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1st Isomorphism thm for dihedral gps
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[QUOTE="fresh_42, post: 6062602, member: 572553"] I don't see why ##f(\Gamma) \neq 1##. We can always bring ##\Gamma \in \operatorname{ker}(f)## into the form ##\Gamma = SR^k\; , \;k\in\mathbb{Z}\; , \;## or ##\Gamma=R^k##. The second case is easy. Now in the first case ##f(\Gamma)=1=sr^k## and thus ##s=r^k##, i.e. ##1=r^{2k}##. Since ##r^{2k}\in D_{2n}##, we now have ##n\,|\,k## or ##\Gamma = S(R^n)^l##. But how do you conclude from ##s=r^{n\cdot l}## that ##S\in \langle R^n\rangle\,?## Theoretically there can be an ##s## on the RHS, e.g. ##\Gamma = SRSR## maps to ##1=srsr##. Sure, this is ##1##, but it shows that in principle more complex words ##\Gamma## can be mapped on ##1## even if they contain an ##S##. Edit: Got it. ##s=r^k## is not possible and therefore ##\Gamma \neq SR^k## and ##\Gamma = R^k## is the only left possibility. [/QUOTE]
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1st Isomorphism thm for dihedral gps
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