Check whether these two groups are isomorphic using their Cayley tables

Question

I have recently been introduced to the topic of Group Theory and I'm trying to do the practice questions. Was wondering if I got the idea of Isomorphism correct.

Problem :

Given two Cayley Tables :

$\begin{array}{c|c|c|c|c|c} \circ & f_1 & f_2 & f_3 & f_4 \\ \hline f_1 & f_1 & f_2 & f_3 & f_4 \\ \hline f_2 & f_2 & f_1 & f_4 & f_3 \\ \hline f_3 & f_3 & f_4 & f_1 & f_2 \\ \hline f_4 & f_4 & f_3 & f_2 & f_1 \\ \hline \end{array}$

$\begin{array}{c|c|c|c|c|c} + & 0 & 1 & 2 & 3 \\ \hline 0 & 0 & 1 & 2 & 3 \\ \hline 1 & 1 & 2 & 3 & 0 \\ \hline 2 & 2 & 3 & 0 & 1 \\ \hline 3 & 3 & 0 & 1 & 2 \\ \hline \end{array}$

are the two groups Isomorphic? Justify your answer.

Answer :

The answer I have come to is that they are not Isomorphic because it is not possible to map the identity element ($f_1$) of the first group to the identity element (0) of the second group.

I understand there are many other requirements for groups to be considered isomorphic but I was wondering if this is a strong enough answer and justification?

Any help would be appreciated, thanks~

Answer 1

As it stands, you have no justification yet. You're basically saying that it's impossible because it's impossible. You need to find a convincing reason that $f_1$ can't be mapped to $0$.

Here's a hint for another direction you could take this: count the orders of the elements in each group.

Answer 2

Isomorphisms preserve elements' order. In the former group (with identity $f_1$), all the nontrivial elements have order $2$. In the latter (with identity $0$), just $2$ has order $2$, whereas $1$ and $3$ have order $4$. Therefore, no one bijection between the two groups, sending nontrivial elements to nontrivial elements, can be order-preserving, so less an isomorphism.