If $G$ is an abelian group, then $H=g$ is a subgroup of $G$
Clash Royale CLAN TAG#URR8PPP
up vote
5
down vote
favorite
$$H= text divides 12$$
We have to prove that $H$ is a subgroup of $G$.
Consider $ain G$ and, the group $langle a rangle$ where $|a|=12$. This group will contain elements orders of which will be divide 12. This is a cyclic subgroup I think this should be equal to H but I am not sure as I don't know if G is finite.
abstract-algebra abelian-groups cyclic-groups
edited Sep 3 at 16:23
Asaf Karagila♦
295k32411739
asked Sep 3 at 8:27
Piyush Divyanakar
3,317222
 |Â
show 1 more comment
up vote
5
down vote
favorite
$$H= text divides 12$$
We have to prove that $H$ is a subgroup of $G$.
Consider $ain G$ and, the group $langle a rangle$ where $|a|=12$. This group will contain elements orders of which will be divide 12. This is a cyclic subgroup I think this should be equal to H but I am not sure as I don't know if G is finite.
abstract-algebra abelian-groups cyclic-groups
edited Sep 3 at 16:23
Asaf Karagila♦
295k32411739
asked Sep 3 at 8:27
Piyush Divyanakar
3,317222
2
If you have $G = Bbb Z_12times Bbb Z_12$, then $H = G$ but $H$ isn't cyclic. So your idea with $langle arangle$ doesn't work. For another reason it doesn't work, if $G = Bbb Z_4$, then once again $H = G$, but no element has order $12$.
– Arthur
Sep 3 at 8:30
2
You can not be sure that such an element $a$ exists in general. For all we know, $H$ could be as small as the $1$-element group $e$.
– Suzet
Sep 3 at 8:31
is the number $12$ significant somehow?
– Alvin Lepik
Sep 3 at 8:32
How else can I approach this?
– Piyush Divyanakar
Sep 3 at 8:32
1
How would you normally prove a statement like "Show that the subset $A$ of a group $B$ is actually a subgroup"? I would guess that you have a list of three properties, including something very much like "$ein A$". Try those, and if you can't do them all, come back here with all your results and we'll help you get the rest.
– Arthur
Sep 3 at 8:34
 |Â
show 1 more comment
up vote
5
down vote
favorite
up vote
5
down vote
favorite
$$H= text divides 12$$
We have to prove that $H$ is a subgroup of $G$.
Consider $ain G$ and, the group $langle a rangle$ where $|a|=12$. This group will contain elements orders of which will be divide 12. This is a cyclic subgroup I think this should be equal to H but I am not sure as I don't know if G is finite.
abstract-algebra abelian-groups cyclic-groups
edited Sep 3 at 16:23
Asaf Karagila♦
295k32411739
asked Sep 3 at 8:27
Piyush Divyanakar
3,317222
$$H= text divides 12$$
We have to prove that $H$ is a subgroup of $G$.
Consider $ain G$ and, the group $langle a rangle$ where $|a|=12$. This group will contain elements orders of which will be divide 12. This is a cyclic subgroup I think this should be equal to H but I am not sure as I don't know if G is finite.
abstract-algebra abelian-groups cyclic-groups
abstract-algebra abelian-groups cyclic-groups
edited Sep 3 at 16:23
Asaf Karagila♦
295k32411739
asked Sep 3 at 8:27
Piyush Divyanakar
3,317222
edited Sep 3 at 16:23
Asaf Karagila♦
295k32411739
asked Sep 3 at 8:27
Piyush Divyanakar
3,317222
edited Sep 3 at 16:23
Asaf Karagila♦
295k32411739
edited Sep 3 at 16:23
Asaf Karagila♦
295k32411739
edited Sep 3 at 16:23
Asaf Karagila♦
295k32411739
295k32411739
asked Sep 3 at 8:27
Piyush Divyanakar
3,317222
asked Sep 3 at 8:27
Piyush Divyanakar
3,317222
asked Sep 3 at 8:27
Piyush Divyanakar
3,317222
3,317222
2
If you have $G = Bbb Z_12times Bbb Z_12$, then $H = G$ but $H$ isn't cyclic. So your idea with $langle arangle$ doesn't work. For another reason it doesn't work, if $G = Bbb Z_4$, then once again $H = G$, but no element has order $12$.
– Arthur
Sep 3 at 8:30
2
You can not be sure that such an element $a$ exists in general. For all we know, $H$ could be as small as the $1$-element group $e$.
– Suzet
Sep 3 at 8:31
is the number $12$ significant somehow?
– Alvin Lepik
Sep 3 at 8:32
How else can I approach this?
– Piyush Divyanakar
Sep 3 at 8:32
1
How would you normally prove a statement like "Show that the subset $A$ of a group $B$ is actually a subgroup"? I would guess that you have a list of three properties, including something very much like "$ein A$". Try those, and if you can't do them all, come back here with all your results and we'll help you get the rest.
– Arthur
Sep 3 at 8:34
 |Â
show 1 more comment
2
If you have $G = Bbb Z_12times Bbb Z_12$, then $H = G$ but $H$ isn't cyclic. So your idea with $langle arangle$ doesn't work. For another reason it doesn't work, if $G = Bbb Z_4$, then once again $H = G$, but no element has order $12$.
– Arthur
Sep 3 at 8:30
2
You can not be sure that such an element $a$ exists in general. For all we know, $H$ could be as small as the $1$-element group $e$.
– Suzet
Sep 3 at 8:31
is the number $12$ significant somehow?
– Alvin Lepik
Sep 3 at 8:32
How else can I approach this?
– Piyush Divyanakar
Sep 3 at 8:32
1
How would you normally prove a statement like "Show that the subset $A$ of a group $B$ is actually a subgroup"? I would guess that you have a list of three properties, including something very much like "$ein A$". Try those, and if you can't do them all, come back here with all your results and we'll help you get the rest.
– Arthur
Sep 3 at 8:34
2
2
If you have $G = Bbb Z_12times Bbb Z_12$, then $H = G$ but $H$ isn't cyclic. So your idea with $langle arangle$ doesn't work. For another reason it doesn't work, if $G = Bbb Z_4$, then once again $H = G$, but no element has order $12$.
– Arthur
Sep 3 at 8:30
If you have $G = Bbb Z_12times Bbb Z_12$, then $H = G$ but $H$ isn't cyclic. So your idea with $langle arangle$ doesn't work. For another reason it doesn't work, if $G = Bbb Z_4$, then once again $H = G$, but no element has order $12$.
– Arthur
Sep 3 at 8:30
2
2
You can not be sure that such an element $a$ exists in general. For all we know, $H$ could be as small as the $1$-element group $e$.
– Suzet
Sep 3 at 8:31
You can not be sure that such an element $a$ exists in general. For all we know, $H$ could be as small as the $1$-element group $e$.
– Suzet
Sep 3 at 8:31
is the number $12$ significant somehow?
– Alvin Lepik
Sep 3 at 8:32
is the number $12$ significant somehow?
– Alvin Lepik
Sep 3 at 8:32
How else can I approach this?
– Piyush Divyanakar
Sep 3 at 8:32
How else can I approach this?
– Piyush Divyanakar
Sep 3 at 8:32
1
1
How would you normally prove a statement like "Show that the subset $A$ of a group $B$ is actually a subgroup"? I would guess that you have a list of three properties, including something very much like "$ein A$". Try those, and if you can't do them all, come back here with all your results and we'll help you get the rest.
– Arthur
Sep 3 at 8:34
How would you normally prove a statement like "Show that the subset $A$ of a group $B$ is actually a subgroup"? I would guess that you have a list of three properties, including something very much like "$ein A$". Try those, and if you can't do them all, come back here with all your results and we'll help you get the rest.
– Arthur
Sep 3 at 8:34
 |Â
show 1 more comment
2 Answers
2
active
oldest
votes
up vote
9
down vote
accepted
Hint: We can rewrite $H=gin Gmid 12g=0$. Therefore, if you can show that the map $gmapsto 12g$ is a homomorphism, then you have that $H$ is the kernel of a homomorphism and therefore a subgroup.
answered Sep 3 at 9:00
Aaron
15.4k22552
I have just started so I haven't covered homorphisms yet so I couldn't think in this direction, but I understand the idea based on my Linear algebra knowledge. Great solve thanks.
– Piyush Divyanakar
Sep 4 at 7:21
@PiyushDivyanakar You can use the rewriting without talking about homomorphisms. Show that if $12g=0$, then $12(-g)=0$, if $12g=12h=0$, then $12(g+h)=0$, and $12(0)=0$. There are more things to check this way, but it is still straight forward.
– Aaron
Sep 4 at 7:24
Yes that's how I solved it. I based it on ArsenBerk's answer.
– Piyush Divyanakar
Sep 4 at 7:26
add a comment |Â
up vote
4
down vote
Instead of seeking a specific example like cyclic groups, you can prove it in a more general way:
HINT: $H le G$ if and only if
1) $H$ is not empty.
2) $H$ is closed under binary operation of $G$.
3) $H$ is closed under inverses.
answered Sep 3 at 8:39
ArsenBerk
7,00221034
2
"actually 2) and 3) implies 1) but we need to ensure $H$ is non-empty." Then 2) and 3) do not imply 1). 2) and 3) together with "1') $H$ is non-empty" imply 1).
– Arthur
Sep 3 at 8:42
Oh, that is really logical.
– ArsenBerk
Sep 3 at 8:43
1
Yes, we do assume that. It's a convention called vacuous truth, and it often does seem counterintuitive to people when they first encounter it. But it works better with all our other logic than any alternative.
– Arthur
Sep 3 at 8:48
1
Well, after I asked, I thought that it is not counter-intuitive because in an empty set, there is no element that breaks closure under binary operation or closure under inverses. I will check "vacuous truth" out, thank you for that :)
– ArsenBerk
Sep 3 at 8:50
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
accepted
Hint: We can rewrite $H=gin Gmid 12g=0$. Therefore, if you can show that the map $gmapsto 12g$ is a homomorphism, then you have that $H$ is the kernel of a homomorphism and therefore a subgroup.
answered Sep 3 at 9:00
Aaron
15.4k22552
I have just started so I haven't covered homorphisms yet so I couldn't think in this direction, but I understand the idea based on my Linear algebra knowledge. Great solve thanks.
– Piyush Divyanakar
Sep 4 at 7:21
@PiyushDivyanakar You can use the rewriting without talking about homomorphisms. Show that if $12g=0$, then $12(-g)=0$, if $12g=12h=0$, then $12(g+h)=0$, and $12(0)=0$. There are more things to check this way, but it is still straight forward.
– Aaron
Sep 4 at 7:24
Yes that's how I solved it. I based it on ArsenBerk's answer.
– Piyush Divyanakar
Sep 4 at 7:26
add a comment |Â
up vote
9
down vote
accepted
Hint: We can rewrite $H=gin Gmid 12g=0$. Therefore, if you can show that the map $gmapsto 12g$ is a homomorphism, then you have that $H$ is the kernel of a homomorphism and therefore a subgroup.
answered Sep 3 at 9:00
Aaron
15.4k22552
I have just started so I haven't covered homorphisms yet so I couldn't think in this direction, but I understand the idea based on my Linear algebra knowledge. Great solve thanks.
– Piyush Divyanakar
Sep 4 at 7:21
@PiyushDivyanakar You can use the rewriting without talking about homomorphisms. Show that if $12g=0$, then $12(-g)=0$, if $12g=12h=0$, then $12(g+h)=0$, and $12(0)=0$. There are more things to check this way, but it is still straight forward.
– Aaron
Sep 4 at 7:24
Yes that's how I solved it. I based it on ArsenBerk's answer.
– Piyush Divyanakar
Sep 4 at 7:26
add a comment |Â
up vote
9
down vote
accepted
up vote
9
down vote
accepted
Hint: We can rewrite $H=gin Gmid 12g=0$. Therefore, if you can show that the map $gmapsto 12g$ is a homomorphism, then you have that $H$ is the kernel of a homomorphism and therefore a subgroup.
answered Sep 3 at 9:00
Aaron
15.4k22552
Hint: We can rewrite $H=gin Gmid 12g=0$. Therefore, if you can show that the map $gmapsto 12g$ is a homomorphism, then you have that $H$ is the kernel of a homomorphism and therefore a subgroup.
answered Sep 3 at 9:00
Aaron
15.4k22552
answered Sep 3 at 9:00
Aaron
15.4k22552
answered Sep 3 at 9:00
Aaron
15.4k22552
answered Sep 3 at 9:00
Aaron
15.4k22552
15.4k22552
I have just started so I haven't covered homorphisms yet so I couldn't think in this direction, but I understand the idea based on my Linear algebra knowledge. Great solve thanks.
– Piyush Divyanakar
Sep 4 at 7:21
@PiyushDivyanakar You can use the rewriting without talking about homomorphisms. Show that if $12g=0$, then $12(-g)=0$, if $12g=12h=0$, then $12(g+h)=0$, and $12(0)=0$. There are more things to check this way, but it is still straight forward.
– Aaron
Sep 4 at 7:24
Yes that's how I solved it. I based it on ArsenBerk's answer.
– Piyush Divyanakar
Sep 4 at 7:26
add a comment |Â
I have just started so I haven't covered homorphisms yet so I couldn't think in this direction, but I understand the idea based on my Linear algebra knowledge. Great solve thanks.
– Piyush Divyanakar
Sep 4 at 7:21
@PiyushDivyanakar You can use the rewriting without talking about homomorphisms. Show that if $12g=0$, then $12(-g)=0$, if $12g=12h=0$, then $12(g+h)=0$, and $12(0)=0$. There are more things to check this way, but it is still straight forward.
– Aaron
Sep 4 at 7:24
Yes that's how I solved it. I based it on ArsenBerk's answer.
– Piyush Divyanakar
Sep 4 at 7:26
I have just started so I haven't covered homorphisms yet so I couldn't think in this direction, but I understand the idea based on my Linear algebra knowledge. Great solve thanks.
– Piyush Divyanakar
Sep 4 at 7:21
I have just started so I haven't covered homorphisms yet so I couldn't think in this direction, but I understand the idea based on my Linear algebra knowledge. Great solve thanks.
– Piyush Divyanakar
Sep 4 at 7:21
@PiyushDivyanakar You can use the rewriting without talking about homomorphisms. Show that if $12g=0$, then $12(-g)=0$, if $12g=12h=0$, then $12(g+h)=0$, and $12(0)=0$. There are more things to check this way, but it is still straight forward.
– Aaron
Sep 4 at 7:24
@PiyushDivyanakar You can use the rewriting without talking about homomorphisms. Show that if $12g=0$, then $12(-g)=0$, if $12g=12h=0$, then $12(g+h)=0$, and $12(0)=0$. There are more things to check this way, but it is still straight forward.
– Aaron
Sep 4 at 7:24
Yes that's how I solved it. I based it on ArsenBerk's answer.
– Piyush Divyanakar
Sep 4 at 7:26
Yes that's how I solved it. I based it on ArsenBerk's answer.
– Piyush Divyanakar
Sep 4 at 7:26
add a comment |Â
up vote
4
down vote
Instead of seeking a specific example like cyclic groups, you can prove it in a more general way:
HINT: $H le G$ if and only if
1) $H$ is not empty.
2) $H$ is closed under binary operation of $G$.
3) $H$ is closed under inverses.
answered Sep 3 at 8:39
ArsenBerk
7,00221034
2
"actually 2) and 3) implies 1) but we need to ensure $H$ is non-empty." Then 2) and 3) do not imply 1). 2) and 3) together with "1') $H$ is non-empty" imply 1).
– Arthur
Sep 3 at 8:42
Oh, that is really logical.
– ArsenBerk
Sep 3 at 8:43
1
Yes, we do assume that. It's a convention called vacuous truth, and it often does seem counterintuitive to people when they first encounter it. But it works better with all our other logic than any alternative.
– Arthur
Sep 3 at 8:48
1
Well, after I asked, I thought that it is not counter-intuitive because in an empty set, there is no element that breaks closure under binary operation or closure under inverses. I will check "vacuous truth" out, thank you for that :)
– ArsenBerk
Sep 3 at 8:50
add a comment |Â
up vote
4
down vote
Instead of seeking a specific example like cyclic groups, you can prove it in a more general way:
HINT: $H le G$ if and only if
1) $H$ is not empty.
2) $H$ is closed under binary operation of $G$.
3) $H$ is closed under inverses.
answered Sep 3 at 8:39
ArsenBerk
7,00221034
2
"actually 2) and 3) implies 1) but we need to ensure $H$ is non-empty." Then 2) and 3) do not imply 1). 2) and 3) together with "1') $H$ is non-empty" imply 1).
– Arthur
Sep 3 at 8:42
Oh, that is really logical.
– ArsenBerk
Sep 3 at 8:43
1
Yes, we do assume that. It's a convention called vacuous truth, and it often does seem counterintuitive to people when they first encounter it. But it works better with all our other logic than any alternative.
– Arthur
Sep 3 at 8:48
1
Well, after I asked, I thought that it is not counter-intuitive because in an empty set, there is no element that breaks closure under binary operation or closure under inverses. I will check "vacuous truth" out, thank you for that :)
– ArsenBerk
Sep 3 at 8:50
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Instead of seeking a specific example like cyclic groups, you can prove it in a more general way:
HINT: $H le G$ if and only if
1) $H$ is not empty.
2) $H$ is closed under binary operation of $G$.
3) $H$ is closed under inverses.
answered Sep 3 at 8:39
ArsenBerk
7,00221034
Instead of seeking a specific example like cyclic groups, you can prove it in a more general way:
HINT: $H le G$ if and only if
1) $H$ is not empty.
2) $H$ is closed under binary operation of $G$.
3) $H$ is closed under inverses.
answered Sep 3 at 8:39
ArsenBerk
7,00221034
edited Sep 3 at 8:44
answered Sep 3 at 8:39
ArsenBerk
7,00221034
answered Sep 3 at 8:39
ArsenBerk
7,00221034
answered Sep 3 at 8:39
ArsenBerk
7,00221034
7,00221034
2
"actually 2) and 3) implies 1) but we need to ensure $H$ is non-empty." Then 2) and 3) do not imply 1). 2) and 3) together with "1') $H$ is non-empty" imply 1).
– Arthur
Sep 3 at 8:42
Oh, that is really logical.
– ArsenBerk
Sep 3 at 8:43
1
Yes, we do assume that. It's a convention called vacuous truth, and it often does seem counterintuitive to people when they first encounter it. But it works better with all our other logic than any alternative.
– Arthur
Sep 3 at 8:48
1
Well, after I asked, I thought that it is not counter-intuitive because in an empty set, there is no element that breaks closure under binary operation or closure under inverses. I will check "vacuous truth" out, thank you for that :)
– ArsenBerk
Sep 3 at 8:50
add a comment |Â
2
"actually 2) and 3) implies 1) but we need to ensure $H$ is non-empty." Then 2) and 3) do not imply 1). 2) and 3) together with "1') $H$ is non-empty" imply 1).
– Arthur
Sep 3 at 8:42
Oh, that is really logical.
– ArsenBerk
Sep 3 at 8:43
1
Yes, we do assume that. It's a convention called vacuous truth, and it often does seem counterintuitive to people when they first encounter it. But it works better with all our other logic than any alternative.
– Arthur
Sep 3 at 8:48
1
Well, after I asked, I thought that it is not counter-intuitive because in an empty set, there is no element that breaks closure under binary operation or closure under inverses. I will check "vacuous truth" out, thank you for that :)
– ArsenBerk
Sep 3 at 8:50
2
2
"actually 2) and 3) implies 1) but we need to ensure $H$ is non-empty." Then 2) and 3) do not imply 1). 2) and 3) together with "1') $H$ is non-empty" imply 1).
– Arthur
Sep 3 at 8:42
"actually 2) and 3) implies 1) but we need to ensure $H$ is non-empty." Then 2) and 3) do not imply 1). 2) and 3) together with "1') $H$ is non-empty" imply 1).
– Arthur
Sep 3 at 8:42
Oh, that is really logical.
– ArsenBerk
Sep 3 at 8:43
Oh, that is really logical.
– ArsenBerk
Sep 3 at 8:43
1
1
Yes, we do assume that. It's a convention called vacuous truth, and it often does seem counterintuitive to people when they first encounter it. But it works better with all our other logic than any alternative.
– Arthur
Sep 3 at 8:48
Yes, we do assume that. It's a convention called vacuous truth, and it often does seem counterintuitive to people when they first encounter it. But it works better with all our other logic than any alternative.
– Arthur
Sep 3 at 8:48
1
1
Well, after I asked, I thought that it is not counter-intuitive because in an empty set, there is no element that breaks closure under binary operation or closure under inverses. I will check "vacuous truth" out, thank you for that :)
– ArsenBerk
Sep 3 at 8:50
Well, after I asked, I thought that it is not counter-intuitive because in an empty set, there is no element that breaks closure under binary operation or closure under inverses. I will check "vacuous truth" out, thank you for that :)
– ArsenBerk
Sep 3 at 8:50
add a comment |Â
Â
draft saved
draft discarded
Â
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2903650%2fif-g-is-an-abelian-group-then-h-g-in-g-mid-g-text-divides-12-is-a%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
If you have $G = Bbb Z_12times Bbb Z_12$, then $H = G$ but $H$ isn't cyclic. So your idea with $langle arangle$ doesn't work. For another reason it doesn't work, if $G = Bbb Z_4$, then once again $H = G$, but no element has order $12$.
– Arthur
Sep 3 at 8:30
2
You can not be sure that such an element $a$ exists in general. For all we know, $H$ could be as small as the $1$-element group $e$.
– Suzet
Sep 3 at 8:31
is the number $12$ significant somehow?
– Alvin Lepik
Sep 3 at 8:32
How else can I approach this?
– Piyush Divyanakar
Sep 3 at 8:32
1
How would you normally prove a statement like "Show that the subset $A$ of a group $B$ is actually a subgroup"? I would guess that you have a list of three properties, including something very much like "$ein A$". Try those, and if you can't do them all, come back here with all your results and we'll help you get the rest.
– Arthur
Sep 3 at 8:34