Show that G=[1,-1,i,-i] is a group under usual multiplication of complex number.

1 Answer

The adjoioning table shows the result of multiplication of elements G

Since for every pair a,b,E, G

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$\ G_1 \Rightarrow$ Since multiplication of complex number is associative, the multiplication associative in G

$\ G_2 \Rightarrow$ From the first coloumn (or row), we see that l is an identity element. Hence, 1 E G is an identity element.

$\ G_3 \Rightarrow$ Since

11=1, _1-1=1, i * -i=1

-i * i=1, inverse exists

for every elements in G

and, we have

$\ 1^{-1} =1, -1^{-1} =-1, i^{-1} =-i, -i^{-i} = i

Hence G is a group under multiplication.

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