Let $F$ be a field. I am trying to show $F^\times$ contains no subgroup which is isomorphic to $\mathbb{Z}_2 \times \mathbb{Z}_4$. For obvious reasons, I am only considering the finite subgroups of $F^\times$. I know one way to do it is to recognize that a subgroup of the multiplicative group is cyclic and $\mathbb{Z}_2 \times \mathbb{Z}_4$ is not, so they cannot be isomorphic. However, to test my understanding I'm trying to justify it a different way.
Are we allowed to say that they aren't isomorphic because $\mathbb{Z}_2 \times \mathbb{Z}_4$ has a zero element and $F^\times$ does not, so no subgroup of $F^\times$ could contain a zero element? Or does that not work because even if a subgroup of $F^\times$ doesn't have a zero element, it can still have an element which is mapped to the zero element of $\mathbb{Z}_2 \times \mathbb{Z}_4$?
The groups $U(2) = \{1,-1\}$ (2nd roots of unity under multiplication) and $\mathbb{Z}_2 = \mathbb{Z}/2\mathbb{Z} = \{0,1\}$ (Integers modulo 2 under addition) are isomorpic via the isomorphism $1 \mapsto 0$ and $-1 \mapsto 1$.
Hence you cannot argue as you do. The identity element must be mapped to the identity element, that is correct, but an identity element can be written as 0 in additive notation, but could be written as $1$ in multiplicative notation. In $U(2)$ the identity element is $1$, and in $\mathbb{Z}/2\mathbb{Z}$ the identity element is $0$. A commutative group written muliplicatively may be isomorphic to a group written additively.