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How to determine number of isomorphic group of order 121?
Since 121 is the square of a prime, there are only two distinct isomorphic groups.
How do you prove Cayley's theorem which states that every group is isomorphic to a permutation group?
Cayley's theorem:Let (G,$) be a group. For each g Є G, let Jg be a permutation of G such thatJg(x) = g$xJ, then, is a function from g to Jg, J: g --> Jg and is an isomorphism from (G,$) onto a permutation group on G.Proof:We already know, from another established theorem that I'm not going to prove here, that an element invertible for an associative composition is cancellable for that composition, therefore Jg is a permutation of G. Given another permutation, Jh = Jg, then h = h$x = Jh(x) = Jg(x) = g$x = g, meaning J is injective. Now for the fun part!For every x Є G, a composition of two permutations is as follows:(Jg ○ Jh)(x) = Jg(Jh(x)) = Jg(h$x) = g$(h$x) = (g$h)$x = Jg$h(x)Therefore Jg ○ Jh = Jg$h(x) for all g, h Є GThat means that the set Ђ = {Jg: g Є G} is a stable subset of the permutation subset of G, written as ЖG, and J is an isomorphism from G onto Ђ. Consequently, Ђ is a group and therefore is a permutation group on G.Q.E.D.
Give an example of a permutation and combination?
If there is a group of 3 coloured balls, then any groups of 2 balls selected from it will be considered as a combination, whereas the different arrangements of every combination will be considered as a permutation
G is finite group if and only if the number of it is subgroups is finite?
False. G may be a finite group without sub-groups.
What is the permutation group for 123?
{123, 132, 213, 231, 312, 321}
