Group Isomorphism is a bijective map between $2$ groups such that it preserves the group homomorphism.
Suppose, we have $2$ groups $(G_1,*)$ and $(G_2, \circ)$ and $f : G_1 \rightarrow G_2$ is a bijection between these $2$ groups such that $f(x * y ) = f(x) \circ f(y)$, where $x,y \in G_1$.
If there exist an isomorphism between $2$ groups then these $2$ groups are called isomorphic groups otherwise these are called non-isomorphic groups. Basically, Isomorphism means same structure, So, if there exist a one-to-one correspondence between elements of $2$ groups then they are called isomorphic groups.
Some properties to test whether groups are isomorphic or not :
1) $|G_1| = |G_2|$ (i.e. both groups should have same order(no. of elements or cardinality ))
2) $G_1$ is abelian $\Leftrightarrow$ $G_2$ is abelian
3) $G_1$ and $G_2$ have same number of elements of every order
Now, comes to the question,
Suppose, we have a group $(G_1,\circ)$ where elements of $G_1$ are : $\{e, \tau_1,\tau_2, \tau_3 \}$. Where,
$e = \begin{Bmatrix} 1 \longrightarrow &1 \\ 2 \longrightarrow &2 \\ 3 \longrightarrow &3 \\ 4 \longrightarrow &4 \end{Bmatrix}$, $\tau_1 = \begin{Bmatrix} 1 \longrightarrow &2 \\ 2 \longrightarrow &1 \\ 3 \longrightarrow &4 \\ 4 \longrightarrow &3 \end{Bmatrix}$, $\tau_2 = \begin{Bmatrix} 1 \longrightarrow &3 \\ 2 \longrightarrow &4 \\ 3 \longrightarrow &1 \\ 4 \longrightarrow &2 \end{Bmatrix}$ and $\tau_3 = \begin{Bmatrix} 1 \longrightarrow &4 \\ 2 \longrightarrow &3 \\ 3 \longrightarrow &2 \\ 4 \longrightarrow &1 \end{Bmatrix}$
Here, $e$ is an identity map.
Here, group operation(law of composition) $" \circ "$ is function composition and here $\tau_1\tau_2 = \tau_3$
Now, if we check the order of the elements : Since, $\tau_1^2 = e$, $\tau_2^2 = e$ and $\tau_3^2 = e$.
So, $o(\tau_1) = 2, o(\tau_2) = 2, o(\tau_3) = 2 $.
Now, consider another group $(G_2, .)$ where $G_2 = \{\pm1, \pm i\}$ and group operation is multiplication.
Here, element $"i"$ has order $4$ because $i^4 = 1(identity \; element)$.
Since, Group $G_1$ does not have any element of order $4$ which group $G_2$ have. So, these $2$ groups can't have the same structure. Hence, these $2$ groups are non-isomorphic.
Now, it is not difficult to make the composition table for these groups.
For $(G_1, \circ)$, composition table is :
$\circ$ |
e |
$\tau_1$ |
$\tau_2$ |
$\tau_3$ |
e |
e |
$\tau_1$ |
$\tau_2$ |
$\tau_3$ |
$\tau_1$ |
$\tau_1$ |
$e$ |
$\tau_3$ |
$\tau_2$ |
$\tau_2$ |
$\tau_2$ |
$\tau_3$ |
$e$ |
$\tau_1$ |
$\tau_3$ |
$\tau_3$ |
$\tau_2$ |
$\tau_1$ |
$e$ |
For $(G_2, .)$, composition table is :
$.$ |
$1$ |
$-1$ |
$i$ |
$-i$ |
$1$ |
$1$ |
$-1$ |
$i$ |
$-i$ |
$-1$ |
$-1$ |
$1$ |
$-i$ |
$i$ |
$i$ |
$i$ |
$-i$ |
$-1$ |
$1$ |
$-i$ |
$-i$ |
$i$ |
$1$ |
$-1$ |