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.

  • $\begingroup$ You are correct that isomorphisms preserve order of elements. For the classification, show that there is only one group of order 4 (up to isomorphism) with an element of order 4, then show that there is only one group (again up to isomorphism) without an element of order 4. $\endgroup$ – James Oct 2 '14 at 15:16
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    $\begingroup$ I fixed the notation, but just so you know: $\bigoplus$ is to $\oplus$ what $\Sigma$ is to $+$ (it's not a binary operation). $\setminus$ (setminus) is the set-theoretic difference, which is different from $/$ (division on the right) and $\backslash$ (\backslash, division on the left, though granted, this difference is a bit of pedantry). $\mathbb Z \setminus 2\mathbb Z$ is not a quotient group, but the set of odd integers. $\mathbb Z \backslash 2\mathbb Z$ doesn't make sense, as $\mathbb Z$ is not a subgroup of $2\mathbb Z$, but the other way around. $\endgroup$ – tomasz Oct 2 '14 at 15:20
  • $\begingroup$ Thank you for correcting the notation, I see how the direction in which the backslash is going would mean different things. As for the \Oplus analogy, thanks again, I did not know that. $\endgroup$ – Jamil_V Oct 2 '14 at 15:24
  • $\begingroup$ @Jamil_V: I actually made a mistake there, as it should be \bigoplus (fixed now). Still, I think you should get the idea. :) $\endgroup$ – tomasz Oct 2 '14 at 15:25
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    $\begingroup$ @Jamil_V not quite, as there is a group of order 4 with elements of order at most 2. $\endgroup$ – James Oct 2 '14 at 17:01

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 $\mathbb{Z}/2\oplus \mathbb{Z}/2$.

  • $\begingroup$ So by assuming that $G$ has order 4 but not an element of order 4, then the linked multiplication table would show that there is only one such group meeting the criteria? $\endgroup$ – Jamil_V Oct 2 '14 at 15:27
  • $\begingroup$ Yes, the link shows that there is only one possible table then - the one of $V_4$. $\endgroup$ – Dietrich Burde Oct 2 '14 at 15:30
  • $\begingroup$ What do you mean by $V_4$? Sorry, I was reading the linked page and didn't see this notation. $\endgroup$ – Jamil_V Oct 2 '14 at 15:35
  • $\begingroup$ Just to be explicit with the solution to the second part, are there 2 groups of order 4 up to isomorphism? One being $G=\{1,a,a^2,a^3\}$ and the other $\mathbb{Z}/2 \oplus \mathbb{Z}/2$? $\endgroup$ – Jamil_V Oct 2 '14 at 16:23
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    $\begingroup$ Yes. The second is also called $V_4$, the Klein four-group. $\endgroup$ – Dietrich Burde Oct 2 '14 at 18:24

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