$\mathbb{Z}/\mathbb{2Z} \bigoplus \mathbb{Z}/\mathbb{2Z}$ not isomorphic to $\mathbb{Z}/\mathbb{4Z}$

Question

Greetings fellow Mathematics Community. I am having some doubts about my solution to the following problem:

Show that $\mathbb{Z}/ \mathbb{2Z} \oplus \mathbb{Z}/ \mathbb{2Z}$ is not isomorphic to $\mathbb{Z}/ \mathbb{4Z}$. Now find all groups of order 4 up to isomorphism.

Here is what I have:

First, I listed the elements (or maybe in this context, the costets) of each group:

$\mathbb{Z}/\mathbb{2Z}$ =

• $0+\mathbb{2Z} = \{0,\pm2,\pm4,\ldots\}$ and
• $1+\mathbb{2Z} = \{\pm1,\pm3,\pm5,\pm7,\ldots\}$

$\mathbb{Z}/ \mathbb{4Z}$=

• $0+\mathbb{4Z} = \{0,\pm4,\pm8,\pm12 \ldots\}$
• $1+\mathbb{4Z} = \{-7,-3,1,5,9,13,\ldots\}$
• $2+\mathbb{4Z} = \{-10,-6,-2,2,6,10,\ldots\}$
• $3+\mathbb{4Z} = \{-13,-9,-5,-1,3,7,11,15\ldots\}$

From this, I note that $|\mathbb{Z}/\mathbb{2Z} \oplus \mathbb{Z}/\mathbb{2Z}| =4 \quad$ and $\quad |\mathbb{Z}/\mathbb{4Z}|=4$

Since they both have the same order, then I would want to say that an element $a \in \mathbb{Z}/\mathbb{2Z}$ has either order 1 or order 2 by Lagrange's Theorem. Similarly, an element $b\in \mathbb{Z}/\mathbb{4Z}$ would have either order 1, order 2, or order 4. If they were isomorphic, that is

$\mathbb{Z}/\mathbb{2Z} \oplus \mathbb{Z}/\mathbb{2Z} \cong \mathbb{Z}/\mathbb{4Z}$, then wouldn't that imply that the orders of elements are preserved under the mapping? Since there is no element of order 4 in the first group, then may I conclude that they are not isomorphic to each other?

As for the second part, how would I find all groups of order 4 up to isomorphism?

Any help is greatly appreciated since I have been trying to understand this problem for a while now.

Answer 1

The argument using the order of the elements is correct. Because $\mathbb{Z}/2\oplus \mathbb{Z}/2$ has no element of order $4$, the group cannot be isomorphic to $\mathbb{Z}/4$. This is already half of the classification of groups of order $4$. If $G$ has an element $a$ of order $4$, then $G=\{1,a,a^2,a^3\}$, which is $\mathbb{Z}/4$. Suppose that $G$ has order $4$, but does not contain an element of order $4$. Then its multiplication table is already uniquely determined, see here. And its group is just $\mathbb{Z}/2\oplus \mathbb{Z}/2$.