On group theory 8 (An easy one)
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On group theory 8 (An easy one)
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
\(\displaystyle{[G:f^{-1}(H)]=n}\) .
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
\(\displaystyle{[G:f^{-1}(H)]=n}\) .
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- Community Team
- Posts: 426
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Re: On group theory 8 (An easy one)
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, \(\displaystyle{[G:f^{-1}(H)]=|G/f^{-1}(H)|=|G/H|=[G:H]=n}\) .
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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