Could somebody help me refine my proof, please? (group theory)

[Boi]

This is an exercise from Herstein's "Topics in algebra":$G$

What follows isn't really a proof, more so a collection of my main ideas, which I believe should lead to a proof. I kindly ask anyone reading to not only check whether these ideas are correct, but also point out which ones ought to be made more rigorous.
Define two sets $L = \{ aH \; | \; a \in G\}$, $C= \{ aHa^{-1}\; | \; a \in G \}$ and a function between them $f : L \to C$ which sends a left coset $aH$ to subgroup $aHa^{-1}$. I heard that before using any function I need to show that it is well-defined, so if $x=x'$ then $f(x)=f(x')$. In my case, this becomes $aH=bH \implies aHa^{-1}=bHb^{-1}$. It was mentioned earlier in the chapter that $aH=bH \iff b^{-1}a \in H$. Using the fact that $H$ is a subgroup, I have $a^{-1}b \in H$. Saying that $aHa^{-1}=bHb^{-1}$ basically means that if you give me an element $h_1 \in H$ then I can give you elements $h_2, \: h_3 \in H$, such that $ah_1a^{-1}=bh_2b^{-1}$ and $bh_2b^{-1}=ah_3a^{-1}$. I can construct these elements quite easily by utilising the fact that $H$ is a subgroup and that both of $a^{-1}b$ and $b^{-1}a$ are in $H$. $h_2=b^{-1}ah_1a^{-1}b$ while $h_3=a^{-1}bh_1b^{-1}a$. This concludes the proof that $f$ is well-defined. Now, $f$ is clearly surjective, since for any $aHa^{-1}$ there exists $aH$ which is sent to it. Then, using the fact that $H$ is of finite index in $G$ I conclude that $L$ is finite. And, because I have a surjective function from $L$ to $C$, the order of $C$ cannot be more than that of $L$. Hence, $C$ must also be finite.
Thanks in advance.
P.S. : I apologise for any grammatical mistakes, incorrect vocabulary or terminology and hope that despite all that, my thread isn't too confusing.

 

[fresh_42]

I haven't found a flaw.

I would only abbreviate the argument of well-definition by
$$aH=bH \Longrightarrow b^{-1}a\, , \,a^{-1}b\in H \Longrightarrow aHa^{-1}=a(a^{-1}b)H(b^{-1}a)a^{-1}=bHb^{-1}$$instead of using single elements $h_i.$

 

[Boi]

I haven't found a flaw.

I would only abbreviate the argument of well-definition by
$$aH=bH \Longrightarrow b^{-1}a\, , \,a^{-1}b\in H \Longrightarrow aHa^{-1}=a(a^{-1}b)H(b^{-1}a)a^{-1}=bHb^{-1}$$instead of using single elements $h_i.$

Oh, is this really okay? I also thought about it, but was a bit afraid about using subgroups just as elements.
Thank you for your reply!

 

[fresh_42]

Oh, is this really okay? I also thought about it, but was a bit afraid about using subgroups just as elements.

It is ok and it is used a lot in group theory. It is an equation of sets and a short form of for example
$$aH=a\cdot \{h\in G\,|\,h\in H\}=\{ah\in G\,|\,h\in H\} .$$$aH=bH$ does not mean $ah=bh$ and one must keep that in mind, however, it means $ah=bh'$ for some $h'\in H.$