Permutation subgroup

**tttcomrader** · September 11th 2007, 07:28 PM

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

How should I start this?

**ThePerfectHacker** · September 11th 2007, 07:40 PM

Originally Posted by tttcomrader

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

How should I start this?

This is referred to an "alternating group" $A_n$ .

How about you start off and I help you if necessary.

Let $G$ be a group and $H$ a non-trivial subset of $G$ . What makes $H$ a subgroup? What conditions must be met?

Now do the same thing here. Show all of these conditions are satisfied.

**tttcomrader** · September 11th 2007, 08:20 PM

Okay, let me think more about this...

Suppose that $H \subset S_{n}$ with $H= \{ a_{1},a_{2}, ... , a_{n} \}$ 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 $\alpha , \beta \in H$ .

Then $\alpha = (a_{1} a_{2})(a_{3}a_{4})$ and $\beta = (b_{1}b_{2})(b_{3}b_{4})$

Now, $\alpha \beta$ 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 $\alpha$ is its own inverse, thus satisfied the conditions for H to be a subgroup of Sn.

Am I doing something right here?

Thanks.

**ThePerfectHacker** · September 11th 2007, 08:30 PM

[QUOTE=tttcomrader;69480]

Am I doing something right here?
.

Almost.

$\alpha$ is even and $\beta$ is even then $\alpha \beta$ is even. You mentioned that but you did not show it. Why is that? Remember you need to show this to show that $H$ is closed.

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