Hi everyone, thank you for interest to solve this problem:

49. Up to isomorphism, how many additive abelian groups G of order 16 have the property that x+x+x+x=0 for each x in G? (A)0 (B)1 (C)2 (D)3 (E)5

order(<x>)=4. So according to the theories about elementary divisor or the invariant factors of G, G could be: (here "+" stands for any additive operation not just the ordinary one)

Z2+Z2+Z2+Z2

Z2+Z2+Z4

Z2+Z8

Z4+Z4

Z16

It looks like there are 5. However, the answer is (D). I think the condition that "x+x+x+x=0 for each x in G" must have some constrain. But, I don't know how to use this constrain to choose the correct groups.

## An Interesting Question_GR0568 #49

### Re: An Interesting Question_GR0568 #49

Are you sure Z16 works? What about Z2 + Z8?

In fact 1 + 1 + 1 + 1 = 4 != 0 in Z16, and (0,1) + (0,1) + (0,1) + (0,1) = (0,4) != 0 in Z2 + Z8.

In fact 1 + 1 + 1 + 1 = 4 != 0 in Z16, and (0,1) + (0,1) + (0,1) + (0,1) = (0,4) != 0 in Z2 + Z8.

### Re: An Interesting Question_GR0568 #49

Thanks for your hint!

After second thought, x+x+x+x=0 not only implies order(x)=4, but also indicates x=0 or order(x)=2. Therefore, the candidate group will be:

Z2+Z2+Z2+Z2, which contains (0,0,0,0) or other elements with the order of 2

Z2+Z2+Z4, which contains (0,0,0) or other element with the order of 2 or 4

Z4+Z4, which contains (0,0) or other elements with the order of 2 or 4

After second thought, x+x+x+x=0 not only implies order(x)=4, but also indicates x=0 or order(x)=2. Therefore, the candidate group will be:

Z2+Z2+Z2+Z2, which contains (0,0,0,0) or other elements with the order of 2

Z2+Z2+Z4, which contains (0,0,0) or other element with the order of 2 or 4

Z4+Z4, which contains (0,0) or other elements with the order of 2 or 4