Given "G=\\{3^n : n \\ element \\ of \\ \\Z\\}" .
Closure: Let any "a\\in G, b\\in G" , so "a = 3^n, b = 3^m" where "n \\in \\Z, m\\in \\Z" .
Now "a \\times b = 3^n \\times 3^m = 3^{n m}" where "n m \\in \\Z" "\\implies ab \\in G" .
Associative: Let "a, b, c \\in G" , so "a = 3^n, b = 3^m, c = 3^k" ,
hence "(a \\times b) \\times c = (3^{nm}) \\times 3^k = 3^{nmk} = 3^n \\times (3^m \\times 3^k) = a \\times (b\\times c)" .
Identity: For all "a \\in G" , "a = 3^m" where "m \\in \\Z" .
There exists "e = 1 = 3^0" where "0 \\in \\Z" , such that "a \\times e = e\\times a = a" .
Inverse: For all "a \\in G" , "a = 3^m" where "m \\in \\Z" .
There exists "b = 3^{-m} : a\\times b = b\\times a = 1" .
Commutative: "a \\times b = 3^{nm} = b \\times a \\ \\ \\forall \\ a, b \\in G" .
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