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Example 8. Consider $$\Bbb Z_2\times\Bbb Z_2 = \{(0,0),(0,1),(1,0),(1,1)\}$$ Although $\Bbb Z_2\times\Bbb Z_2$ and $\Bbb Z_4$ both contain four elements, it is easy to see that they are not isomorphic since for every element $(a,b)$ in $\Bbb Z_2\times\Bbb Z_2$, $$(a,b) + (a,b) = (0,0),$$ but $\Bbb Z_4$ is cyclic.

I understand the composition $\Bbb Z_2\times\Bbb Z_2 = \{(0,0),(0,1),(1,0),(1,1)\}$ and that $\Bbb Z_4=\{0,1,2,3\}$. I understand also that "for every element $(a,b)$ in $\Bbb Z_2\times\Bbb Z_2$, $(a,b) + (a,b) = (0,0)$".

But my questions remain: why does this show that they are not isomorphic?and how does the argument "but $\Bbb Z_4$ is cyclic" help answering whether or not they are isomorphic.?

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You should think of isomorphic groups as being really the "same". For instance:

Fact: If $G$ and $H$ are isomorphic groups, and $g\in G$ corresponds to $h\in H$ under this isomorphism, then $g$ and $h$ have the same order.

This is what is being used in the proof you're asking about. So, since $(a,b)+(a,b)=(0,0)$ for all $(a,b)\in\Bbb Z_2\times\Bbb Z_2$, this precisely says that every element of $\Bbb Z_2\times\Bbb Z_2$ has order $2$. On the other hand, since $\Bbb Z_4$ is cyclic, there is an element of order $4$. Referring to the fact above, we see the two groups cannot be isomorphic.

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Well, $(a,b)+(a,b)=(2a,2b)=(0,0)$ since $2a=0$ for all $a\in \mathbb{Z}_2$. It shows that there are no elements of order greater than $2$ in $\mathbb{Z}_2\times \mathbb{Z}_2$. By definition, $\mathbb{Z}_4$ does have an element of order $4$.

You can show that isomorphisms of groups preserve orders of elements. (Any property that can be intrinsically defined, should be preserved by isomorphisms).

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In $\mathbb{Z}_2 \times \mathbb{Z}_2$, the operation is $(a,b)+(c,d)=\left((a+c)\mod2),(b+d)\mod2\right)$

That's why every element has order $2$. But in $\mathbb{Z}_4$, there is an element of order $4$ namely $1$.

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  • $\begingroup$ is it supposed to be (b+d) mod 2 ? $\endgroup$ – gegu Apr 9 '17 at 16:56
  • $\begingroup$ @gegu yes!! Thank you $\endgroup$ – tattwamasi amrutam Apr 9 '17 at 16:57

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