Proving A_4 ≠ S_4: Is Order of Elements Enough?

[latentcorpse]

How do I go about showing A_4 \not\cong S_4

So far the only argument I've been able to come up with is that the order of the elements of the two groups differ. Is this sufficient to conclude the two groups aren't isomorphic - it just doesn't seem that rigorous to me.

 

[Dick]

A group isomorphism is a 1-1 and onto mapping between two groups. Of course that implies they have the same order. There's nothing nonrigorous about saying that.

 

[crazyjimbo]

If \phi is an isomorphism between groups then \phi(g^n) = \phi(g)^n so \phi must preserve orders. If you can demonstrate there is an element in one group with an order which doesn't appear in the other group then that means there can be no isomorphism.

 

[latentcorpse]

my bad,
the question was to show A4 isn't isomorphic to D6.
surely i can just apply the argument given in post 3 to this case though?

 

[Dick]

my bad,
the question was to show A4 isn't isomorphic to D6.
surely i can just apply the argument given in post 3 to this case though?

Absolutely.