Group definition (checks)

  • Thread starter ChrisVer
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  • #1
ChrisVer
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Main Question or Discussion Point

I was wondering, if we take a "group" [itex]G[/itex] (so multiplication is defined among the elements) it forms a group if it has the following properties:
Closure
Contains the identity element
Contains the inverse elements
follows associativity.

I was wondering if associativity is not a must though... like it can be contained in the previous properties.

Take for example [itex] g_1, g_2, g_3 \in G[/itex], then:
[itex]g_1 g_2 g_3 = g \in G [/itex] (closure).

Let's take the :

[itex]g_1 (g_2 g_3) = g_1 g_k = g_h [/itex]
and
[itex](g_1 g_2) g_3 = g_m g_3 = g_i [/itex]
associativity holds if [itex]g_h = g_i [/itex] or doesn't if [itex]g_h \ne g_i [/itex]. Let's take the last assumption, that is [itex]g_h \ne g_i [/itex]
Since the elements [itex]g_{1,2,3}[/itex] are taken arbitrarily, I can take [itex]g_{2}=g_{3}^{-1} [/itex] as well as [itex]g_1=e[/itex]. If I do that, I'm getting that [itex]g_i =g_h[/itex] which is a contradiction.

Is that a correct thinking?
 

Answers and Replies

  • #2
phyzguy
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I was wondering, if we take a "group" [itex]G[/itex] (so multiplication is defined among the elements) it forms a group if it has the following properties:
Closure
Contains the identity element
Contains the inverse elements
follows associativity.

I was wondering if associativity is not a must though... like it can be contained in the previous properties.

Take for example [itex] g_1, g_2, g_3 \in G[/itex], then:
[itex]g_1 g_2 g_3 = g \in G [/itex] (closure).

Let's take the :

[itex]g_1 (g_2 g_3) = g_1 g_k = g_h [/itex]
and
[itex](g_1 g_2) g_3 = g_m g_3 = g_i [/itex]
associativity holds if [itex]g_h = g_i [/itex] or doesn't if [itex]g_h \ne g_i [/itex]. Let's take the last assumption, that is [itex]g_h \ne g_i [/itex]
Since the elements [itex]g_{1,2,3}[/itex] are taken arbitrarily, I can take [itex]g_{2}=g_{3}^{-1} [/itex] as well as [itex]g_1=e[/itex]. If I do that, I'm getting that [itex]g_i =g_h[/itex] which is a contradiction.

Is that a correct thinking?
No. You have shown that associativity will hold as a consequence of the other properties for some elements of G, but in order to be a group, associativity must hold for all elements of G. If I can find any g1, g2, g3 for which [itex]g_h \ne g_i[/itex], then G is not a group.
 
  • #3
Erland
Science Advisor
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A set with a binary operation which has an identity element and inverse elements needs not be associative. As an example, consider the set {1,a,b} with the following composition table for the binary operation:

* | 1 a b
1 | 1 a b
a | a 1 b
b | b b 1

1 is the identity element, 1-1=1, a-1=a, b-1=b, but associativity doesn't hold, since e.g. a(bb)=a1=a but (ab)b=bb=1.
 
  • #4
Stephen Tashi
Science Advisor
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Since the elements [itex]g_{1,2,3}[/itex] are taken arbitrarily,
The notation for the elements is arbitrary, but if you say that a product involving [itex] g_1,g_2,g_3 [/itex] can represent an arbitrary product of 3 elements in the group then the assumption [itex] (g_1)(g_2 g_3) \ne (g_1 g_2) g_3 [/itex] becomes a claim about each combination of 3 elements in the group. For an indirect proof, you would only be allowed the assumption that there exist 3 elements whose multiplication is not associative - rather than the assumption that each set of 3 elements has a multiplication that is non-associative.
 

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