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Show that the set {[1], [3], [4], [5], [9]} forms an abelian group under multiplication modulo 11.

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  • Multiplication modulo n $(._{\large n}):a._{\large n} b=r ; 0\leq r< n$,where r is the least non-negative remainder when $ab$ is divided by n.
  • Congruence modulo n :Let $a, b \in Z$ and $n$ be a fixed positive integer.
  • We say that “a is congruent to b modulo n” $\Leftrightarrow (a − b)$ is divisible by n Symbolically,
  • $a \equiv b (mod\; n) \Leftrightarrow (a − b)$ is divisible by n.
Step 1:
Let $G=\{[1],[3],[4],[5],[9]\}$
Consider $(G,._{11})$
Drawing up the multiplication table,we have
Step 2:
Closure :All the products in the table belong to $G$.Closure is satisfied.
Associativity : Modular multiplication is associative.Since multiplication of integers is associative.
Existence of identity :From the table,it is evident that [1] is the identity element.
Existence of inverse: From the table it can be seen that the inverse of [1] is [1],$[3]^{-1} $ is [4],$[4]^{-1}$ is [3],$[5]^{-1}$ is [9] and $[9]^{-1}$ is [5].All the elements of $G$ have inverse in G.
The four group axioms being satisfied.,$(G,._{11})$ is a group.
Step 3:
Commutative property : Modular multiplication is commutative since multiplication of integers is commutative.
$\therefore (G,._{11})$ is an abelian group.
answered Sep 16, 2013 by sreemathi.v

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