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A subgroup $H$ is conjugate to a subgroup $K$ of a group $G$ if there exists an inner automorphism $i_g$ of $G$ such that $i_g[H]=K$ Show that conjugacy is an equivalence relation on the collection of subgroups of $G$

My main question is here that the different approach I've taken from the textbook. I've taken a quite a similar approach, but textbook has refused this. I'm asking this question here, because I believe if my simple approach was acceptable textbook would have implemented that instead?

Symmetry : If $H$ is conjugate to K then $\exists g \epsilon G. : gHg^{-1}=K \implies H =g Kg^{-1}$ So indeed $K$ is conjugate to $H$

Howeer textbook adopted a different approach :

Suppose that $i_g[H]=H$ so that for each $k \epsilon K$ we have $k=ghg^{-1}$ for exactly one $h \epsilon H$. Then $h=g^{-1}kg = g^{-1}k(g^{-1})^{-1},$ and we see that $i_g[K]=H$ so $K$ is alo conjugate to $H$

My question is, why the textbook couldn't just do these steps between sets as I did, instead felt to do it between elements of the set? Is my approach wrong in this case, Can't we play with an equal relation between sets?

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  • $\begingroup$ Your implication is incorrect. You need to invert $g$ to make it correct. $\endgroup$ – Tobias Kildetoft Apr 22 '17 at 10:46
  • $\begingroup$ @TobiasKildetoft "invert"?, I see, myMAİN question is not about that. $\endgroup$ – Xenidia Apr 22 '17 at 10:51
  • $\begingroup$ I think you've lost lots of $g^{-1}$'s in both your proof and the book's version. And muddled at least one $K$ into an $H$. But anyway it's not at all important, maybe the author thought students would find her way easier? $\endgroup$ – ancientmathematician Apr 22 '17 at 12:50

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