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For the following pair of groups, explain why they are not isomorphic: $ \mathbb{Z}_4 \times \mathbb{Z}_4$ and $\mathbb{Z}_4 \times \mathbb{Z}_2 \times \mathbb{Z}_2$

I have started my exercises on isomorphisms and have run into this problem. In the answer section it says that these two groups are not isomorphic because they have different amounts of elements of order $2$.

I have listed the elements of each group and the effect does not have the same number of elements of order $2$. But how could this procedure be done without having to resort to listing the elements?

On the other hand, given one of these situations, what number $n$ to take? That is, can I by trial and error with the possible orders that divide the order of the group?

Shaun
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Hopmaths
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    You don't need to count all the elements of order $2$. The second group has a subgroup isomorphic to $\mathbb Z_2\times\mathbb Z_2\times\mathbb Z_2$. The first one must contain an isomorphic copy of this which has to be generated by the pairs $(2,0)$, $(0,2)$, and $(2,2)$ which correspond to the triples $(1,0,0)$, $(0,1,0)$, respectively $(0,0,1)$. But the sum of any two pairs equals the third, which is not the case for triples. – user26857 Feb 01 '21 at 18:24

2 Answers2

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The fact that $\Bbb Z/4\Bbb Z\times\Bbb Z/4\Bbb Z\not\cong\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ is implied by $\Bbb Z/4\Bbb Z\not\cong\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$ (which is much easier to check: one of these groups has no element of order $>2$ while the other does) and related to the Fundamental Theorem of finite(ly generated) Abelian groups. Long story short: one can decompose (non-cyclic) abelian groups into direct products of cyclic groups of prime power order. However, splitting prime powers in this process produces non-isomorphic abelian groups.

As I am aware, this might be a bit over the top for this simple question. A simpler thing to observe is the following: $$\operatorname{ord}((a_1,\dots,a_n))=\operatorname{lcm}(\operatorname{ord}(a_1),\dots,\operatorname{ord}(a_n))$$ It is not hard to see that $\Bbb Z/4\Bbb Z$ has $2$ elements of order $4$, $1$ of order $2$ and $1$ of order $1$ while $\Bbb Z/2\Bbb Z$ has $1$ element of order $2$ and $1$ of order $1$. Combining these gives a much more efficient way of computing the number of elements of a given order.

The process always works in the sense that if you find different numbers of elements of a given order (dividing the group order) they cannot be isomorphic as isomorphisms preserve the order of elements. However, the converse is not necessarily true, that is there are groups of the same order with the same number of elements for every order which are not isomorphic (see here for a counterexample).

Let's consider the group $\Bbb Z/4\Bbb Z\times\Bbb Z/2\Bbb Z\times\Bbb Z/2\Bbb Z$. Write $\Bbb Z/4\Bbb Z=\{0,1,2,3\}$ and $\Bbb Z/2\Bbb Z=\{0,1\}$. As $\operatorname{lcm}(n_1,n_2,n_3)\ge\max\{n_1,n_2,n_3\}$ we conclude that for $(a_1,a_2,a_3)$ having order $2$ we have $a_1\ne1,3$. This leaves $2\cdot2\cdot2=8$ possibilites. However, if $a_1=a_2=a_3=0$ then $\operatorname{ord}(a_1,a_2,a_3)=1<2$. But this is the only exception to be found. Hence, there are exactly $7$ elements of order $2$.

For $\Bbb Z/4\Bbb Z\times\Bbb Z/4\Bbb Z$ we are immediately down to at most $4$ possibilites ($3$ remain in the end). This shows that they cannot be isomorphic.

mrtaurho
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It follows from the Fundamental Theorem of Finite(ly-generated) Abelian Groups. But this is a little glib on my part.

Yes, counting the number of elements of order two is sufficient. Trial & error will work since the groups are small enough.

Shaun
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