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    Does Cayley's Theorem imply all groups are countable?

    Note that under union with 0 this is in fact a group, if x is in G and y is in R and |x-y|<epsilon then x is equivalent to y if you consider y to be an element of G. In other words, once you set the precision, to consider such a y as an element of G also, the part of y that is smaller than...
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    Does Cayley's Theorem imply all groups are countable?

    What I'm saying for the real numbers, is that every time you try to fix some labeling of them, there exist some elements in the real line not in this labeling. How then can a transitive action, which just means that the action has a single orbit containing all permutations of the group, be defined?
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    Does Cayley's Theorem imply all groups are countable?

    I understand that it may be infinite. But i just don't see how it is possible to have a group acting on itself transitively if it is uncountable. If it's uncountable it can't be indexed by any subset of the natural numbers, so a set of permutations which permutes every element to every other...
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    Does Cayley's Theorem imply all groups are countable?

    My question is exactly what is stated in the title: Does Cayley's theorem imply that all groups are countable? I don't see how a well defined transitive action of an uncountable group on itself. How could you possibly find a set of permutations sending a single element to every other element in...
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    Consequence of the First isomorphism theorem

    If n is the order of G, Aut(G) is isomorphic to the symmetric group on n letters. It's order is n!. A quotient of G by some normal subgroup is less than or equal to the order of G. Say, the order G/Z(G) is m. Then m<=n. How can this quotient possibly be of order n! unless n is less than or equal...
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    Question about normal subgroups/Lattice Isomorphism Theorem

    I was just brushing up on some Algebra for the past couple of days. I realize that the lattice isomorphism theorem deals with the collection of subgroups of a group containing a normal subgroup of G. Now, in general, if N is a normal subgroup of G, all of the subgroups of larger order than N do...
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