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

[Boi]

This is an exercise from Herstein's "Topics in algebra":[imath]G[/imath]

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 [imath]L = \{ aH \; | \; a \in G\}[/imath], [imath]C= \{ aHa^{-1}\; | \; a \in G \}[/imath] and a function between them [imath]f : L \to C[/imath] which sends a left coset [imath]aH[/imath] to subgroup [imath]aHa^{-1}[/imath]. I heard that before using any function I need to show that it is well-defined, so if [imath]x=x'[/imath] then [imath]f(x)=f(x')[/imath]. In my case, this becomes [imath]aH=bH \implies aHa^{-1}=bHb^{-1}[/imath]. It was mentioned earlier in the chapter that [imath]aH=bH \iff b^{-1}a \in H[/imath]. Using the fact that [imath]H[/imath] is a subgroup, I have [imath]a^{-1}b \in H[/imath]. Saying that [imath]aHa^{-1}=bHb^{-1}[/imath] basically means that if you give me an element [imath]h_1 \in H[/imath] then I can give you elements [imath]h_2, \: h_3 \in H[/imath], such that [imath]ah_1a^{-1}=bh_2b^{-1}[/imath] and [imath]bh_2b^{-1}=ah_3a^{-1}[/imath]. I can construct these elements quite easily by utilising the fact that [imath]H[/imath] is a subgroup and that both of [imath]a^{-1}b[/imath] and [imath]b^{-1}a[/imath] are in [imath]H[/imath]. [imath]h_2=b^{-1}ah_1a^{-1}b[/imath] while [imath]h_3=a^{-1}bh_1b^{-1}a[/imath]. This concludes the proof that [imath]f[/imath] is well-defined. Now, [imath]f[/imath] is clearly surjective, since for any [imath]aHa^{-1}[/imath] there exists [imath]aH[/imath] which is sent to it. Then, using the fact that [imath]H[/imath] is of finite index in [imath]G[/imath] I conclude that [imath]L[/imath] is finite. And, because I have a surjective function from [imath]L[/imath] to [imath]C[/imath], the order of [imath]C[/imath] cannot be more than that of [imath]L[/imath]. Hence, [imath]C[/imath] 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
[math] 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} [/math]instead of using single elements [imath] h_i. [/imath]

 

[Boi]

I haven't found a flaw.

I would only abbreviate the argument of well-definition by
[math] 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} [/math]instead of using single elements [imath] h_i. [/imath]

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
[math] aH=a\cdot \{h\in G\,|\,h\in H\}=\{ah\in G\,|\,h\in H\} .[/math][imath] aH=bH [/imath] does not mean [imath] ah=bh [/imath] and one must keep that in mind, however, it means [imath] ah=bh' [/imath] for some [imath] h'\in H. [/imath]