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is the intersection of subgroups a subgroup of each subgroup
A group with no proper non-trivial subgroupsSubgroups that are isomorphic to each other, and contain a common element are the same subgroupIf a group has no maximal subgroups then all elements are non-generators? Frattini subgroup characterizationLet $P$, $Q$ be two Sylow p-subgroups of $G$, is it true that $N_P(Q)=Qcap P$?Subgroups of $G=(mathbb{Z}_{12},+)$join of pronormal subgroupsProperty of normally embedded subgroupsParity of order of intersection of cyclic and noncyclic subgroupsListing elements of the subgroups and generatorsIntersection of two subgroups
$begingroup$
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
$endgroup$
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 56 mins ago
Shaun
10.3k113686
10.3k113686
asked 6 hours ago
JustWanderingJustWandering
592
592
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add a comment |
2 Answers
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$begingroup$
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 $cdot$ 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 |
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2 Answers
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2 Answers
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$begingroup$
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 $cdot$ 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 $cdot$ 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 $cdot$ 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 $cdot$ 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 5 hours ago
answered 6 hours ago
rolandcyprolandcyp
2,309422
2,309422
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 3 hours ago
janmarqzjanmarqz
6,25741630
6,25741630
add a comment |
add a comment |
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