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.

  • 3
    $\begingroup$ Because $Z_{100} \equiv Z_4 \times Z_{25}$. $4$, $25$ are not invariant factors, because $4$ does not divide $25$. $\endgroup$
    – Milten
    Nov 1, 2019 at 15:00

3 Answers 3


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.

  • 2
    $\begingroup$ 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. $\endgroup$ 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 $\mathbb Z_4\times\mathbb Z_{25}$.


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