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?
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3The most obvious way is to show $g_1,g_2 \in G_2$. (Does this count as "trivial"?) – Douglas S. Stones Oct 29 '12 at 07:29
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OK, 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 $\Gamma_i s$ – Kaave Oct 29 '12 at 08:02
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1@Kaave: Are you thinking of $\Gamma$'s as vector spaces or just some sets? – Mikasa Oct 29 '12 at 09:30
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$\Gamma 's$ are just sets. – Kaave Oct 29 '12 at 11:28
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1If the $,\Gamma',$s are just sets then it suffices to show $,\Gamma_1\subset \Gamma_2,$ – DonAntonio Oct 29 '12 at 11:57
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Technically 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 $G_2$. – Chris Godsil Oct 29 '12 at 12:45