On group theory 8 (An easy one)

Groups, Rings, Domains, Modules, etc, Galois theory
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Papapetros Vaggelis
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Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

On group theory 8 (An easy one)

#1

Post by Papapetros Vaggelis »

Let \(\displaystyle{\left(G,\cdot\right)}\) be a group and \(\displaystyle{f:G\longrightarrow G}\)

be a homomorphism which is onto \(\displaystyle{G}\) .

If \(\displaystyle{H\leq G}\) such that \(\displaystyle{[G:H]=n\in\mathbb{N}}\), then prove that

\(\displaystyle{[G:f^{-1}(H)]=n}\) .
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: On group theory 8 (An easy one)

#2

Post by Papapetros Vaggelis »

Here is a solution.

We define \(\displaystyle{g:G/f^{-1}(H)\longrightarrow G/H}\) by

\(\displaystyle{g(x\,f^{-1}(H))=f(x)\,H}\) . We observe that

\(\displaystyle{g(x\,f^{-1}(H))=f(x)\,H\in G/H\,,\forall\,x\,f^{-1}(H)\in G/f^{-1}(H)}\) and if

\(\displaystyle{x\,f^{-1}(H)=y\,f^{-1}(H)}\), then \(\displaystyle{x^{-1}\,y\in f^{-1}(H)}\), so :

\(\displaystyle{f(x^{-1}\,y)\in H\iff (f(x))^{-1}\,f(y)\in H\iff f(x)\,H=f(y)\,H\iff g(x\,f^{-1}(H))=g(y\,f^{-1}(H))}\) .

By this way, we proved that the function \(\displaystyle{g}\) is well defined and one to one.

Let \(\displaystyle{y\,H\in G/H}\). Since \(\displaystyle{y\in G}\) and \(\displaystyle{f}\) is onto \(\displaystyle{G}\),

we have that \(\displaystyle{y=f(x)}\) for some \(\displaystyle{x\in G}\). Then,

\(\displaystyle{x\,f^{-1}(H)\in G/f^{-1}(H)}\) and \(\displaystyle{g(x\,f^{-1}(H))=f(x)\,H=y\,H}\)

which means that the function \(\displaystyle{g}\) is onto \(\displaystyle{G/H}\) .

So, \(\displaystyle{[G:f^{-1}(H)]=|G/f^{-1}(H)|=|G/H|=[G:H]=n}\) .
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