Prove the set of even permutation in $\displaystyle S_{n}$ forms a subgroup of $\displaystyle S_{n}$.
How should I start this?
This is referred to an "alternating group" $\displaystyle A_n$.
How about you start off and I help you if necessary.
Let $\displaystyle G$ be a group and $\displaystyle H$ a non-trivial subset of $\displaystyle G$. What makes $\displaystyle H$ a subgroup? What conditions must be met?
Now do the same thing here. Show all of these conditions are satisfied.
Okay, let me think more about this...
Suppose that $\displaystyle H \subset S_{n}$ with $\displaystyle H= \{ a_{1},a_{2}, ... , a_{n} \}$ with a being permutations that can be expressed as even number of 2-cycles, in other words, H contains all of the even permutations in Sn.
Now pick $\displaystyle \alpha , \beta \in H$.
Then $\displaystyle \alpha = (a_{1} a_{2})(a_{3}a_{4})$ and $\displaystyle \beta = (b_{1}b_{2})(b_{3}b_{4})$
Now, $\displaystyle \alpha \beta $ is in H because it is only another expression of even permutations. (Do I need to justify this? Or do we already know that? I know that by the problems that I have done before)
Then $\displaystyle \alpha$ is its own inverse, thus satisfied the conditions for H to be a subgroup of Sn.
Am I doing something right here?
Thanks.
[QUOTE=tttcomrader;69480]
Almost.Am I doing something right here?
.
$\displaystyle \alpha$ is even and $\displaystyle \beta$ is even then $\displaystyle \alpha \beta$ is even. You mentioned that but you did not show it. Why is that? Remember you need to show this to show that $\displaystyle H$ is closed.