Is this an alright proof?
First note that Z(n) has exactly one d-subgroup, where d is any divisor of n.
Using the fundamental theorem of finite abelian groups, G can be isomorphic to either:
Z540, which has only one 3-subgroup,
Z3 x Z180, which has three 3-subgroups (namely, the subgroups...
Why does it make sense ( when considering Z4)to form the factor group
Z4 / (2Z4) where kZn = {0, k mod n, 2k mod n, ..., nk mod n}?
I believe that this above factor group is isomorphic to Z2, but how can I prove this?
This isn't homework, it was proposed by a professor of mine and I'm dying here because the hint makes no sense to me
Let G be an abelian group of order 540, what is the largest possible number of subgroups of order 3 such a group G can have?
He said to classify abelian groups of order 27...