Mixing
Proving two subgroups with same cardinality are identical if one is normal
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I need to prove the following statement:
Let $G$ be a finite group, $H, K subset G$ two subgroups, $gcd(Card(H), Card(G/H))=1, Card(H)=Card(K), H$ normal in G. Then $H=K.$
We know that Card($G$) = Card($H$) [G:H], so Card($G$) = Card($H$) Card($G/H$). According to one of the isomorphism theorems, $KH<G, (Kcap H) triangleleft H $, and $f: K rightarrow G/H$ is a group homomorphism with ker $ f= Kcap H$ and $K/(K cap H) cong KH/H.$
Somehow I can not connect the dots. Can somebody help me.
Many thanks.
abstract-algebra
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user249018
194117
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up vote
2
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favorite
I need to prove the following statement:
Let $G$ be a finite group, $H, K subset G$ two subgroups, $gcd(Card(H), Card(G/H))=1, Card(H)=Card(K), H$ normal in G. Then $H=K.$
We know that Card($G$) = Card($H$) [G:H], so Card($G$) = Card($H$) Card($G/H$). According to one of the isomorphism theorems, $KH<G, (Kcap H) triangleleft H $, and $f: K rightarrow G/H$ is a group homomorphism with ker $ f= Kcap H$ and $K/(K cap H) cong KH/H.$
Somehow I can not connect the dots. Can somebody help me.
Many thanks.
abstract-algebra
asked yesterday
user249018
194117
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I need to prove the following statement:
Let $G$ be a finite group, $H, K subset G$ two subgroups, $gcd(Card(H), Card(G/H))=1, Card(H)=Card(K), H$ normal in G. Then $H=K.$
We know that Card($G$) = Card($H$) [G:H], so Card($G$) = Card($H$) Card($G/H$). According to one of the isomorphism theorems, $KH<G, (Kcap H) triangleleft H $, and $f: K rightarrow G/H$ is a group homomorphism with ker $ f= Kcap H$ and $K/(K cap H) cong KH/H.$
Somehow I can not connect the dots. Can somebody help me.
Many thanks.
abstract-algebra
asked yesterday
user249018
194117
I need to prove the following statement:
Let $G$ be a finite group, $H, K subset G$ two subgroups, $gcd(Card(H), Card(G/H))=1, Card(H)=Card(K), H$ normal in G. Then $H=K.$
We know that Card($G$) = Card($H$) [G:H], so Card($G$) = Card($H$) Card($G/H$). According to one of the isomorphism theorems, $KH<G, (Kcap H) triangleleft H $, and $f: K rightarrow G/H$ is a group homomorphism with ker $ f= Kcap H$ and $K/(K cap H) cong KH/H.$
Somehow I can not connect the dots. Can somebody help me.
Many thanks.
abstract-algebra
abstract-algebra
asked yesterday
user249018
194117
asked yesterday
user249018
194117
asked yesterday
user249018
194117
asked yesterday
user249018
194117
asked yesterday
user249018
194117
194117
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add a comment |
2 Answers
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Let $k in K$ and consider $kH in G/H$. Denote $n = |K| = |H|$ and $|k| = m$. Then since the order of $k$ divides the order of $K$, we have $m mid n$.
Notice that $(kH)^m = H$ in $G/H$, and so $|kH| = r$ must divide $m$ which divides $n$. Since $kH in G/H$, we also have that $r$ divides $|G/H|$. If $r >1$ then we have that $n$ and $|G/H|$ share a common factor greater than $1$, namely $r$. But this is a contradiction since we have assumed $gcd(n, |G/H|) = 1$.
So it must be that the order of $kH$ is $1$, which then implies that $k in H$. Since $k in K$ was arbitrary we have that $K leq H$. But this implies that $H = K$ since they have the same order.
answered yesterday
matt stokes
524210
Thanks. I dont understand the implication |$G/H|$ could devide |$K$|.
– user249018
yesterday
@user249018 Should be fixed now.
– matt stokes
yesterday
add a comment |
up vote
0
down vote
Matt stokes has already given a good answer, but I think it's a bit indirect, using the orders of elements of the image of $K$ in $G/H$. Instead we can directly consider the size of the image of $K$ in $G/H$ and do the following.
Let $phi : Gto G/H$ be the quotient map. Consider $phi(K)$. By the first isomorphism theorem, we have $phi(K)cong K/Kcap H$, so $|phi(K)|=|K/Kcap H|$, and $|phi(K)|$ divides $|K|=|H|$, however $phi(K)$ is also a subgroup of $G/H$, so $|phi(K)|$ divides $|G/H|$. Hence $|phi(K)|$ divides $operatorname{gcd}(|H|,|G/H|)=1$. Thus $|phi(K)|=1$, so $Ksubseteq H$, and since $|K|=|H|$, this implies $K=H$.
answered yesterday
jgon
9,69211538
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2 Answers
2
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oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $k in K$ and consider $kH in G/H$. Denote $n = |K| = |H|$ and $|k| = m$. Then since the order of $k$ divides the order of $K$, we have $m mid n$.
Notice that $(kH)^m = H$ in $G/H$, and so $|kH| = r$ must divide $m$ which divides $n$. Since $kH in G/H$, we also have that $r$ divides $|G/H|$. If $r >1$ then we have that $n$ and $|G/H|$ share a common factor greater than $1$, namely $r$. But this is a contradiction since we have assumed $gcd(n, |G/H|) = 1$.
So it must be that the order of $kH$ is $1$, which then implies that $k in H$. Since $k in K$ was arbitrary we have that $K leq H$. But this implies that $H = K$ since they have the same order.
answered yesterday
matt stokes
524210
Thanks. I dont understand the implication |$G/H|$ could devide |$K$|.
– user249018
yesterday
@user249018 Should be fixed now.
– matt stokes
yesterday
add a comment |
up vote
0
down vote
Let $k in K$ and consider $kH in G/H$. Denote $n = |K| = |H|$ and $|k| = m$. Then since the order of $k$ divides the order of $K$, we have $m mid n$.
Notice that $(kH)^m = H$ in $G/H$, and so $|kH| = r$ must divide $m$ which divides $n$. Since $kH in G/H$, we also have that $r$ divides $|G/H|$. If $r >1$ then we have that $n$ and $|G/H|$ share a common factor greater than $1$, namely $r$. But this is a contradiction since we have assumed $gcd(n, |G/H|) = 1$.
So it must be that the order of $kH$ is $1$, which then implies that $k in H$. Since $k in K$ was arbitrary we have that $K leq H$. But this implies that $H = K$ since they have the same order.
answered yesterday
matt stokes
524210
Thanks. I dont understand the implication |$G/H|$ could devide |$K$|.
– user249018
yesterday
@user249018 Should be fixed now.
– matt stokes
yesterday
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $k in K$ and consider $kH in G/H$. Denote $n = |K| = |H|$ and $|k| = m$. Then since the order of $k$ divides the order of $K$, we have $m mid n$.
Notice that $(kH)^m = H$ in $G/H$, and so $|kH| = r$ must divide $m$ which divides $n$. Since $kH in G/H$, we also have that $r$ divides $|G/H|$. If $r >1$ then we have that $n$ and $|G/H|$ share a common factor greater than $1$, namely $r$. But this is a contradiction since we have assumed $gcd(n, |G/H|) = 1$.
So it must be that the order of $kH$ is $1$, which then implies that $k in H$. Since $k in K$ was arbitrary we have that $K leq H$. But this implies that $H = K$ since they have the same order.
answered yesterday
matt stokes
524210
Let $k in K$ and consider $kH in G/H$. Denote $n = |K| = |H|$ and $|k| = m$. Then since the order of $k$ divides the order of $K$, we have $m mid n$.
Notice that $(kH)^m = H$ in $G/H$, and so $|kH| = r$ must divide $m$ which divides $n$. Since $kH in G/H$, we also have that $r$ divides $|G/H|$. If $r >1$ then we have that $n$ and $|G/H|$ share a common factor greater than $1$, namely $r$. But this is a contradiction since we have assumed $gcd(n, |G/H|) = 1$.
So it must be that the order of $kH$ is $1$, which then implies that $k in H$. Since $k in K$ was arbitrary we have that $K leq H$. But this implies that $H = K$ since they have the same order.
answered yesterday
matt stokes
524210
edited yesterday
answered yesterday
matt stokes
524210
answered yesterday
matt stokes
524210
answered yesterday
matt stokes
524210
524210
Thanks. I dont understand the implication |$G/H|$ could devide |$K$|.
– user249018
yesterday
@user249018 Should be fixed now.
– matt stokes
yesterday
add a comment |
Thanks. I dont understand the implication |$G/H|$ could devide |$K$|.
– user249018
yesterday
@user249018 Should be fixed now.
– matt stokes
yesterday
Thanks. I dont understand the implication |$G/H|$ could devide |$K$|.
– user249018
yesterday
Thanks. I dont understand the implication |$G/H|$ could devide |$K$|.
– user249018
yesterday
@user249018 Should be fixed now.
– matt stokes
yesterday
@user249018 Should be fixed now.
– matt stokes
yesterday
add a comment |
up vote
0
down vote
Matt stokes has already given a good answer, but I think it's a bit indirect, using the orders of elements of the image of $K$ in $G/H$. Instead we can directly consider the size of the image of $K$ in $G/H$ and do the following.
Let $phi : Gto G/H$ be the quotient map. Consider $phi(K)$. By the first isomorphism theorem, we have $phi(K)cong K/Kcap H$, so $|phi(K)|=|K/Kcap H|$, and $|phi(K)|$ divides $|K|=|H|$, however $phi(K)$ is also a subgroup of $G/H$, so $|phi(K)|$ divides $|G/H|$. Hence $|phi(K)|$ divides $operatorname{gcd}(|H|,|G/H|)=1$. Thus $|phi(K)|=1$, so $Ksubseteq H$, and since $|K|=|H|$, this implies $K=H$.
answered yesterday
jgon
9,69211538
add a comment |
up vote
0
down vote
Matt stokes has already given a good answer, but I think it's a bit indirect, using the orders of elements of the image of $K$ in $G/H$. Instead we can directly consider the size of the image of $K$ in $G/H$ and do the following.
Let $phi : Gto G/H$ be the quotient map. Consider $phi(K)$. By the first isomorphism theorem, we have $phi(K)cong K/Kcap H$, so $|phi(K)|=|K/Kcap H|$, and $|phi(K)|$ divides $|K|=|H|$, however $phi(K)$ is also a subgroup of $G/H$, so $|phi(K)|$ divides $|G/H|$. Hence $|phi(K)|$ divides $operatorname{gcd}(|H|,|G/H|)=1$. Thus $|phi(K)|=1$, so $Ksubseteq H$, and since $|K|=|H|$, this implies $K=H$.
answered yesterday
jgon
9,69211538
add a comment |
up vote
0
down vote
up vote
0
down vote
Matt stokes has already given a good answer, but I think it's a bit indirect, using the orders of elements of the image of $K$ in $G/H$. Instead we can directly consider the size of the image of $K$ in $G/H$ and do the following.
Let $phi : Gto G/H$ be the quotient map. Consider $phi(K)$. By the first isomorphism theorem, we have $phi(K)cong K/Kcap H$, so $|phi(K)|=|K/Kcap H|$, and $|phi(K)|$ divides $|K|=|H|$, however $phi(K)$ is also a subgroup of $G/H$, so $|phi(K)|$ divides $|G/H|$. Hence $|phi(K)|$ divides $operatorname{gcd}(|H|,|G/H|)=1$. Thus $|phi(K)|=1$, so $Ksubseteq H$, and since $|K|=|H|$, this implies $K=H$.
answered yesterday
jgon
9,69211538
Matt stokes has already given a good answer, but I think it's a bit indirect, using the orders of elements of the image of $K$ in $G/H$. Instead we can directly consider the size of the image of $K$ in $G/H$ and do the following.
Let $phi : Gto G/H$ be the quotient map. Consider $phi(K)$. By the first isomorphism theorem, we have $phi(K)cong K/Kcap H$, so $|phi(K)|=|K/Kcap H|$, and $|phi(K)|$ divides $|K|=|H|$, however $phi(K)$ is also a subgroup of $G/H$, so $|phi(K)|$ divides $|G/H|$. Hence $|phi(K)|$ divides $operatorname{gcd}(|H|,|G/H|)=1$. Thus $|phi(K)|=1$, so $Ksubseteq H$, and since $|K|=|H|$, this implies $K=H$.
answered yesterday
jgon
9,69211538
answered yesterday
jgon
9,69211538
answered yesterday
jgon
9,69211538
answered yesterday
jgon
9,69211538
9,69211538
add a comment |
add a comment |
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