Why is $\Bbb Z/4\Bbb Z\times \Bbb Z/12\Bbb Z\times \Bbb Z/40\Bbb Z$ not isomorphic to $\Bbb Z/8\Bbb Z\times \Bbb Z/10\Bbb Z\times \Bbb Z/24\Bbb Z$?

Question

Why is $\Bbb Z/4\Bbb Z\times \Bbb Z/12\Bbb Z\times \Bbb Z/40\Bbb Z$ not isomorphic to $\Bbb Z/8\Bbb Z\times \Bbb Z/10\Bbb Z\times \Bbb Z/24\Bbb Z$?

I was thinking to show that they have distinct number of subgroups of the same order, or maybe they have different number of elements of a given order. Since I know they are abelian, have the same order, they are not cyclic, they have subgroups and elements of order for every divisor of 1920, so it is the only idea I have left. Or maybe using the Fundamental Theorem for Finite Abelian Groups?

But, the problem is I don't know how to recognize the number of subgroups or elements of a given order, of course, efficiently. Because I could waste my day looking for all the subgroups and elements of a given order but that is not the idea.

I would be thankful if someone can guide me in the right way.

Answer 1

One simple way to see this is that the second group has an element $(0,5,0)$ of order $2$ and no element $g$ such that $g+g=(0,5,0).$

There is no element like this in the first group.

If $(x,y,z)\in\mathbb Z_4\times\mathbb Z_{12}\times\mathbb Z_{40}$ is of order $2,$ then $2x$ is divisible by $4,$ $2y$ has to be divisible by $12$ and $2z$ has to be divisible by $40.$ But that means $x,y,z$ have to be even, and there is an element $g$ such that $g+g=(x,y,z).$

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I found this element by factorizing the groups, and found the second group had a factor $\mathbb Z/2\mathbb Z,$ and $(0,5,0)$ corresponds to the $1$ in $\mathbb Z/2\mathbb Z.$

In general, a finite abelian group has an element like this if and only if the fundamental theorem factorization has a factor of $\mathbb Z/2\mathbb Z.$

More generally, if $A$ is a finite abelian group with a factor $\mathbb Z/p^k\mathbb Z,$ then $A$ has an element $a$ of order $p^k$ such that $pb=a$ has no solution $b.$

This can often be used to find examples in one group but not another, but it doesn't work all the time, for example, comparing:$$\mathbb Z/p\mathbb Z\times\left(\mathbb Z/p^2\mathbb Z\right)^2\\ \left(\mathbb Z/p\mathbb Z\right)^3\times\mathbb Z/p^2\mathbb Z.$$

Now the difference between these groups is in the number of elements of order $p$ which do not have $p$th roots.

Answer 2

We have that the first group is isomorphic to

$$\Bbb Z_4\times(\Bbb Z_3\times \Bbb Z_4)\times (\Bbb Z_5\times \Bbb Z_8),$$

while the second group is isomorphic to

$$\Bbb Z_8\times(\Bbb Z_2\times \Bbb Z_5)\times (\Bbb Z_3\times\Bbb Z_8).$$

This uses the following lemma:

Lemma: For any $a,b\in\Bbb N$, we have $$\gcd(a,b)=1\iff\Bbb Z_{ab}\cong \Bbb Z_a\times \Bbb Z_b.$$

For a proof, see here.

Therefore, by the Fundamental Theorem of Finite Abelian Groups, they are not isomorphic.