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Question 847208: hello,
if I have (z,+) as a group ,
h={2x,x belongs z)
h={-8,-4,-2,0,2,4,8}=2z
2z C z
0 belongs to 2z(0=2*0) and h!= empty set
how can i show that 2z is a subgroup of z ?
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! Ah, abstract algebra. Good times.
First off, let's think about how we can show how one set is a subgroup of another.
We must show closure, identity, and an inverse element.
Let's recap. We have h = {2x | where x is an integer}
We want to show that 2z is a subgroup of z under the additive operator.
Closure under addition:
Let 2x and 2y be integers.
Then 2x + 2y = 2(x+y) which exists in 2z.
Identity:
We know that in Z the identity element is 0. Since we have that 0 = 2*0, 0 exists in 2z.
Inverse:
Suppose we have some 2x in Z. The additive inverse for 2x is -2x and -2x = 2*(-x) which means -2x exists in 2z
Hence 2z is a subgroup of z under addition.
Note the upshot is we try to find elements that have the form 2 * (something).
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