- #1

ChrisVer

Gold Member

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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

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?