# Why is $\langle \mathbb{Z}_4, + \rangle$ not isomorphic to $\langle \mathbb{Z}_2 \times \mathbb{Z}_2, + \rangle$?

I'm having some trouble here, specifically with the idea of $\langle \mathbb{Z}_2 \times \mathbb{Z}_2, + \rangle$ as a group. Can anyone help me out with some explanations?

Moreover, I generally haven't wrapped my head around groups like that. Can anyone shed some light on exactly how to think about, for example, $\langle \mathbb{Z}_2 \times \mathbb{Z}_3, + \rangle$?

• one is cyclic, the other not – janmarqz Feb 3 '14 at 18:51

Think of an element of $\Bbb Z_n \times \Bbb Z_m$ as having two parts, written $\langle p,q\rangle$. The left part $p$ is an element of $\Bbb Z_n$ and the right part $q$ is an element of $\Bbb Z_m$. The two parts never interact. To add $\langle a, b\rangle$ to $\langle c, d\rangle$, you add the two parts separately, and get $\langle a+c, b+d\rangle$. The left-side addition is done $\Bbb Z_n$-style, because $a$ and $c$ are elements of $\Bbb Z_n$, and the right-side addition is done $\Bbb Z_m$-style.

$\def\V{\Bbb Z_2\times \Bbb Z_2}\V$ has four elements, which are $\langle 0,0\rangle, \langle 0,1\rangle, \langle 1,0\rangle,$ and $\langle 1,1\rangle$. Its identity element is $\langle 0,0\rangle$. But $\V$ is not $\Bbb Z_4$, and the easiest way to see this is that every element $x$ of $\V$ has the property that $x+x$ is the identity element $\langle 0,0\rangle$: \begin{align} \langle 0,0\rangle + \langle 0,0\rangle & = \langle 0,0\rangle\\ \langle 0,1\rangle + \langle 0,1\rangle & = \langle 0,0\rangle\\ \langle 1,0\rangle + \langle 1,0\rangle & = \langle 0,0\rangle\\ \langle 1,1\rangle + \langle 1,1\rangle & = \langle 0,0\rangle \end{align} all the additions being done $\Bbb Z_2$-style.

But $\Bbb Z_4$ is different; it has two elements, 1 and 3, that do not have the property that $x+x=0$. They have $1+1 = 3+3 = 2$ instead, and in $\Bbb Z_4, 2\ne 0$. $\V$ has nothing like this.

Or looked at in the opposite direction, $\Bbb Z_4$ has an operation, namely the operation of adding 1, which you must do four times before you get back to where you started; it is analogous to giving an object a quarter turn. After four quarter turns, and no fewer, the object has returned to its original position. $\V$ has nothing like this; every operation in $\V$ gets you back to where you started after at no more than two repetitions.

• All the answers were good, but this is by far the most detailed. Cheers! – Newb Feb 3 '14 at 19:06
• Great answer! +1 – mathematics2x2life Feb 3 '14 at 20:33

$\mathbb Z_2\times\mathbb Z_2$ is the set of ordered pairs $(a,b)$ with $a,b\in\mathbb Z_2$, i.e. $\{(0,0),(1,0),(0,1),(1,1)\}$. The group operation is coordinate-wise $\mathbb Z_2$-addition.

It cannot be isomorphic to $\mathbb Z_4$ because all nonzero elements in $\mathbb Z_2\times\mathbb Z_2$ have order 2, while in $\mathbb Z_4$ there are elements of order 4. In particular, $\mathbb Z_2\times\mathbb Z_2$ is not cyclic.

• $(0,0)$ has order $1$. – Martin Brandenburg Feb 3 '14 at 18:56
• Corrected, thanks. – Martin Argerami Feb 3 '14 at 19:04

These finite groups are great because you can list all of their elements.

For instance, here are the elements of $\mathbb{Z}_4$: $\{0,1,2,3\}$.

And here those of $\mathbb{Z}_2\times \mathbb{Z}_2$: $\{ (0,0), (0,1), (1,0), (1,1)\}$.

(I'm omitting in both cases to put some $\overline{2}$ all above those numbers in order to point out that they are not actual natural numbers, but classes. But this way I can write faster.)

So, this group $\mathbb{Z}_4$ has a particular guy, $1$, inside with this property:

$$1 + 1 = 2, \ 1 + 1 + 1 = 3, \ 1 + 1 + 1 + 1 = 4 = 0 \ .$$

Can you find anyone inside $\mathbb{Z}_2\times \mathbb{Z}_2$ with a similar behaviour? Namely, that four times itself produces the neutral element? [EDIT: Thanks to MJD's remark. And not just two times?] Nope? You can't? -Then, they cannot be isomorphic.

• Every element of $\Bbb Z_2 \times \Bbb Z_2$ has the property that four times itself produces the neutral element. – MJD Feb 3 '14 at 19:07
• But not that you need to add precisely four times to produce the neutral element. – Agustí Roig Feb 3 '14 at 19:07
• But you did not say that. – MJD Feb 3 '14 at 19:08
• You're right. I've corrected my answer. Thank you. – Agustí Roig Feb 3 '14 at 19:09

Recall that $$\mathbb Z_{mn} \cong \mathbb Z_m \times \mathbb Z_n\;\;\text{ if and only if }\;\;\gcd(m, n) = 1$$

In your first case, take $m = n = 2.$

In the case of $\mathbb{Z}_2 \times \mathbb{Z}_3,\;$ we DO have that $\mathbb Z_6 = \mathbb Z_{2 \times 3} \cong \mathbb{Z}_2 \times \mathbb{Z}_3,\;$ since $\;\gcd(2, 3) = 1$.

• Nobody who asks for $\mathbb{Z}/2 \times \mathbb{Z}/2 \not\cong \mathbb{Z}/4$ knows what you "recall". – Martin Brandenburg Feb 3 '14 at 19:08