Prove the set of even permutation in $\displaystyle S_{n}$ forms a subgroup of $\displaystyle S_{n}$.

How should I start this?

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- Sep 11th 2007, 07:28 PMtttcomraderPermutation subgroup
Prove the set of even permutation in $\displaystyle S_{n}$ forms a subgroup of $\displaystyle S_{n}$.

How should I start this? - Sep 11th 2007, 07:40 PMThePerfectHacker
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. - Sep 11th 2007, 08:20 PMtttcomrader
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. - Sep 11th 2007, 08:30 PMThePerfectHacker
[QUOTE=tttcomrader;69480]

Quote:

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.