tables of cyclic subgroups and conjugates

Question

$G = S_5$, I need to construct tables for $H$ and $aHa^{-1}$ ($H =$ cyclic subgroup $(142)(35),$ and $a = (2354) \in G$) and see what can be inferred. In my attempt $H$ = $\{(142)(35), (124)(35), (142), (124), (35), ()\}$, and $ aHa^{-1} = \{(132)(45), (123)(45), (132), (123), (45), (1)\}$, and those seem to be isomorphic. Any input would be appreciated.

Answer 1

Conjugate subgroups are isomorphic. The fixing $a \in G$ and defining $\phi_a : h \mapsto a^{-1}ha$ is an automorphism (specifically, an inner automorphism) of $G$ - i.e., a bijective homomorphism $G \rightarrow G$.

Thus, restricting this map to a subgroup $H \leq G$ gives an isomorphism between $H$ and $\phi_a(H)=a^{-1}Ha$.

(Note, I've used the convention that the conjugation action is given by $g^a = a^{-1}ga$ since this works nicer with group actions, but the conclusion is the same as in your question - just replace $a$ with $a^{-1}$ if it's not clear.)