# Group acting on set A by its generators

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 $Q_8$.

• 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? – Eric Wofsey Sep 17 '15 at 3:12
• I will re structure the question – Dude Sep 17 '15 at 3:21
• The answer is no unless the action is faithful. – David Hill Sep 17 '15 at 3:23
• I have edited the question @DavidHill – Dude Sep 17 '15 at 3:25

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.