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Question Number 158143 by alcohol last updated on 31/Oct/21

Commented by TheHoneyCat last updated on 31/Oct/21
$G_3 and G_2 are clearly isomorphical buy looking at their respectiv tables. we just need to map as follow: { (1,⇆,1),(5,⇆,3),(7,⇆,5),((11),⇆,7) :} G_1 cannot be isomorphic to either one because 3×_(10) 3=9≢1 [10] while ∀n∈{1,5,7,11} n×_(12) n=1 and ∀n∈{1,3,5,7} n×_8 n=1$
$G_{3} and G_{2} are clearly isomorphical buy looking$ $at their respectiv tables .$ $we just need to map as follow :$ ${\begin{cases} 1 & ⇆ & 1 \\ 5 & ⇆ & 3 \\ 7 & ⇆ & 5 \\ 11 & ⇆ & 7 \end{cases}$ $G_{1} cannot be isomorphic to either one because$ $3 \times_{10} 3 = 9 ≢ 1 [10]$ $while \forall n \in {1, 5, 7, 11} n \times_{12} n = 1$ $and \forall n \in {1, 3, 5, 7} n \times_{8} n = 1$
Commented by TheHoneyCat last updated on 31/Oct/21

$By definition of a group, each element has an$ $inverse . Let x^{- 1} be that of x for any x in the$ $group (y o u i s u n i q u e s i n c e$ $a x = x a = e \land b x = x b = e \Rightarrow a x a = b x a \Rightarrow a = b)$ $So naturaly :$ $a^{3} = a \Rightarrow a^{2} \times a = a \Rightarrow a^{2} \times a \times a^{- 1} = a \times a^{- 1} = e$ $\Rightarrow a^{2} = e$
Commented by TheHoneyCat last updated on 31/Oct/21

$G_{1} :$ $\| \begin{matrix} \times_{10} & 1 & 3 & 7 & 9 \\ 1 & 1 & 3 & 7 & 9 \\ 3 & 3 & 9 & 1 & 7 \\ 7 & 7 & 1 & 9 & 3 \\ 9 & 9 & 7 & 3 & 1 \end{matrix} \|$ $G_{2} :$ $\| \begin{matrix} \times_{12} & 1 & 5 & 7 & 11 \\ 1 & 1 & 5 & 7 & 11 \\ 5 & 5 & 1 & 11 & 7 \\ 7 & 7 & 11 & 1 & 5 \\ 11 & 11 & 7 & 5 & 1 \end{matrix} \|$ $G_{3} :$ $\| \begin{matrix} \times_{8} & 1 & 3 & 5 & 7 \\ 1 & 1 & 3 & 5 & 7 \\ 3 & 3 & 1 & 7 & 5 \\ 5 & 5 & 7 & 1 & 3 \\ 7 & 7 & 5 & 3 & 1 \end{matrix} \|$
Commented by TheHoneyCat last updated on 31/Oct/21
oups, I forgot to put the last one in blue...
Commented by phanphuoc last updated on 31/Oct/21

$y o u c a n p r o o f : g r o u p G s t \forall x \in G : x^{n} = e \to G a b e l$
Commented by TheHoneyCat last updated on 31/Oct/21
it did say "∀x" but "∃a" so I just did not assume anything
	Answered by TheHoneyCat last updated on 31/Oct/21
	$As for the last question: since a^3 =a ⇔ a^2 =1 we can just check the digonnal for ones so: a^3 =_G_1 a ⇔ a∈{1,9} a^3 =_G_2 a ⇔ a ∈ G_2 and a^3 =_G_3 a ⇔ a ∈ G_3$
	$As for the last question :$ $since a^{3} = a \Leftrightarrow a^{2} = 1$ $we can just check the digonnal for ones$ $so :$ $a^{3} =_{G_{1}} a \Leftrightarrow a \in {1, 9}$ $a^{3} =_{G_{2}} a \Leftrightarrow a \in G_{2}$ $and a^{3} =_{G_{3}} a \Leftrightarrow a \in G_{3}$
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