All the other answers are more than satisfactory. I have some lecture notes on basic set theory which also answer these questions, so I might as well post links to them:

To see that the cardinals do not form a set, see Fact 20 on page 10 of

http://math.uga.edu/~pete/settheorypart1.pdf

This first handout is super-naive set theory (countable choice is assumed without comment) which is meant to be accessible to an undergraduate math major. In particular I don't say "cardinal" there but rather spell things out in a more explicit way.

This seems a little simpler than the Burali-Forti paradox (which on the other hand is manifestly choice-free), for which see Section 1.4 of

http://math.uga.edu/~pete/settheorypart3.pdf

To see that there is an X of any given infinite cardinality (where X is: a field, a noncommutative group, etc.) see

http://math.uga.edu/~pete/settheorypart4.pdf

My argument for the case of fields -- which implies that of groups by taking the additive group of the field -- uses (only) that for any infinite cardinal $\kappa$, $\kappa \times \aleph_0 = \kappa$. I'm pretty sure that this is a lot weaker than AC.

If you know the Skolem-Lowenheim theorem in model theory, then it is silly to do all of these cases individually: a consistent theory in a language of cardinality $\kappa$ admits models of any infinite cardinality which is greater than or equal to $\kappa$. This is equivalent to AC (see Bell and Slomson), but the special case of countable languages is presumably not.

All the other answers are more than satisfactory. I have some lecture notes on basic set theory which also answer these questions, so I might as well post links to them:

To see that the cardinals do not form a set, see Fact 20 on page 10 of

http://alpha.math.uga.edu/~pete/settheorypart1.pdf

This first handout is super-naive set theory (countable choice is assumed without comment) which is meant to be accessible to an undergraduate math major. In particular I don't say "cardinal" there but rather spell things out in a more explicit way.

This seems a little simpler than the Burali-Forti paradox (which on the other hand is manifestly choice-free), for which see Section 1.4 of

http://alpha.math.uga.edu/~pete/settheorypart3.pdf

To see that there is an X of any given infinite cardinality (where X is: a field, a noncommutative group, etc.) see

http://alpha.math.uga.edu/~pete/settheorypart4.pdf

My argument for the case of fields -- which implies that of groups by taking the additive group of the field -- uses (only) that for any infinite cardinal $\kappa$, $\kappa \times \aleph_0 = \kappa$. I'm pretty sure that this is a lot weaker than AC.

If you know the Skolem-Lowenheim theorem in model theory, then it is silly to do all of these cases individually: a consistent theory in a language of cardinality $\kappa$ admits models of any infinite cardinality which is greater than or equal to $\kappa$. This is equivalent to AC (see Bell and Slomson), but the special case of countable languages is presumably not.

All the other answers are more than satisfactory. I have some lecture notes on basic set theory which also answer these questions, so I might as well post links to them:

To see that the cardinals do not form a set, see Fact 20 on page 10 of

http://math.uga.edu/~pete/settheorypart1.pdf

This first handout is super-naive set theory (countable choice is assumed without comment) which is meant to be accessible to an undergraduate math major. In particular I don't say "cardinal" there but rather spell things out in a more explicit way.

This seems a little simpler than the Burali-Forti paradox, for which see Section 1.4 of

http://math.uga.edu/~pete/settheorypart3.pdf

To see that there is an X of any given infinite cardinality (where X is: a field, a noncommutative group, etc.) see

http://math.uga.edu/~pete/settheorypart4.pdf

My argument for the case of fields -- which implies that of groups by taking the additive group of the field -- uses (only) that for any infinite cardinal $\kappa$, $\kappa \times \aleph_0 = \kappa$. I'm pretty sure that this is a lot weaker than AC.

If you know the Skolem-Lowenheim theorem in model theory, then it is silly to do all of these cases individually: a consistent theory in a language of cardinality $\kappa$ admits models of any infinite cardinality which is greater than or equal to $\kappa$. This is equivalent to AC (see Bell and Slomson), but the special case of countable languages is presumably not.

All the other answers are more than satisfactory. I have some lecture notes on basic set theory which also answer these questions, so I might as well post links to them:

To see that the cardinals do not form a set, see Fact 20 on page 10 of

http://math.uga.edu/~pete/settheorypart1.pdf

This first handout is super-naive set theory (countable choice is assumed without comment) which is meant to be accessible to an undergraduate math major. In particular I don't say "cardinal" there but rather spell things out in a more explicit way.

This seems a little simpler than the Burali-Forti paradox (which on the other hand is manifestly choice-free), for which see Section 1.4 of

http://math.uga.edu/~pete/settheorypart3.pdf

To see that there is an X of any given infinite cardinality (where X is: a field, a noncommutative group, etc.) see

http://math.uga.edu/~pete/settheorypart4.pdf

My argument for the case of fields -- which implies that of groups by taking the additive group of the field -- uses (only) that for any infinite cardinal $\kappa$, $\kappa \times \aleph_0 = \kappa$. I'm pretty sure that this is a lot weaker than AC.

If you know the Skolem-Lowenheim theorem in model theory, then it is silly to do all of these cases individually: a consistent theory in a language of cardinality $\kappa$ admits models of any infinite cardinality which is greater than or equal to $\kappa$. This is equivalent to AC (see Bell and Slomson), but the special case of countable languages is presumably not.