I have two permutation groups $G_1=\langle g_1,g_2\rangle$ and $G_2=\langle h_1,h_2\rangle$ acting on $\Gamma_1$ and $\Gamma_2$ respectively. I want to prove that $G_1 \leq G_2$. What are more nontrivial methods to do so?
What are various approaches of proving that a permutation group is a subgroup of another?
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finite-groups
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3The most obvious way is to show $g_1,g_2 \in G_2$. (Does this count as "trivial"?) – 2012-10-29
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0OK, but how we can show that $g_1,g_2 \in G_2$? We should give a mapping from $\Gamma_1$ to $\Gamma_2$. But what if it's not possible? One can do the same for vector spaces over $\Gamma_i s$ – 2012-10-29
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1@Kaave: Are you thinking of $\Gamma$'s as vector spaces or just some sets? – 2012-10-29
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0$\Gamma 's$ are just sets. – 2012-10-29
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1If the $\,\Gamma'\,$s are just sets then it suffices to show $\,\Gamma_1\subset \Gamma_2\,$ – 2012-10-29
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0Technically the permutation group $G_1$ cannot be a subgroup of the permutation group $G_2$ unless $\Gamma_1=\Gamma_2$. If $\Gamma_1\subseteq\Gamma_2$ then $G_1$ can be isomorphic to a subgroup of $G_2$. – 2012-10-29