Show $S_2 \cong Z_2$ How do I show  $S_2 \cong Z_2$? I'd say it makes good sense when thinking about both groups are of order two, and they both are abelian, but I'm not sure how to tackle the problem. 

Recall that $\mathbb{Z}_{2}=\left\lbrace [0], [1]\right\rbrace$, and that $S_{2}=\left\lbrace (1), (1 \:2)\right\rbrace$. Also recall that $\mathbb{Z}_{2}$ is a group with respect to the operation $\oplus$ and that $S_{2}$ is a group with respect to composition ($\circ$). Let $\xi:\mathbb{Z}_{2} \to S_{2}$ be defined in the following manner.
$\xi([0])=(1)$ and $\xi([1])=(1\:2)$
Then $\xi$ preserves group structure, and the following is apparent.
$\xi([0]\oplus [1])=\xi([0+1])=\xi([1])$
$=(1\:2)$
$=(1\:2) \circ (1)$
$=(1)\circ (1\:2)$
$=\xi([0]) \circ \xi([1])$
Therefore, $\xi$ is a homomorphism. And since the mapping is also one-to-one and onto, it is an isomorphism. Therefore, $\mathbb{Z}_{2} \approx S_{2}$.
