If I act using right or left action by generators of some group G on some set A will I get a subgroup of $S_A$ isomorphic to G ? If so can someone provide me with a reason.

For example:

Use the left regular representation of $Q_8$ to produce two elements of $S_8$ which generate a subgroup of $S_8$ isomorphic to the quaternion group $Q_8$.

We know $Q_8 = <i,j>$, so suppose I use left representation of $Q_8$ by acting on $Q_8$ by i and j, then will I have a isomorphic image of $S_8$, which generate subgroup of $S_8$ to the quaternion $Q_8$.

  • $\begingroup$ Can you explain what you mean more explicitly? What is a "right or left action by generators of $G$"? What is $S_A$ supposed to be, exactly? $\endgroup$ – Eric Wofsey Sep 17 '15 at 3:12
  • $\begingroup$ I will re structure the question $\endgroup$ – Dude Sep 17 '15 at 3:21
  • $\begingroup$ The answer is no unless the action is faithful. $\endgroup$ – David Hill Sep 17 '15 at 3:23
  • $\begingroup$ I have edited the question @DavidHill $\endgroup$ – Dude Sep 17 '15 at 3:25

It depends on how you define your action. You can define a trivial action that takes an element of G, an element of A, and gives you that same element of A.

If you require your action to be faithful, this is probably closer to what you had in mind.

If $g \in G$, then $g \star (g \star a) = (g \ast g) \star a$. If $G$ has generators $g_1,g_2,...g_n$, then $\forall h \in G, h = g_1^{k_1}\ast g_2^{k_2}\ast ... \ast g_n^{k_n}$ for some integers $k_1,k_2,...,k_n$.

If every $h\in G$ corresponds to some $\sigma_h \in S_A$ , $\sigma_h(a)= h \star a,$ this $\sigma$ can be produced by $(g_1^{k_1} \star(g_2^{k_2} \star(...(g_n^{k_n} \star a)...))) $.

The map $ h \mapsto \sigma_h $ is a homomorphism; $h_1 \ast h_2 \mapsto \sigma_{{h_1}{h_2}}$, and $\sigma_{{h_1}{h_2}}(a)= (h_1\ast h_2) \star a = h_1 \star (h_2 \star a) = \sigma_{h_1}(\sigma_{h_2}(a))= \sigma_{h_1} \circ \sigma_{h_2} (a),$ as desired. Since the two sets are bijective, we have our isomorphism.

  • $\begingroup$ I have edited the question $\endgroup$ – Dude Sep 17 '15 at 3:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.