Help verify my understanding of groups?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
1
down vote

favorite
1












I just learned the definition of groups today an would like help in verifying my understanding. Please point out any mistakes or misconceptions in my examples.



Definition: A group is a pair $G = (G,*)$ consisting of a set of elements $G$, and a binary operation $*$ on G such that:



• $G$ has an identity element



• The operation is associative



• Each element $g in G$ has an inverse



An example:



The pair $(Bbb Z, +)$ is a group.



Explanation:



$Bbb Z$ $= ...-2, -1, 0, 1, 2 ...$ is the set $(G)$ and $+$ is the associative operation. The element $0 in Bbb Z$ is an identity because $$a+0=0+a=a$$ for any element $a in Bbb Z$. Every element $a in Bbb Z$ also has an additive inverse such that $$a + (-a)=(-a) + a = 0$$



Non-example:



$(Bbb N, -)$ is not a group.



Explanation:



$Bbb N$ $ = 0, 1, 2,...$ is the set $(G)$ and $-$ is the associative operation.



There is no identity element for this set because $$a-0=0$$ but $$0-aneq a$$



There is also no inverse element because $$a - (-a) neq 0$$



EDIT



A major mistake was made...



$(Bbb N, -)$ is not a group because subtraction is not associative and it isn't a binary operation of the natural numbers.




abstract-algebra group-theory




share|cite|improve this question


















  • 2




    Subtraction is not an associative operation. Note that $(9 - 2) - 3 neq 9 - (2 - 3)$. Also, subtraction is not a binary operation on the natural numbers since $1 - 2 notin mathbbN$.
    – N. F. Taussig
    Aug 12 at 6:39










  • I see, that's a little embarrassing...that's what I get for doing math at 2:40 in the morning.
    – CaptainAmerica16
    Aug 12 at 6:40







  • 2




    In group theory, you should see $a-b$ as $a+c$ where $c=-b$.
    – nicomezi
    Aug 12 at 6:43






  • 1




    Thanks for the feedback.
    – CaptainAmerica16
    Aug 12 at 7:05














up vote
1
down vote

favorite
1












I just learned the definition of groups today an would like help in verifying my understanding. Please point out any mistakes or misconceptions in my examples.



Definition: A group is a pair $G = (G,*)$ consisting of a set of elements $G$, and a binary operation $*$ on G such that:



• $G$ has an identity element



• The operation is associative



• Each element $g in G$ has an inverse



An example:



The pair $(Bbb Z, +)$ is a group.



Explanation:



$Bbb Z$ $= ...-2, -1, 0, 1, 2 ...$ is the set $(G)$ and $+$ is the associative operation. The element $0 in Bbb Z$ is an identity because $$a+0=0+a=a$$ for any element $a in Bbb Z$. Every element $a in Bbb Z$ also has an additive inverse such that $$a + (-a)=(-a) + a = 0$$



Non-example:



$(Bbb N, -)$ is not a group.



Explanation:



$Bbb N$ $ = 0, 1, 2,...$ is the set $(G)$ and $-$ is the associative operation.



There is no identity element for this set because $$a-0=0$$ but $$0-aneq a$$



There is also no inverse element because $$a - (-a) neq 0$$



EDIT



A major mistake was made...



$(Bbb N, -)$ is not a group because subtraction is not associative and it isn't a binary operation of the natural numbers.




abstract-algebra group-theory




share|cite|improve this question


















  • 2




    Subtraction is not an associative operation. Note that $(9 - 2) - 3 neq 9 - (2 - 3)$. Also, subtraction is not a binary operation on the natural numbers since $1 - 2 notin mathbbN$.
    – N. F. Taussig
    Aug 12 at 6:39










  • I see, that's a little embarrassing...that's what I get for doing math at 2:40 in the morning.
    – CaptainAmerica16
    Aug 12 at 6:40







  • 2




    In group theory, you should see $a-b$ as $a+c$ where $c=-b$.
    – nicomezi
    Aug 12 at 6:43






  • 1




    Thanks for the feedback.
    – CaptainAmerica16
    Aug 12 at 7:05












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I just learned the definition of groups today an would like help in verifying my understanding. Please point out any mistakes or misconceptions in my examples.



Definition: A group is a pair $G = (G,*)$ consisting of a set of elements $G$, and a binary operation $*$ on G such that:



• $G$ has an identity element



• The operation is associative



• Each element $g in G$ has an inverse



An example:



The pair $(Bbb Z, +)$ is a group.



Explanation:



$Bbb Z$ $= ...-2, -1, 0, 1, 2 ...$ is the set $(G)$ and $+$ is the associative operation. The element $0 in Bbb Z$ is an identity because $$a+0=0+a=a$$ for any element $a in Bbb Z$. Every element $a in Bbb Z$ also has an additive inverse such that $$a + (-a)=(-a) + a = 0$$



Non-example:



$(Bbb N, -)$ is not a group.



Explanation:



$Bbb N$ $ = 0, 1, 2,...$ is the set $(G)$ and $-$ is the associative operation.



There is no identity element for this set because $$a-0=0$$ but $$0-aneq a$$



There is also no inverse element because $$a - (-a) neq 0$$



EDIT



A major mistake was made...



$(Bbb N, -)$ is not a group because subtraction is not associative and it isn't a binary operation of the natural numbers.




abstract-algebra group-theory




share|cite|improve this question














I just learned the definition of groups today an would like help in verifying my understanding. Please point out any mistakes or misconceptions in my examples.



Definition: A group is a pair $G = (G,*)$ consisting of a set of elements $G$, and a binary operation $*$ on G such that:



• $G$ has an identity element



• The operation is associative



• Each element $g in G$ has an inverse



An example:



The pair $(Bbb Z, +)$ is a group.



Explanation:



$Bbb Z$ $= ...-2, -1, 0, 1, 2 ...$ is the set $(G)$ and $+$ is the associative operation. The element $0 in Bbb Z$ is an identity because $$a+0=0+a=a$$ for any element $a in Bbb Z$. Every element $a in Bbb Z$ also has an additive inverse such that $$a + (-a)=(-a) + a = 0$$



Non-example:



$(Bbb N, -)$ is not a group.



Explanation:



$Bbb N$ $ = 0, 1, 2,...$ is the set $(G)$ and $-$ is the associative operation.



There is no identity element for this set because $$a-0=0$$ but $$0-aneq a$$



There is also no inverse element because $$a - (-a) neq 0$$



EDIT



A major mistake was made...



$(Bbb N, -)$ is not a group because subtraction is not associative and it isn't a binary operation of the natural numbers.





abstract-algebra group-theory





share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 12 at 6:47

























asked Aug 12 at 6:35









CaptainAmerica16

171110




171110







  • 2




    Subtraction is not an associative operation. Note that $(9 - 2) - 3 neq 9 - (2 - 3)$. Also, subtraction is not a binary operation on the natural numbers since $1 - 2 notin mathbbN$.
    – N. F. Taussig
    Aug 12 at 6:39










  • I see, that's a little embarrassing...that's what I get for doing math at 2:40 in the morning.
    – CaptainAmerica16
    Aug 12 at 6:40







  • 2




    In group theory, you should see $a-b$ as $a+c$ where $c=-b$.
    – nicomezi
    Aug 12 at 6:43






  • 1




    Thanks for the feedback.
    – CaptainAmerica16
    Aug 12 at 7:05












  • 2




    Subtraction is not an associative operation. Note that $(9 - 2) - 3 neq 9 - (2 - 3)$. Also, subtraction is not a binary operation on the natural numbers since $1 - 2 notin mathbbN$.
    – N. F. Taussig
    Aug 12 at 6:39










  • I see, that's a little embarrassing...that's what I get for doing math at 2:40 in the morning.
    – CaptainAmerica16
    Aug 12 at 6:40







  • 2




    In group theory, you should see $a-b$ as $a+c$ where $c=-b$.
    – nicomezi
    Aug 12 at 6:43






  • 1




    Thanks for the feedback.
    – CaptainAmerica16
    Aug 12 at 7:05







2




2




Subtraction is not an associative operation. Note that $(9 - 2) - 3 neq 9 - (2 - 3)$. Also, subtraction is not a binary operation on the natural numbers since $1 - 2 notin mathbbN$.
– N. F. Taussig
Aug 12 at 6:39




Subtraction is not an associative operation. Note that $(9 - 2) - 3 neq 9 - (2 - 3)$. Also, subtraction is not a binary operation on the natural numbers since $1 - 2 notin mathbbN$.
– N. F. Taussig
Aug 12 at 6:39












I see, that's a little embarrassing...that's what I get for doing math at 2:40 in the morning.
– CaptainAmerica16
Aug 12 at 6:40





I see, that's a little embarrassing...that's what I get for doing math at 2:40 in the morning.
– CaptainAmerica16
Aug 12 at 6:40





2




2




In group theory, you should see $a-b$ as $a+c$ where $c=-b$.
– nicomezi
Aug 12 at 6:43




In group theory, you should see $a-b$ as $a+c$ where $c=-b$.
– nicomezi
Aug 12 at 6:43




1




1




Thanks for the feedback.
– CaptainAmerica16
Aug 12 at 7:05




Thanks for the feedback.
– CaptainAmerica16
Aug 12 at 7:05















active

oldest

votes











Your Answer




StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: true,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);








 

draft saved


draft discarded


















StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2880041%2fhelp-verify-my-understanding-of-groups%23new-answer', 'question_page');

);

Post as a guest






















active

oldest

votes













active

oldest

votes









active

oldest

votes






active

oldest

votes










 

draft saved


draft discarded


























 


draft saved


draft discarded














StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2880041%2fhelp-verify-my-understanding-of-groups%23new-answer', 'question_page');

);

Post as a guest
































































-

這個網誌中的熱門文章

How to combine Bézier curves to a surface?

圖片
Clash Royale CLAN TAG #URR8PPP up vote 2 down vote favorite My aim is to smooth the terrain in a video game. Therefore I contrived an algorithm that makes use of Bézier curves of different orders. But this algorithm is defined in a two dimensional space for now. To shift it into the third dimension I need to somehow combine the Bézier curves. Given two curves in each two dimensions, how can I combine them to build a surface? I thought about something like a curve of a curve. Or maybe I could multiply them somehow. How can I combine them to cause the wanted behavior? Is there a common approach? In the image you can see the input, on the left, and the output, on the right, of the algorithm that works in a two dimensional space. For the real application there is a three dimensional input. The algorithm relays only on the adjacent tiles. Given these it will define the control points of the mentioned Bézier curve. Additionally it marks one side of the curve as inside t
閱讀完整內容

Carbon dioxide

圖片
"CO2" redirects here. For other uses, see CO2 (disambiguation). Carbon dioxide Names Other names Carbonic acid gas Carbonic anhydride Carbonic oxide Carbon oxide Carbon(IV) oxide Dry ice (solid phase) Identifiers CAS Number 124-38-9   Y 3D model (JSmol) Interactive image Interactive image 3DMet B01131 Beilstein Reference 1900390 ChEBI CHEBI:16526   Y ChEMBL ChEMBL1231871   N ChemSpider 274   Y ECHA InfoCard 100.004.271 EC Number 204-696-9 E number E290 (preservatives) Gmelin Reference 989 KEGG D00004   Y MeSH Carbon+dioxide PubChem CID 280 RTECS number FF6400000 UNII 142M471B3J   Y UN number 1013 (gas), 1845 (solid) InChI InChI=1S/CO2/c2-1-3   Y Key: CURLTUGMZLYLDI-UHFFFAOYSA-N   Y InChI=1/CO2/c2-1-3 Key: CURLTUGMZLYLDI-UHFFFAOYAO SMILES O=C=O C(=O)=O Properties Chemical formula C O 2 Molar mass 44.01 g·mol −1 Appearance Colorless gas Odor Low concentrations: none High concentrations: sharp; acidic [1] Density 1562 kg/m 3 (solid at 1 atm and −78.5 °C) 1101 kg/m 3 (liquid at
閱讀完整內容

Why am i infinitely getting the same tweet with the Twitter Search API?

圖片
up vote 0 down vote favorite My code workes perfectly for other queries, but for one query I am infinitely getting the same tweet. I can't understand what the problem is. API call: query='Allergic asthma OR Nonallergic asthma OR Occupational asthma OR EIB OR Exercise-induced bronchoconstriction OR Nocturnal asthma OR Cough-variant asthma' new_tweets = api.search(q=searchQuery, count=100, since_id=since_id, lang='en',tweet_mode='extended') if not new_tweets: print("No more tweets found") break for tweet in new_tweets: if True: print(jsonpickle.encode(tweet._json, unpicklable=False)) my_file = open('searchTweets.txt','a') my_file.write(jsonpickle.encode(tweet._json, unpicklable=False)+'n') my_file.close() else: continue Output in file: https://pastebin.com/JmRCsTeE These are the JSONs of the same Tweet. Other queries that work: Breast cancer OR Prostate cancer OR Colon cancer OR Lung cancer Type 2 dia
閱讀完整內容