Automorphism Group/Examples/Klein Four-Group

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Example of Automorphism Group

The automorphism group of the Klein $4$-group is the Symmetric Group on 3 Letters $S_3$.


Proof


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Let $f:V\rightarrow V$ be a permutation that fixes the identity element.

By definition, this map is bijective.

$V$ has the property that the product of any 2 distinct non-identity elements is the 3rd non-identity element.

This implies:

$\map f {vv'} = \map f v \cdot \map f {v'}$

for all $v, v' \in V$ with $v, v' \ne e$.

For products involving the identity element:

$\map f {e v} = \map f e \cdot \map f v = e \cdot \map f v = \map f v$

for all $v \in V$.

Also, since all non-identity elements of $V$ have order $2$:

$\map f {v^2} = \map f {v v} = \map f v \cdot \map f v = e$

for all $v \in V$.

Hence, any permutation of the elements of $V$ that fixes the identity element is an automorphism of $V$.

Since these permutations exhaust all possible automorphisms of $V$, they are the elements of $\Aut V$.

Since the $3! = 6$ elements of $\Aut V$ represent all permutations of $3$ objects, $\Aut V$ is isomorphic to $S_3$.

$\blacksquare$


Sources

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