(27.) A subgroup H is conjugate to a subgroup K of a group G
(viz. p. 141 $K \le G$ is a conjugate subgroup of $H$), if $i_g[H] = gHg^{-1} =K$ for some $g \in G$.
Show that conjugacy is an equivalence relation on the collection of subgroups of G.(28.) Characterize the normal subgroups of a group G in terms of the cells where they appear in the partition given by the conjugacy relation in exercise (27.)
Answer. We see that the normal subgroups of G are precisely the subgroups in the one-element cells of the conjugacy partition of the subgroups of G.
(29.) Referring to Exercise 27, find all subgroups of $S_3$ conjugate to $\{id, (1,3)(2) \}$.
(27.) Answer on p. 50 says $gHg^{-1} = K$ means for each $k \in K, k = ghg^{-1}$ for exactly one $h \in H$. Why is $h$ unique here? As I asked here, $gH = Hg \iff gh_1 = h_2g$ where $g_1$ can $\neq g_2$ ?
(28.) Is there a picture please for the answer to (28.) to help me understand?
(29.) References respectively: http://www.sfu.ca/~jtmulhol/math302/notes/302notes.pdf p. 126 and
Source http://www.auburn.edu/~huanghu/math5310/alg-hw-ans-13 i think.pdf
(29.) wants us to find all $K \le S_3$ such that $g\{id, (1,3)(2) \}g^{-1} = K$ for all $g \in S_3$.
Hence why does the solution fret about only 3 elements of $S_3$ for $g$? What about the other 3?
Best Answer
Here is the idea. Two elements of $S_n$ are conjugate if and only if they have the same cycle type, and we are looking for all elements of $S_3$ that are conjugate to $(1 3)(2)$. Since conjugation by a fixed element is a group automorphism, the elements we find will be the generators of the subgroups that are conjugate to your subgroup. So, the subgroups conjugate to your subgroup are $\{id, (13)(2)\}$, $\{id, (12)(3)\}$, and $\{id, (23)(1)\}$.