# Nonisomorphic Groups of order $100$

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.

• Because $Z_{100} \equiv Z_4 \times Z_{25}$. $4$, $25$ are not invariant factors, because $4$ does not divide $25$. Nov 1, 2019 at 15:00

You are wrong, $$\mathbb Z_{100}$$ is an option. But it appears twice in your list: Note that $$4$$ and $$25$$ do not share any prime factors, so by the Chinese Remainder Theorem, $$\mathbb Z_4 \times \mathbb Z_{25} \cong \mathbb Z_{100}.$$ As pointed out in the comments, an elementary way to see this is as follows. The element $$(1, 1) \in \mathbb Z_4 \times \mathbb Z_{25}$$ has order $$100$$, because if $$n$$ is a multiple of $$4$$ and a multiple of $$25$$, then it is a multiple of $$\text{lcm}(4, 25) = 100$$. So $$(n, n)$$ is zero in $$\mathbb Z_4 \times \mathbb Z_{25}$$ for no $$n$$ below $$100$$. Thus, there are at least $$100$$ elements in this group.
• An easy way for OP to see this is that the element $(1,1)$ would have order $100$ since $4$ and $25$ don't share any factors. Nov 1, 2019 at 15:01
$$\Bbb Z_{100}$$ is isomorphic to $$\Bbb Z_4\times\Bbb Z_{25}$$.
Personally, I prefer to let each factor have prime power order when solving an exercise like this. That way it's much easier to check both that I have all options and that I have no repeats. In this case it would be $$\Bbb Z_4\times\Bbb Z_{25}\\ \Bbb Z_2\times\Bbb Z_2\times\Bbb Z_{25}\\ \Bbb Z_4\times\Bbb Z_5\times\Bbb Z_5\\ \Bbb Z_2\times\Bbb Z_{2}\times \Bbb Z_5\times\Bbb Z_5$$
Since $$4$$ and $$25$$ are relatively prime, $$\mathbb Z_4\times \mathbb Z_{25}\simeq \mathbb Z_{100}$$, which is why you have an extra one on your list. Of course $$\mathbb Z_{100}$$ is a valid example, as it is an abelian group of order $$100$$.
The way to ensure you don't have extras is to make sure all of the factors have prime power order, so while $$\mathbb Z_{100}$$ is correct, it would not show up on your list as such, but rather as $$\mathbb Z_4\times\mathbb Z_{25}$$.