Is the Alternating group $A_5$ isomorphic to the external direct product $A_4 \oplus \mathbb{Z}_5$?

Question

I have been thinking about this question for a while.

The Alternating group $A_5$ is isomorphic to the icosahedron group which has order 60. $A_5$ is a simple group which means the only normal subgroups are the trivial subgroup and itself.

The group defined by the external direct product $A_4 \oplus \mathbb{Z}_5$ is not simple but it does have order 60. Can we still define an Isomorphism from a simple group to a non simple group?

I might be over thinking this question but are all groups of order 60 isomorphic to each other?

Answer 1

Without solvability/simplicity, one can use the center. The center of $A_5$ is the trivial group $\{1\}$ while the center of $A_4\times C_5$ is at least $C_5$.

Answer 2

The group $A_4\times C_5$ is solvable, hence any homomorphic image of it is again solvable. Since $A_5$ is not solvable, there is no such isomorphism.

Answer 3

No, they are not isomorphic. In fact, the group $A_5$ is simple, so it cannot be expressed as the internal direct product (or even semidirect product) of two nontrivial proper subgroups. It could, however, be a Zappa–Szép product, but I'm not sure.

Answer 4

The fact of being simple (and any other property that you can express without referring to the actual name of the elements or what they stand for in a particular case) is invariant under isomorphism.

As for your last question, God no, even if you consider only abelian groups, where the situation is extremely simple, there are more than one group of a given order $n$ as soon as $n$ is not a square-free number.