(1) First, is it possible for someone to check if my proof of the following statement is correct?
Theorem: H is a subgroup of G, and $a \in G$. Prove that $aH=Ha$ if and only if $H=aHa^{-1}$.
$Proof$:
"$\rightarrow$"
Assume $aH=Ha$. Then there exist $h,h' \in H$ such that $ah=h'a$.
Now we want to show $H \subset aHa^{-1}$.
Let $x \in H$, and let $e$ be the identity in G. Then $x=eje^{-1}$, hence $x \in aHa^{-1}$ (Is this part of the proof correct? That is, can I pick a=e?)
Furthermore we want to show $aHa^{-1}\subset H $.
Let $x \in aHa^{-1}$. Then $x=aha^{-1}$ for some $h \in H$. We then have $x=(h'a)a^{-1}=h' \in H.$ (because $ah=h'a$).
Thus we have proved $H=aHa^{-1}$.
"$\leftarrow$"
Assume $H=aHa^{-1}$. Then there exist $h,h' \in H$ such that $h=ah'a^{-1}$.
We want to show $aH\subset Ha $.
Let $x \in aH$. Then $x=ah'$. For some $h' \in H$. Then we have $x=ah'a^{-1}a=ha \in Ha$.
Now we want to show $Ha \subset aH$.
Let $x \in Ha$. Then $x=ha$. For some $h \in H$. Then we have $x=(ah'a^{-1})a=ah' \in aH$.
Thus we have proved $aH=Ha$.
(2) In my textbook: Gallian's Contemporary Abstract Algebra, 9TH edition, on page 140, he proves it the following way:
Note that: $aH=Ha$ iff $(aH)a^{-1}=(Ha)a^{-1}=H$. That is, iff $aHa^{-1}=H$.
Is this correct? What kind of manipulations are those?
How can you manipulate a set like that by multiplying on the right and on the left?
If so, can you do the following? $a^{-1}\{ah_1, \dots, ah_n\} = \{a^{-1}ah_1, \dots, a^{-1}ah_n\}=\{h_1, \dots, h_n\}$
And if yes, by what rule of sets are you allowed to do this? Is this formal?
Best Answer
For the first question, the answer is no, you can't pick $a$ to be $e$ (assuming I understand correctly what you are saying). In general you can't pick $a$, it is given in the beginning. Also, writing $x=eje^{-1}$ yields $x=j$ so it does not help. The second part of the $\rightarrow$ is right along with $\leftarrow$.
For the second question, yes, in group theory you can do that; we define $aH$ to be the set $\{ah:h\in H\}$. With this definition we have $$a^{-1}\{ah_1,\dots,ah_n\}=\{a^{-1}(ah_1),\dots,a^{-1}(ah_n)\}=\{h_1,\dots,h_n\}$$ as you have written (note that one must use the associative property to see that $a^{-1}(ah)=(a^{-1}a)h=h$).
We not only define $aH$ but $KH$ also to be $\{k\cdot h:k\in K,h\in H\}$. You will use that in the second isomorphism theorem.
EDIT: A hint for the first part would be to rethink the relation $aH=Ha$ and try to get more. Specifically, for $x\in H$, $ax$ will be in $aH=Ha$. See where that takes you.