I was reviewing an ISO standard and a reference was made to "air equilibrated water."
Anyone know what it is? I'm guessing it's water left standing open that comes into equilibrium with the air around it, but I don't really know.
So there could be an element of G that equals its own inverse. So squaring the product insures that this is reduced to e as well? Since they equal each other they should square to identity I think. Is this it?
'If the group has elements of order 2' I don't really understand that. The terminology in this book I understand (so far) is if the group is of order 2 that means it is a finite group with two elements.
Things are sometimes squared to get rid of a negative sign. But if the elements are...
This is not homework. Self-study. And I'm really enjoying it. But, as I'm going through this book ("A Book of Abstract Algebra" by Charles C. Pinter) every so often I run into a problem or concept I don't understand.
Let G be a finite abelian group, say G = (e,a1, a2, a3,...,an).
Prove...
OK, it took some scratch work but I think I understand it. Basically the kernal consists of the elements of the domain that by way of the mapping, show up as identity elements in the range.
Thanks again for the help.
Thanks, yes, I didn't mention the homomorphism part.
I understand your reply but I'm looking at {e,a}.
a^2 = 1 so a = 1 I would think and so does e = 1.
Are you saying because the kernal has two elements and they both aren't e that shows the kernal isn't always equal to the identity? I...
I'm having a bit of a problem understanding the kernal of a subgroup. It appears to always be equal to the identity element. But that doesn't seem to make much sense. Anyone care to clarify this for me?