Question-158143

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 :

{\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 \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}