is a finite set, we deduce that $G=HK$.
P.S. We indeed have the following proposition. See a comment by xsnl below for a sketch of a proof.
Proposition. Let $H$ and $K$ be subgroups of an arbitrary (finite or infinite) group $G$ such that $[G:H]$ and $[G:K]$ are finite. Suppose further that $gcdbig([G:H],[G:K]big)=1$. Then, $G=HK$.
answered Aug 3 at 15:43
Batominovski
22.6k22776
1
It follows from finite case. $[G, H cap K] leq [G, H][G, K]$, in particular, is finite. $H cap K$ can be not normal, but taking intersection of conjugates we have finite quotient of $G$ with kernel contained in both $H$ and $K$. Images of $H$ and $K$ generate it, so $G = HK$.
– xsnl
Aug 3 at 16:39
@xsnl Great argument! If you put that as an answer here, I will upvote it. This deserves to be an answer.
– Batominovski
Aug 3 at 16:41
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
I assume that $G$ is a finite group (due to the tag "finite-groups" and the quote "$|G|=n$"). Note (for example, from here) that
$$|HK|,|Hcap K|=|H|,|K|,.$$
That is,
$$frac=fracKHcap K=[K:Hcap K]inmathbbZ,.$$
Similarly,
$$fracK=fracHcap K=[H:Hcap K]inmathbbZ,.$$
That is, $|H|$ and $|K|$ both divides $|HK|$. This mean $textlcmbig(|H|,|K|big)$ divides $|HK|$, or
$$|HK|geq textlcmbig(|H|,|K|big),.$$
Consequently,
$$fracleq fractextlcmbig(=gcdleft(frac,fracKright),.$$
From the given condition
$$gcdleft(frac,fracKright)=gcdbig([G:H],[G:K]big)=1,,$$
we conclude that
$$fracleq 1text or |G|leq |HK|,.$$
Since $HKsubseteq G$ and $G$ is a finite set, we deduce that $G=HK$.
P.S. We indeed have the following proposition. See a comment by xsnl below for a sketch of a proof.
Proposition. Let $H$ and $K$ be subgroups of an arbitrary (finite or infinite) group $G$ such that $[G:H]$ and $[G:K]$ are finite. Suppose further that $gcdbig([G:H],[G:K]big)=1$. Then, $G=HK$.
answered Aug 3 at 15:43
Batominovski
22.6k22776
I assume that $G$ is a finite group (due to the tag "finite-groups" and the quote "$|G|=n$"). Note (for example, from here) that
$$|HK|,|Hcap K|=|H|,|K|,.$$
That is,
$$frac=fracKHcap K=[K:Hcap K]inmathbbZ,.$$
Similarly,
$$fracK=fracHcap K=[H:Hcap K]inmathbbZ,.$$
That is, $|H|$ and $|K|$ both divides $|HK|$. This mean $textlcmbig(|H|,|K|big)$ divides $|HK|$, or
$$|HK|geq textlcmbig(|H|,|K|big),.$$
Consequently,
$$fracleq fractextlcmbig(=gcdleft(frac,fracKright),.$$
From the given condition
$$gcdleft(frac,fracKright)=gcdbig([G:H],[G:K]big)=1,,$$
we conclude that
$$fracleq 1text or |G|leq |HK|,.$$
Since $HKsubseteq G$ and $G$ is a finite set, we deduce that $G=HK$.
P.S. We indeed have the following proposition. See a comment by xsnl below for a sketch of a proof.
Proposition. Let $H$ and $K$ be subgroups of an arbitrary (finite or infinite) group $G$ such that $[G:H]$ and $[G:K]$ are finite. Suppose further that $gcdbig([G:H],[G:K]big)=1$. Then, $G=HK$.
answered Aug 3 at 15:43
Batominovski
22.6k22776
edited Aug 3 at 16:54
answered Aug 3 at 15:43
Batominovski
22.6k22776
answered Aug 3 at 15:43
Batominovski
22.6k22776
answered Aug 3 at 15:43
Batominovski
22.6k22776
22.6k22776
1
It follows from finite case. $[G, H cap K] leq [G, H][G, K]$, in particular, is finite. $H cap K$ can be not normal, but taking intersection of conjugates we have finite quotient of $G$ with kernel contained in both $H$ and $K$. Images of $H$ and $K$ generate it, so $G = HK$.
– xsnl
Aug 3 at 16:39
@xsnl Great argument! If you put that as an answer here, I will upvote it. This deserves to be an answer.
– Batominovski
Aug 3 at 16:41
add a comment |Â
1
It follows from finite case. $[G, H cap K] leq [G, H][G, K]$, in particular, is finite. $H cap K$ can be not normal, but taking intersection of conjugates we have finite quotient of $G$ with kernel contained in both $H$ and $K$. Images of $H$ and $K$ generate it, so $G = HK$.
– xsnl
Aug 3 at 16:39
@xsnl Great argument! If you put that as an answer here, I will upvote it. This deserves to be an answer.
– Batominovski
Aug 3 at 16:41
1
1
It follows from finite case. $[G, H cap K] leq [G, H][G, K]$, in particular, is finite. $H cap K$ can be not normal, but taking intersection of conjugates we have finite quotient of $G$ with kernel contained in both $H$ and $K$. Images of $H$ and $K$ generate it, so $G = HK$.
– xsnl
Aug 3 at 16:39
It follows from finite case. $[G, H cap K] leq [G, H][G, K]$, in particular, is finite. $H cap K$ can be not normal, but taking intersection of conjugates we have finite quotient of $G$ with kernel contained in both $H$ and $K$. Images of $H$ and $K$ generate it, so $G = HK$.
– xsnl
Aug 3 at 16:39
@xsnl Great argument! If you put that as an answer here, I will upvote it. This deserves to be an answer.
– Batominovski
Aug 3 at 16:41
@xsnl Great argument! If you put that as an answer here, I will upvote it. This deserves to be an answer.
– Batominovski
Aug 3 at 16:41
add a comment |Â
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1
What is MDC? Is it GCD?
– Batominovski
Aug 3 at 14:40
@Batominovski yup, forgot tô translate, Will do
– user528821
Aug 3 at 14:41
You write LCD, but do you mean GCD? I mean if lcm$(k,l)=1$, then $k=1=l$.
– Nicky Hekster
Aug 3 at 14:48
1
I don't understand why you think the intersection of $H$ and $K$ is trivial. For example, take $G=mathbbZ_2timesmathbbZ_3times mathbbZ_5$, $H=mathbbZ_2timesmathbbZ_3times 0$, and $K=0timesmathbbZ_3timesmathbbZ_5$.
– Batominovski
Aug 3 at 14:50
1
@NickyHekster fixed it
– user528821
Aug 3 at 14:50