Is set of even integers is abelian under addition ? : Group Theory
Problem :
If G is the set of even integers i.e. G = { ...,-4,-2,0,2,4,...} then prove that G is an abelian group
with usual addition .
Or
Show that the set of all even integers forms an abelian group under ordinary '+' .
Solution :
Suppose G is the set of even integers i.e. G = { ...,-4,-2,0,2,4,...}.
Now we prove (G,+) is abelian.
Closure Property :
Let a, b 𝞊 G.
Suppose a = 2x and b = 2y for x , y 𝞊 Z.
Now a + b = 2x+2y
= 2( x+y) | since x , y are integers then so is x+y |
a+b 𝞊 G for all a,b 𝞊 G.
Hence '+' is closed in G.
Associative Law :
Let a,b,c 𝞊 G.
Suppose a = 2x, b = 2y, c = 2z for x,y,z 𝞊 Z.
Now a+ (b+c) = 2x + (2y + 2z)
= 2x + ( 2 ( y+z ) )
= 2 ( x + (y+z) )
= 2 ( (x+y ) +z )
= 2 ( x+y ) + 2z
= ( 2x+2y ) + 2z
a+(b+c) = (a+b)+c for all a,b,c 𝞊 G.
'+' is associative in G.
Existence of Identity :
Let a 𝞊 G.
Suppose a = 2x for x 𝞊 Z.
we have 0 𝞊 G.
Also a+0 = 2x + 0 = 2x = a
Similarly 0+a = 0+2x = 2x =a
a+0=0+a=a for all a𝞊G.
'0' is the identity element in G.
Existence of inverse :
Let a 𝞊 G .
Suppose a= 2x for x 𝞊 Z.
Since x 𝞊 Z, -x 𝞊 Z
-a = -2x 𝞊 G.
Also a+(-a) = 2x + (-2x ) = 2 ( x+(-x) ) = 2 (x-x ) = 2 (0) = 0
i.e. a + ( -a ) = 0
Similarly (-a) + a = 0
a+(-a) = 0 = (-a) + a
'-a' is the inverse of a in G.
Each element in G having inverse in G.
Commutative Law :
Let a, b 𝞊 G.
Suppose a = 2x and b = 2y for x,y 𝞊 Z.
Now a+b = 2x+2y
= 2(x+y)
= 2(y+x)
= 2y + 2x
a+b = b+a for all a,b 𝞊 G.
( G , + ) is an abelian group.
*** Hence The Proof ***
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