2. Prove that the set An of even permutations of Sn is a subgroup of Sn- Hint: C

Question

2. Prove that the set An of even permutations of Sn is a subgroup of Sn- Hint: Consider the following example. In S3, we have three even permutations, namely (13)(12), = (13)(12) and the identity . Now, = (13) (12) (12)(13). Just based on this, one can conclude Bis even as it can be written as a product of 4 transpositions (and 4 is even). Also note that -1-(12)(13) (I reversed the order of the transpositions in the original expression of .) Based on this, one can conclude -1 is even as it can be written as a product of 2 transpositions (and 2 is even). Now generalize this argument and show that if = t! . . .t2k E An and = 81 82m An then o An and -1 E An, that is, show that and -1 can be written as an even product of transpositions

Explanation / Answer

Let An = the set of even permutations in Sn.

To Show that An is a subgroup of Sn. An is called the alternating group of degree n.

CLOSURE :: The product of two even permutations will again have an even number of factors, if a and b are product of even numbers of transpositions each, then their product ab is also product of even number of transposition, so An is closed.

INVERSES :: Notice that (a b)-1 = (a b). Thus the inverse of a product of even permutation (using the general formula for the inverse of a product) is the product of the same transpositions in reverse order, hence is even again

if, a = t1 t2 ...t2k

then, a-1 = ( t1 t2 ...t2k )-1 = ( (t2k )-1 ... (t2)-1 (t1)-1 ) belongs to An

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