Intereting Posts

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About the proof that lebesgue measure is a premeasure.
A necessary condition for a multi-complex-variable holomorphic function.
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Looking for insightful explanation as to why right inverse equals left inverse for square invertible matrices
Let $A_1,A_2,..,A_n$ be the vertices of n sides of a regular polygon such that $1/A_1.1/A_2=1/A_1.1/A_3+1/A_1.1/A_4$ then value of $n$ must be?
Classic Hand shake question
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Sobolev spaces – about smooth aproximation
Cube stack problem
Set Addition vs. Set Union
Can the inverse of a function be the same as the original function?
Is local isomorphism totally determined by local rings?
Prove $(a_1+b_1)^{1/n}\cdots(a_n+b_n)^{1/n}\ge \left(a_1\cdots a_n\right)^{1/n}+\left(b_1\cdots b_n\right)^{1/n}$

Let’s consider $\mathfrak{G}:=((C_2 \times C_2) \rtimes_{\phi} C_3) \rtimes_{\nu} C_2$ (which I do believe is $\mathcal{S}_4$, please confirm or argue against) and $G:=((C_2 \times C_2) \rtimes_{\mu} C_2) \rtimes_{\xi} C_3$.

Under which hypothesis does there exist an isomorphism $\chi: \mathfrak{G} \rightarrow G$?

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- Are there real world applications of finite group theory?
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- Galois Groups of Finite Extensions of Fixed Fields
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First, let’s take a look at $\mathfrak{G}$ and determine under which $\phi$ and $\nu$ we get $\mathfrak{G}\cong S_4$.

The automorphism group of the Klein $V$ group is $S_3$, since $(1,0)$, $(0,1)$, and $(1,1)$ are interchangable. If we assume that $\phi$ is nontrivial, then $C_3$ embeds into $\text{Aut}(V)$ by acting as a $3$-cycle on these elements. The resulting group is $A_4$.

The automorphism group of $A_4$ has nine elements of order $2$. Three of them will yield $A_4\rtimes C_2\cong A_4 \times C_2$; these are the inner automorphisms induced from the $V$ group. Every other nontrivial automorphism of order $2$ will produce $A_4\rtimes C_2\cong S_4$.

So you see that if we assume that

- $\phi$ is nontrivial, and
- $\nu$ embeds $C_2$ into $\text{Out}((C_2\times C_2)\rtimes_\phi C_3 )$,

then $\mathfrak{G}=((C_2\times C_2) \rtimes_\phi C_3 )\rtimes C_2\cong S_4$.

Now let’s look at $G$. Every automorphism of $V$ of order $2$ produces $D_4$ when we take $(C_2\times C_2)\rtimes C_2$, which is clear from the geometric interpretation of $D_4$ (consider elements of $C_2\times C_2$ as coordinates). However, $\text{Aut}(D_4)$ is isomorphic to $D_4$, and thus has no elements of order $3$. So the only possibility for $\xi$ is the trivial homomorphism. Thus $G$ can never be isomorphic to $S_4$.

So let’s go back and think. How could we have chosen $\phi$ and $\nu$ so that $\mathfrak{G}\cong G$? Aside from choosing $\phi,\nu,$ and $\mu$ to be trivial, the only way for this to happen would be to choose $\phi$ to be trivial and $\nu$ so that $C_2$ centralizes $C_3$ and acts faithfully on $V$, or to choose $\mu,\nu$ trivial and $\phi,\xi$ nontrivial.

So $\mathfrak{G}\cong G$ if and only if $\mathfrak{G}\cong G \cong (C_2)^3\times C_3$, $\mathfrak{G}\cong G \cong D_4\times C_3$, or $\mathfrak{G}\cong G \cong A_4\times C_2$.

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