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. 

Hint: You need to exhibit an isomorphism between the two groups, which is a homomorphism that is also a bijection. There are only two bijections $S_2\to Z_2$, and you can check the definition to see that one of them is a homomorphism.
