Pullback of Quotient Group Isomorphism/Examples/C6 with A4

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Example of Pullback of Quotient Group Isomorphism

Let $G = C_6$ be the cyclic group of order $6$:

$G = \gen x$

Let $H = A_4$ be the alternating group on $4$ letters.

Let $e_G$ and $e_H$ denote the identity elements of $G$ and $H$ respectively.


Let $N$ be the subgroup of $G$:

$N = \gen {x^3}$

which from Subgroup of Abelian Group is Normal is normal.

Let $K$ be the subgroup of $H$:

$\set {e_H, \tuple {1 2} \tuple {3 4}, \tuple {1 3} \tuple {2 4}, \tuple {1 4} \tuple {2 3} }$

which from Klein Four-Group is Normal in A4 is normal.


Let $\theta: G / N \to H / K$ be the mapping defined as:

$\forall M \in G / N: \map \theta M = \begin{cases} K & : M = N \\ \tuple {1 2 3} K & : M = x N \\ \tuple {1 3 2} K & : M = x^2 N \end{cases}$


The pullback of $G$ and $H$ by $\theta$ is:

$G \times^\theta H = \set {\tuple {e_G, e_H}, \tuple {e_G, \tuple {1 2} \tuple {3 4} }, \tuple {e_G, \tuple {1 3} \tuple {2 4} }, \tuple {e_G, \tuple {1 4} \tuple {2 3} }, \\ \tuple {x^3, e_H}, \tuple {x^3, \tuple {1 2} \tuple {3 4} }, \tuple {x^3, \tuple {1 3} \tuple {2 4} }, \tuple {x^3, \tuple {1 4} \tuple {2 3} }, \\ \tuple {x, \tuple {1 2 3} }, \tuple {x, \tuple {1 3 4} }, \tuple {x, \tuple {2 4 3} }, \tuple {x, \tuple {1 4 2} }, \\ \tuple {x^4, \tuple {1 2 3} }, \tuple {x^4, \tuple {1 3 4} }, \tuple {x^4, \tuple {2 4 3} }, \tuple {x^4, \tuple {1 4 2} }, \\ \tuple {x^2, \tuple {1 3 2} }, \tuple {x^2, \tuple {1 4 3} }, \tuple {x^2, \tuple {2 3 4} }, \tuple {x^2, \tuple {1 2 4} }, \\ \tuple {x^5, \tuple {1 3 2} }, \tuple {x^5, \tuple {1 4 3} }, \tuple {x^5, \tuple {2 3 4} }, \tuple {x^5, \tuple {1 2 4} } }$


Proof


Sources