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.

  • $\begingroup$ If you know that both groups have two elements, then you already know they're isomorphic. $\endgroup$ – Ink Jun 12 '13 at 22:48

Map the identities to each other. There cannot be any other choice.

  • $\begingroup$ Thank you for your assistance. I truly appreciate! $\endgroup$ – erik7970 Jun 12 '13 at 22:44
  • $\begingroup$ No problem. Glad to help! $\endgroup$ – Gil Jun 12 '13 at 22:46

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 $\mathbb Z_p$.

  • $\begingroup$ I'm slightly confused. (1,2) is the identity for D2? $\endgroup$ – erik7970 Jun 12 '13 at 22:29
  • $\begingroup$ No, the identity is "do nothing rotation": points remain.The generator of $D_2$ is the rotation $(1\;2) = (2\;1)$ $\endgroup$ – Namaste Jun 12 '13 at 22:34
  • $\begingroup$ That's much clearer now. Thank you so very much! $\endgroup$ – erik7970 Jun 12 '13 at 22:44
  • $\begingroup$ You could also note that there is exactly one group of order $p$ where $p$ is prime, and this is why there is one group of order $2$, one group of order $3$, but more than one group of order $4$. $\endgroup$ – JavaMan Jun 12 '13 at 23:15
  • $\begingroup$ @amWhy: Very nice write up! +1 $\endgroup$ – Amzoti Jun 13 '13 at 0:37


$$\mathbb Z_2=\{\overline{1},-\overline{1}\}\quad;\quad \mathbb D_2=\{\mathrm{id},\mathrm{r}\}$$ where $\mathrm{r}$ is a reflection so can you see that we can identify $\overline{1}\leftrightarrow \mathrm{id}$ and $-\overline{1}\leftrightarrow \mathrm{r}$ ?

  • $\begingroup$ Yes, that makes so much more sense. Thank you. $\endgroup$ – erik7970 Jun 12 '13 at 22:32
  • $\begingroup$ You're welcome. $\endgroup$ – user63181 Jun 12 '13 at 22:35

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