# Are the groups $\Bbb{Z}_8 \times \Bbb{Z}_{10} \times \Bbb{Z}_{24}$ and $\Bbb{Z}_4 \times \Bbb{Z}_{12} \times \Bbb{Z}_{40}$ isomorphic?

This question is taken from "A first course in Abstract Algebra" by Fraleigh 7th edition, section 11 question 18:

Are the groups $$\mathbb{Z}_8 \times \mathbb{Z}_{10} \times \mathbb{Z}_{24}$$ and $$\mathbb{Z}_4 \times \mathbb{Z}_{12} \times \mathbb{Z}_{40}$$ isomorphic?

The solution manual says no. My question is why not?

We have $$\mathbb{Z}_8 \approx \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2},$$

and $$\mathbb{Z}_{10} \approx \mathbb{Z}_{5} \times \mathbb{Z}_{2},$$

and $$\mathbb{Z}_{24} \approx \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3}.$$

Thus $$\mathbb{Z}_8 \times \mathbb{Z}_{10} \times \mathbb{Z}_{24} \approx \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{5} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3}$$

Similiarly, $$\mathbb{Z}_4 \times \mathbb{Z}_{12} \times \mathbb{Z}_{40} \approx \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{3} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{2} \times \mathbb{Z}_{5}$$

These last 2 expressions are the same except for a reordering. Where is the mistake in my reasoning?

Let's focus on a very simple case, comparing $$\mathbb{Z}_{4}$$ and $$\mathbb{Z}_{2} \times \mathbb{Z}_{2}$$. Are these the same? They have the same order, but they are actually not isomorphic! To see why, note that $$1$$ has order $$4$$ in $$\mathbb{Z}_{4}$$ (you need to add it to itself $$4$$ times to get to $$0$$). However, every element of $$\mathbb{Z}_{2} \times \mathbb{Z}_{2}$$ has order at most $$2$$. For example, $$(1,0) + (1,0) = (2,0) = (0,0)$$, or $$(1,1) + (1,1) = (2,2) = (0,0)$$.
What IS true is that $$\mathbb{Z}_{pq} = \mathbb{Z}_{p} \times \mathbb{Z}_{q}$$ for distinct primes $$p$$ and $$q$$. The same thing holds if you replace $$p$$ with $$p^n$$ and $$q$$ with $$q^k$$. This actually encapsulates the entire pattern: you can split up $$\mathbb{Z}_{ab}$$ into $$\mathbb{Z}_{a} \times\mathbb{Z}_{b}$$ if and only if $$a$$ and $$b$$ are coprime, i.e. having no prime factors in common.
The mistake is at the first line: $$\mathbb{Z}_8$$ is not isomorphic to $$\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$$.
You can see this very simply: all elements of $$\mathbb{Z}_2\times\mathbb{Z}_2\times\mathbb{Z}_2$$ have order $$2$$, on the other hand $$\mathbb{Z}_8$$ has elements of higher order.
• In general, $\mathbb{Z}_{mn}$ is isomorphic to $\mathbb{Z}_m\times\mathbb{Z}_n$ if and only if $gcd(m,n)=1$. For example, it is true that $\mathbb{Z}_{24}\cong\mathbb{Z}_8\times\mathbb{Z}_3$ and that $\mathbb{Z}_{10}\cong\mathbb{Z}_2\times\mathbb{Z}_5$. Jun 26, 2020 at 21:28
These groups are not isomorphic: the first group contains the direct product of two cyclic groups of order $$8$$, and the second group does not.