Show $\mathbb{Z}_2 \cong \mathbb{D}_2$ Hello once again everyone. A homework exercise I'm working on is asking me to show $\mathbb{Z}_2 \cong \mathbb{D}_2$ (the Dihedral group). I know that $\mathbb{Z}_2$ is cyclic and generated by $1$, but I've tried show that $\mathbb{D}_2$ with elements $(1,2)$ and $(2,1) $ is generated by $(1,2)$ or $(2,1)$ under addition or multiplication, but I can't find out how to show this in order to establish an isomorphism between the $\mathbb{Z}_2$ and $\mathbb{D}_2$. Can someone perhaps provide me a hint? As always, thank you so much for your support. 
 A: Map the identities to each other. There cannot be any other choice.
A: We have that $$(\mathbb Z_2, +) = \{0, 1\}\;\; \text{under addition modulo}\;2,$$ and we have that $$(D_2, \circ) = \{\text{id}\,, (1\;2)\} \;\text{under reflection (a permutation)}$$
Note that in writing the non-identity element in $D_2$ in cyclic notation, as you have done, $(1\;2) = (2\;1)$: the reflection/group operation is given by permuting points $\;1 \longleftrightarrow 2,\;$ and $\text{id}\;$ is simply the "do nothing" identity element.
The generator of $\mathbb Z_2$ is indeed $1$, and the generator of $D_2$ is the reflection $(1\;2)$. 
Map identity to identity, generator to generator, and you have your isomorphism.
$$0\in \mathbb Z_2 \mapsto \text{id}\in D_2$$
$$1 \in \mathbb Z_2 \mapsto (1\;2) \in D_2$$

Note: you will soon learn that there is exactly one group of order two, up to isomorphism. That is, every group of order 2 is isomorphic to $\mathbb Z_2.\;$ For many reasons: An isomorphism always maps identity to identity, generator to generator, and so there is only one way to map two elements to two elements. 
But we also know that there is only one group, up to isomorphism, of order $3$: As JavaMan comments below: All groups of order $3$ are isomorphic to $\mathbb Z_3.\;$ Indeed, for any prime $p$, there is, up to isomorphism, exactly one group: if $p$ is some prime, then every group of order $p$ is cyclic and isomorphic to $\mathbb Z_p$.
A: Hint
$$\mathbb Z_2=\{\overline{1},-\overline{1}\}\quad;\quad \mathbb D_2=\{\mathrm{id},\mathrm{r}\}$$
where $\mathrm{r}$ is a reflection so can you see that we can identify $\overline{1}\leftrightarrow \mathrm{id}$ and $-\overline{1}\leftrightarrow \mathrm{r}$ ?
