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1

$begingroup$

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

share|cite|improve this question

edited 53 mins ago

Shaun

10.3k113686

asked 6 hours ago

JustWanderingJustWandering

592

$endgroup$

add a comment | 

1

$begingroup$

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

share|cite|improve this question

edited 53 mins ago

Shaun

10.3k113686

asked 6 hours ago

JustWanderingJustWandering

592

$endgroup$

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

share|cite|improve this question

edited 53 mins ago

Shaun

10.3k113686

asked 6 hours ago

JustWanderingJustWandering

592

$endgroup$

share|cite|improve this question

edited 53 mins ago

Shaun

10.3k113686

asked 6 hours ago

JustWanderingJustWandering

592

share|cite|improve this question

edited 53 mins ago

Shaun

10.3k113686

asked 6 hours ago

JustWanderingJustWandering

592

edited 53 mins ago

Shaun

10.3k113686

edited 53 mins ago

Shaun

10.3k113686

asked 6 hours ago

JustWanderingJustWandering

592

asked 6 hours ago

JustWanderingJustWandering

592

2 Answers
2

active

oldest

votes

3

$begingroup$

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:

1. 
   $e in H$,


2. if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,


3. 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 $G$.

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

rolandcyprolandcyp

2,309422

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

janmarqzjanmarqz

6,25741630

$endgroup$

add a comment | 

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3

$begingroup$

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:

1. 
   $e in H$,


2. if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,


3. 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 $G$.

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

rolandcyprolandcyp

2,309422

$endgroup$

add a comment | 

3

$begingroup$

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:

1. 
   $e in H$,


2. if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,


3. 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 $G$.

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

rolandcyprolandcyp

2,309422

$endgroup$

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:

1. 
   $e in H$,


2. if $h_1,h_2 in H$ then $h_1 odot h_2 in H$,


3. 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 $G$.

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

rolandcyprolandcyp

2,309422

$endgroup$

share|cite|improve this answer

edited 5 hours ago

answered 6 hours ago

rolandcyprolandcyp

2,309422

answered 6 hours ago

rolandcyprolandcyp

2,309422

answered 6 hours ago

rolandcyprolandcyp

2,309422

1

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

janmarqzjanmarqz

6,25741630

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 3 hours ago

janmarqzjanmarqz

6,25741630

$endgroup$

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.

share|cite|improve this answer

answered 3 hours ago

janmarqzjanmarqz

6,25741630

$endgroup$

share|cite|improve this answer

answered 3 hours ago

janmarqzjanmarqz

6,25741630

answered 3 hours ago

janmarqzjanmarqz

6,25741630

answered 3 hours ago

janmarqzjanmarqz

6,25741630