This is from Dummit and Foote, exercise 3.4.
Give the number of nonisomophic abelian groups of order 100.
What I did was factor $100$ into invariant factors $$ 1, \; 100$$ $$ 1 ,\; 2 ,\; 50 , \;100$$ $$ 1 ,\; 4 ,\; 25 , \;100$$ $$ 1 ,\; 5 ,\; 20 , \;100$$ $$ 1 ,\; 10 ,\; 10 , \;100$$
This gives me 5 options products of cyclic groups of order 100.
So I get $$ \mathbb{Z}_{100} $$ $$ \mathbb{Z}_{2} \times \mathbb{Z}_{50}$$ $$ \mathbb{Z}_{4} \times \mathbb{Z}_{25}$$ $$ \mathbb{Z}_{5} \times \mathbb{Z}_{20}$$ $$ \mathbb{Z}_{10} \times \mathbb{Z}_{10}.$$
But the answer should be 4, and I have 5 groups here. Comparing with the examples in the book, I think that $ \mathbb{Z}_{100} $ is not an option.
Why is $ \mathbb{Z}_{100} $ not an option? It's abelian, and it has order $100$.
I was thinking maybe cyclic groups only have prime order, but $ \mathbb{Z}_{4} $ is cyclic, abelian, and not of prime order, so that's not true.