is the intersection of subgroups a subgroup of each subgroup
$begingroup$
Suppose $G$ is a group, take $H,K$ as subgroups of $G$ so $H,Kleqslant G$. I know that $Hcap Kleqslant G$ but is it the case that $Hcap Kleqslant H$ and $Hcap Kleqslant K$?
I am guessing this does not hold but why?
Also I tried with the case that $H=langle g rangle,K=langle h rangle$ where $g$ and $h$ are the elements in $G$ ($langle h rangle$ means the minimum subgroup that contains the element $h$ if you haven't seen this notation before). I used the subspace test and I think that $Hcap Kleqslant H$ and $Hcap Kleqslant K$ hold unless I make a mistake somewhere.
Much thanks in advance!
abstract-algebra group-theory
$endgroup$
add a comment |
$begingroup$
Suppose $G$ is a group, take $H,K$ as subgroups of $G$ so $H,Kleqslant G$. I know that $Hcap Kleqslant G$ but is it the case that $Hcap Kleqslant H$ and $Hcap Kleqslant K$?
I am guessing this does not hold but why?
Also I tried with the case that $H=langle g rangle,K=langle h rangle$ where $g$ and $h$ are the elements in $G$ ($langle h rangle$ means the minimum subgroup that contains the element $h$ if you haven't seen this notation before). I used the subspace test and I think that $Hcap Kleqslant H$ and $Hcap Kleqslant K$ hold unless I make a mistake somewhere.
Much thanks in advance!
abstract-algebra group-theory
$endgroup$
$begingroup$
A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
$endgroup$
– M. Winter
12 hours ago
add a comment |
$begingroup$
Suppose $G$ is a group, take $H,K$ as subgroups of $G$ so $H,Kleqslant G$. I know that $Hcap Kleqslant G$ but is it the case that $Hcap Kleqslant H$ and $Hcap Kleqslant K$?
I am guessing this does not hold but why?
Also I tried with the case that $H=langle g rangle,K=langle h rangle$ where $g$ and $h$ are the elements in $G$ ($langle h rangle$ means the minimum subgroup that contains the element $h$ if you haven't seen this notation before). I used the subspace test and I think that $Hcap Kleqslant H$ and $Hcap Kleqslant K$ hold unless I make a mistake somewhere.
Much thanks in advance!
abstract-algebra group-theory
$endgroup$
Suppose $G$ is a group, take $H,K$ as subgroups of $G$ so $H,Kleqslant G$. I know that $Hcap Kleqslant G$ but is it the case that $Hcap Kleqslant H$ and $Hcap Kleqslant K$?
I am guessing this does not hold but why?
Also I tried with the case that $H=langle g rangle,K=langle h rangle$ where $g$ and $h$ are the elements in $G$ ($langle h rangle$ means the minimum subgroup that contains the element $h$ if you haven't seen this notation before). I used the subspace test and I think that $Hcap Kleqslant H$ and $Hcap Kleqslant K$ hold unless I make a mistake somewhere.
Much thanks in advance!
abstract-algebra group-theory
abstract-algebra group-theory
edited 18 hours ago
Shaun
10.4k113686
10.4k113686
asked yesterday
JustWanderingJustWandering
692
692
$begingroup$
A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
$endgroup$
– M. Winter
12 hours ago
add a comment |
$begingroup$
A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
$endgroup$
– M. Winter
12 hours ago
$begingroup$
A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
$endgroup$
– M. Winter
12 hours ago
$begingroup$
A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
$endgroup$
– M. Winter
12 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:
Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:
$e in H$,
- if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,
- for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.
These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $odot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.
Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!
In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.
$endgroup$
add a comment |
$begingroup$
The subgroup test is:
$H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.
Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.
$endgroup$
add a comment |
$begingroup$
The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.
Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
begin{equation}
H text{ is a subgroup of } G
Leftrightarrow
H text{ is a subgroup of } K ,.
end{equation}
Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.
Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.
$endgroup$
add a comment |
Your Answer
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:
Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:
$e in H$,
- if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,
- for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.
These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $odot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.
Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!
In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.
$endgroup$
add a comment |
$begingroup$
It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:
Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:
$e in H$,
- if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,
- for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.
These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $odot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.
Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!
In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.
$endgroup$
add a comment |
$begingroup$
It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:
Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:
$e in H$,
- if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,
- for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.
These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $odot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.
Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!
In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.
$endgroup$
It is indeed true that $H cap K$ is a subgroup of both $H$ and $K$. For the sake of clarity, we recall the definition of a subgroup below:
Definition. Let $(G, odot)$ be group with identity $e$ and let $H$ be a subset of $G$. We say that $H$ is a subgroup of $G$ if each of the following hold:
$e in H$,
- if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,
- for all $h in H$, its inverse element $h^{-1}$ with respect to $odot$ is also in $H$.
These axioms make it so that $(H, odot)$ is a group in its own right with the very same group operation $odot$ and identity $e$. We also point out that these properties have less to do with the set $G$ than the operation $odot$ that $G$ comes equipped with.
Now, let $G$ be a group and let $H,K$ be subgroups of $G$. You have already verified that $H cap K$ is a subgroup of $G$. Why must it also be a subgroup of $H$ (and $K$)? First, it's clear that $H cap K subseteq H, H cap K subseteq K$ and that $H cap K ni e$. Moreover, because $H cap K$ is a subgroup of $G$, it is satisfies properties 2. and 3. above. Thus, by replacing $G$ with $H$ or $K$ in the definition above, it's immediate that $H cap K$ is a subgroup of $H$ (and $K$) as well!
In short, as long as $H$ is a subset of a group $G$ and $H$ satisfies the properties listed above, it will be a subgroup of $G$.
edited 8 hours ago
answered 23 hours ago
rolandcyprolandcyp
2,319422
2,319422
add a comment |
add a comment |
$begingroup$
The subgroup test is:
$H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.
Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.
$endgroup$
add a comment |
$begingroup$
The subgroup test is:
$H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.
Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.
$endgroup$
add a comment |
$begingroup$
The subgroup test is:
$H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.
Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.
$endgroup$
The subgroup test is:
$H$ is a subgroup of $G$ if and only if for each $x,yin H$ we've $xy^{-1}in H$.
Applied to a collection of subgroups ${H_s}$, let $x,yinbigcap_sH_s$ be a pair of elements. Then $x,yin H_s$ for all $S$ and since $H_s<G$ then $xy^{-1}in H_s$ for every index $s$, so $xy^{-1}inbigcap_sH_s$, hence $bigcap_sH_s$ is a subgroup too.
answered 21 hours ago
janmarqzjanmarqz
6,25741630
6,25741630
add a comment |
add a comment |
$begingroup$
The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.
Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
begin{equation}
H text{ is a subgroup of } G
Leftrightarrow
H text{ is a subgroup of } K ,.
end{equation}
Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.
Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.
$endgroup$
add a comment |
$begingroup$
The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.
Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
begin{equation}
H text{ is a subgroup of } G
Leftrightarrow
H text{ is a subgroup of } K ,.
end{equation}
Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.
Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.
$endgroup$
add a comment |
$begingroup$
The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.
Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
begin{equation}
H text{ is a subgroup of } G
Leftrightarrow
H text{ is a subgroup of } K ,.
end{equation}
Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.
Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.
$endgroup$
The intersection of two (or any quantity of) subgroups is always a subgroup of all the intersecting subgroups. Indeed, this is a particular case of the following general remark.
Remark. A subgroup $H leqslant G$ is also a subgroup of any subgroup $K leqslant G$ containing $H$. More precisesly, if $G$ is a group and $H subseteq K leqslant G$ then:
begin{equation}
H text{ is a subgroup of } G
Leftrightarrow
H text{ is a subgroup of } K ,.
end{equation}
Recall that a subset $H subseteq G$ is a subgroup of $(G,cdot)$ if and only if $H$ is a group with (the restriction to $H$ of) the operation in $G$. But, if $H subseteq K leqslant G$, then the restriction of $G$ to $H$ is the same as the restriction of $K$ to $H$. Hence, the condition for $H leqslant G$ is exactly the same as the condition $H leqslant K$.
Your question just corresponds to the case: $H cap K subseteq H,K leqslant G$.
answered 12 hours ago
suitangisuitangi
45928
45928
add a comment |
add a comment |
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$begingroup$
A subgroup is just a subset which behaves as a group under the same operation. The intersection of two subsets is a subset of both. The operation does never change, so subgroups behave like subsets in these aspects.
$endgroup$
– M. Winter
12 hours ago