Answer to Question #193809 in Abstract Algebra for Sadhna Singh

Question

a) Using Cayley’s theorem, find the permutation group to which a cyclic group of

order 12 is isomorphic. (4)

b) Let τ be a fixed odd permutation in . S10 Show that every odd permutation in S is 10

a product of τ and some permutation in . A10 (2)

c) List two distinct cosets of < r > in , D10 where r is a reflection in . D10 (2)

d) Give the smallest n ∈ N for which An is non-abelian. Justify your answer.

Answer 1

a) Given a cyclic group G = <a> of order 12 generated by the element 'a’ of G, we may take the 'permutation group’ as the cyclic subgroup H of S$_{12}$ generated by the cycle (1,2,3,4,5,6,7,8,9,10,11,12) of length 12. This (or a conjugate of it) is what we obtain as per the proof of the Cayley's Theorem.

We may also find isomorphic copies of G as subgroups of groups S$_n$ for all n $\ge$7. For example the product (1,2,3,4)•(5,6,7) of disjoint cycles in S$_7$ has order LCM(4,3)=12 and hence generates a cyclic subgroup of S$_7$ , isomorphic to G.

d)

$A_n$ has order $n!/2$  and is not abelian if $n\ge4$

Answer: $n=4$

c)

Let $D_{10}=\{x^iy^i|x^{10}=y^2=1,xy=yx^9\}$

two cosets: $\{x,x^6\},\ \{x^2,x^7\}$

b)

$A_{10}$ the subset of even permutations in $S_{10}$.

$\tau$ is product of odd number of transpositions in $S_{10}$ ; permutation in $A_{10}$ is product of even number of transpositions in $S_{10}$ . So, result of product of $\tau$ and permutation in $A_{10}$ is:

(odd number of transpositions in $S_{10}$ )+(even number of transpositions in $S_{10}$ )=

=(odd number of transpositions in $S_{10}$ ), that is odd permutation in $S_{10}$.