Question: Let G be a non-abelian group such that the size of G is 10. Prove G is isomorphic to D_10.
I have started by saying the elements of G have order 1,2,5 or 10. And then showing how there are no elements of order 10 as that would make it abelian. I then show there is an element x of...
Ok, I have noticed that <a> = <a^2> = <a^3> = <a^4> for all a in A_6 where a is a 5-cycle. So this means that the number of elements of order 5 must be divided by 4. Hence the answer is 144/4 = 36 subgroups of size 5 in A_6.
Is this correct?
Thanks a lot. I have found what the subgroup is now: {e, (23)(56), (13)(46), (12)(45), (123)(456), (132)(465)}. I need to justify my answer so I need to show that this is isomorphic to D_6 and therefore is a subgroup. I'll have a go.
I need to find just one non-abelian subgroup of size 6 in A_6.
I have started by noting that the subgroup must be isomorphic to D_6 and then tried to use the permutations in D_6 that sends corners to corners. I then came across the problem that the reflection elements in D_6 consist of...
I need to find the number of subgroups of size 5 in A_6.
I have started by noting that as the subgroup size is 5, a prime, the subgroups must be cyclic. I have worked out that there are 144 elements of order 5 in A_6, but this can't be equal to the number of subgroups (i found two subgroups...