Proving a subgroup of $S_4$ is not isomorphic to $(\mathbb{Z}_4, +)$ by contradiction First time poster.
I'm trying to prove that the subgroup $\{f_1, f_2, f_3, f_4\}$ in $S_4$ is not isomorphic to $(\mathbb{Z}_4, +)$ by contradiction. In this particular subgroup, the diagonal from the top left to the bottom right in the Cayley Table is the identity element $f_1$. However, $1+1$ in the Cayley table for the integers modulo $4$ is $2$. I've attempted to prove that these two sets are not isomorphic however I do not feel confident with my proof. Any feedback welcome. 
Proof
Suppose $\{f_1, f_2, f_3, f_4\}=H$ is isomorphic to $(\mathbb{Z}_4, +)$.
Then there exists $f$ which is an isomorphic function from $(\mathbb{Z}_4, +)$ to $H$.
Since $f$ is an isomorphic function, $f(1)^2=f_1$ because for all $f_n \in H$, $f_n^2=f_1$ ($n \in \{1,2,3,4\}$).
So, $|H|=2$ as $f(1)$ generates $H$. This is a contradiction because $|H|=4$
Thus, the two given sets aren't isomorphic.
 A: If the diagonal in the Cayley Table is the identity element, then each element multiplied by itself in your subgroup gives the identity element. Thus, each of $f_{2}, f_{3}, f_{4}$ satisfy
$$
f_{2}^{2} = f_{1}, f_{3}^{2} = f_{1}, f_{4}^{2} = f_{1}.
$$
By definition, these have order $2$ since they give you the identity when you square them. However, you've correctly identified that $1 \in \mathbb{Z}_{4}$ does not have order $2$, since $1 + 1 = 2 \neq 0$. Since the order of any element must be preserved under any group isomorphism, no such isomorphism can exist because there is no element in your subgroup that you can send $1$ to.

Your proof is good, since you've done pretty much the same thing: you've identified that $1$ generates $\mathbb{Z}_{4}$ and that by giving it an image you get a contradiction. However, you don't need to go so far --- once you've identified that there are mismatched orders between the groups, that is enough.
If you're not allowed to work with the knowledge that isomorphic groups preserve element orders, then your proof is definitely appropriate.
