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Give the composition tables (Cayley Tables) of the two non-isomorphic groups of order $4$ with elements $e, a, b, c$ where $c$ is the identity element. Use the order $e, a, b, c$ for the rows and columns.
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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 :

$.$ $1$ $-1$ $i$ $-i$
$1$ $1$ $-1$ $i$ $-i$
$-1$ $-1$ $1$ $-i$ $i$
$i$ $i$ $-i$ $-1$ $1$
$-i$ $-i$ $i$ $1$ $-1$

 

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