Klein Four-Group and Group of Cyclic Group of Order 4 are not Isomorphic

Theorem

The Klein $4$-group $K_4$ and the cyclic group of order $4$ $C_4$ are not isomorphic.

Proof 1

Recall the Cayley table for $K_4$:

$\begin{array}{c|cccc}

 & e & a & b & c \\

\hline e & e & a & b & c \\ a & a & e & c & b \\ b & b & c & e & a \\ c & c & b & a & e \\ \end{array}$

From Finite Cyclic Group is Isomorphic to Integers under Modulo Addition, $C_4$ can be exemplified using the additive group of integers modulo $4$.

Recall the Cayley table for $\struct {\Z_4, +_4}$:

$\begin{array}{r|rrrr}

\struct {\Z_4, +_4} & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 \\ \hline \eqclass 0 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 \\ \eqclass 1 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4 \\ \eqclass 2 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4 \\ \eqclass 3 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 \\ \end{array}$

Note that all elements of $K_4$ are self-inverse.

However, for example:

$\eqclass 1 4 +_4 \eqclass 1 4 = \eqclass 2 4$

and so $\eqclass 1 4$ is not self-inverse.

Thus there can be no bijection between $K_4$ and $\struct {\Z_4, +_4}$ such that:

$\forall a, b \in K_4: \map \phi a +_4 \map \phi b = \map \phi {a b}$

Hence the result.

$\blacksquare$

Proof 2

Note that both $C_4$ and $K_4$ are of order $4$.

Also note that both $C_4$ and $K_4$ are abelian.

By definition, $C_4$ has elements of order $4$.

From Order of Isomorphic Image of Group Element, the image of an element of $C_4$ under an isomorphism to $K_4$ would also be of order $4$.

But $K_4$ has no elements of order $4$.

Hence $C_4$ and $K_4$ are not isomorphic.

$\blacksquare$