- #1
ChrisVer
Gold Member
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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?
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?